Solutions

Prerequisites

Before attempting this problem, you should be comfortable with:

Binary Search Tree (BST) properties - Understanding that inorder traversal yields sorted values and how to search in O(log n) time
Tree traversal (DFS) - Recursively visiting all nodes to collect values or search for complements
Hash Sets - Storing values for O(1) lookup when checking if a complement exists
Two Pointers technique - Using pointers on sorted lists to efficiently find pairs that sum to a target

1. Brute Force

Intuition

The simplest approach is to collect all values from both trees and check every possible pair. If any pair sums to the target, we have found our answer.

Algorithm

Traverse the first tree and store all node values in a list.
Traverse the second tree and store all node values in another list.
For each value a in the first list and each value b in the second list, check if a + b == target.
Return true if a valid pair is found, false otherwise.

class Solution:
    def twoSumBSTs(self, root1: Optional[TreeNode], root2: Optional[TreeNode], target: int) -> bool:
        def dfs(curr_node, node_list):
            if not curr_node:
                return
            node_list.append(curr_node.val)
            dfs(curr_node.left, node_list)
            dfs(curr_node.right, node_list)
        
        node_list1, node_list2 = [], []
        dfs(root1, node_list1)
        dfs(root2, node_list2)
        
        for a in node_list1:
            for b in node_list2:
                if a + b == target:
                    return True
        return False

class Solution {
    private void dfs(TreeNode currNode, List<Integer> nodeList) {
        if (currNode == null) {
            return;
        }
        node_list.add(currNode.val);
        dfs(currNode.left, nodeList);
        dfs(currNode.right, nodeList);
    }

    public boolean twoSumBSTs(TreeNode root1, TreeNode root2, int target) {
        List<Integer> nodeList1 = new ArrayList<>();
        List<Integer> nodeList2 = new ArrayList<>();
        dfs(root1, nodeList1);
        dfs(root2, nodeList2);

        for (int a : nodeList1) {
            for (int b : nodeList2) {
                if (a + b == target) {
                    return true;
                }
            }
        }
        return false;
    }
}

class Solution {
public:
    void dfs(TreeNode* currNode, vector<int>& nodeList) {
        if (currNode == nullptr) {
            return;
        }
        nodeList.push_back(currNode->val);
        dfs(currNode->left, nodeList);
        dfs(currNode->right, nodeList);
    }

    bool twoSumBSTs(TreeNode* root1, TreeNode* root2, int target) {
        vector<int> nodeList1, nodeList2;
        dfs(root1, nodeList1);
        dfs(root2, nodeList2);

        for (int a : nodeList1) {
            for (int b : nodeList2) {
                if (a + b == target) {
                    return true;
                }
            }
        }
        return false;
    }
};

class Solution {
    /**
     * @param {TreeNode} root1
     * @param {TreeNode} root2
     * @param {number} target
     * @return {boolean}
     */
    twoSumBSTs(root1, root2, target) {
        const dfs = (currNode, nodeList) => {
            if (!currNode) {
                return;
            }
            nodeList.push(currNode.val);
            dfs(currNode.left, nodeList);
            dfs(currNode.right, nodeList);
        };

        const nodeList1 = [], nodeList2 = [];
        dfs(root1, nodeList1);
        dfs(root2, nodeList2);

        for (const a of nodeList1) {
            for (const b of nodeList2) {
                if (a + b === target) {
                    return true;
                }
            }
        }
        return false;
    }
}

public class Solution {
    private void Dfs(TreeNode currNode, List<int> nodeList) {
        if (currNode == null) {
            return;
        }
        nodeList.Add(currNode.val);
        Dfs(currNode.left, nodeList);
        Dfs(currNode.right, nodeList);
    }

    public bool TwoSumBSTs(TreeNode root1, TreeNode root2, int target) {
        List<int> nodeList1 = new List<int>();
        List<int> nodeList2 = new List<int>();
        Dfs(root1, nodeList1);
        Dfs(root2, nodeList2);

        foreach (int a in nodeList1) {
            foreach (int b in nodeList2) {
                if (a + b == target) {
                    return true;
                }
            }
        }
        return false;
    }
}

func twoSumBSTs(root1 *TreeNode, root2 *TreeNode, target int) bool {
    var dfs func(currNode *TreeNode, nodeList *[]int)
    dfs = func(currNode *TreeNode, nodeList *[]int) {
        if currNode == nil {
            return
        }
        *nodeList = append(*nodeList, currNode.Val)
        dfs(currNode.Left, nodeList)
        dfs(currNode.Right, nodeList)
    }

    var nodeList1, nodeList2 []int
    dfs(root1, &nodeList1)
    dfs(root2, &nodeList2)

    for _, a := range nodeList1 {
        for _, b := range nodeList2 {
            if a+b == target {
                return true
            }
        }
    }
    return false
}

class Solution {
    private fun dfs(currNode: TreeNode?, nodeList: MutableList<Int>) {
        if (currNode == null) {
            return
        }
        nodeList.add(currNode.`val`)
        dfs(currNode.left, nodeList)
        dfs(currNode.right, nodeList)
    }

    fun twoSumBSTs(root1: TreeNode?, root2: TreeNode?, target: Int): Boolean {
        val nodeList1 = mutableListOf<Int>()
        val nodeList2 = mutableListOf<Int>()
        dfs(root1, nodeList1)
        dfs(root2, nodeList2)

        for (a in nodeList1) {
            for (b in nodeList2) {
                if (a + b == target) {
                    return true
                }
            }
        }
        return false
    }
}

class Solution {
    func twoSumBSTs(_ root1: TreeNode?, _ root2: TreeNode?, _ target: Int) -> Bool {
        func dfs(_ currNode: TreeNode?, _ nodeList: inout [Int]) {
            guard let currNode = currNode else {
                return
            }
            nodeList.append(currNode.val)
            dfs(currNode.left, &nodeList)
            dfs(currNode.right, &nodeList)
        }

        var nodeList1 = [Int](), nodeList2 = [Int]()
        dfs(root1, &nodeList1)
        dfs(root2, &nodeList2)

        for a in nodeList1 {
            for b in nodeList2 {
                if a + b == target {
                    return true
                }
            }
        }
        return false
    }
}

Time & Space Complexity

Time complexity: $O(m \cdot n)$
Space complexity: $O(m+n)$

Where $m$ and $n$ are the number of nodes in the two trees.

2. Binary Search

Intuition

Since the second tree is a BST, we can leverage its sorted structure. For each node in the first tree, we compute the complement (target - node.val) and use binary search to check if that complement exists in the second tree.

Algorithm

Traverse the first BST using dfs.
For each node with value v, compute complement = target - v.
Perform binary search in the second BST to find the complement:
- If current value equals complement, return true.
- If current value is greater, search left.
- Otherwise, search right.
If no pair is found after checking all nodes, return false.

class Solution:
    def twoSumBSTs(self, root1: Optional[TreeNode], root2: Optional[TreeNode], target: int) -> bool:
        def binarySearch(root2, target2):
            if not root2:
                return False
            if root2.val == target2:
                return True
            elif root2.val > target2:
                return binarySearch(root2.left, target2)
            else:
                return binarySearch(root2.right, target2)

        def dfs(root, target):
            if not root:
                return False
            if binarySearch(root2, target - root.val):
                return True
            return dfs(root.left, target) or dfs(root.right, target)

        return dfs(root1, target)

class Solution {
    private boolean binarySearch(TreeNode root2, int target2) {
        if (root2 == null) {
            return false;
        }
        if (root2.val == target2) {
            return true;
        } else if (root2.val > target2) {
            return binarySearch(root2.left, target2);
        } else {
            return binarySearch(root2.right, target2);
        }
    }

    private boolean dfs(TreeNode root1, TreeNode root2, int target) {
        if (root1 == null) {
            return false;
        }
        if (binarySearch(root2, target - root1.val)) {
            return true;
        }
        return dfs(root1.left, root2, target) || dfs(root1.right, root2, target);
    }

    public boolean twoSumBSTs(TreeNode root1, TreeNode root2, int target) {
        return dfs(root1, root2, target);
    }
}

class Solution {
public:
    bool binarySearch(TreeNode* root2, int target2) {
        if (root2 == nullptr) {
            return false;
        }
        if (root2->val == target2) {
            return true;
        } else if (root2->val > target2) {
            return binarySearch(root2->left, target2);
        } else {
            return binarySearch(root2->right, target2);
        }
    }

    bool dfs(TreeNode* root1, TreeNode* root2, int target) {
        if (root1 == nullptr) {
            return false;
        }
        if (binarySearch(root2, target - root1->val)) {
            return true;
        }
        return dfs(root1->left, root2, target) || dfs(root1->right, root2, target);
    }

    bool twoSumBSTs(TreeNode* root1, TreeNode* root2, int target) {
        return dfs(root1, root2, target);
    }
};

class Solution {
    /**
     * @param {TreeNode} root1
     * @param {TreeNode} root2
     * @param {number} target
     * @return {boolean}
     */
    twoSumBSTs(root1, root2, target) {
        const binarySearch = (root2, target2) => {
            if (!root2) {
                return false;
            }
            if (root2.val === target2) {
                return true;
            } else if (root2.val > target2) {
                return binarySearch(root2.left, target2);
            } else {
                return binarySearch(root2.right, target2);
            }
        };

        const dfs = (root1, target) => {
            if (!root1) {
                return false;
            }
            if (binarySearch(root2, target - root1.val)) {
                return true;
            }
            return dfs(root1.left, target) || dfs(root1.right, target);
        };

        return dfs(root1, target);
    }
}

public class Solution {
    private bool BinarySearch(TreeNode root2, int target2) {
        if (root2 == null) {
            return false;
        }
        if (root2.val == target2) {
            return true;
        } else if (root2.val > target2) {
            return BinarySearch(root2.left, target2);
        } else {
            return BinarySearch(root2.right, target2);
        }
    }

    private bool Dfs(TreeNode root1, TreeNode root2, int target) {
        if (root1 == null) {
            return false;
        }
        if (BinarySearch(root2, target - root1.val)) {
            return true;
        }
        return Dfs(root1.left, root2, target) || Dfs(root1.right, root2, target);
    }

    public bool TwoSumBSTs(TreeNode root1, TreeNode root2, int target) {
        return Dfs(root1, root2, target);
    }
}

func twoSumBSTs(root1 *TreeNode, root2 *TreeNode, target int) bool {
    var binarySearch func(root2 *TreeNode, target2 int) bool
    binarySearch = func(root2 *TreeNode, target2 int) bool {
        if root2 == nil {
            return false
        }
        if root2.Val == target2 {
            return true
        } else if root2.Val > target2 {
            return binarySearch(root2.Left, target2)
        } else {
            return binarySearch(root2.Right, target2)
        }
    }

    var dfs func(root1 *TreeNode, target int) bool
    dfs = func(root1 *TreeNode, target int) bool {
        if root1 == nil {
            return false
        }
        if binarySearch(root2, target-root1.Val) {
            return true
        }
        return dfs(root1.Left, target) || dfs(root1.Right, target)
    }

    return dfs(root1, target)
}

class Solution {
    private fun binarySearch(root2: TreeNode?, target2: Int): Boolean {
        if (root2 == null) {
            return false
        }
        return when {
            root2.`val` == target2 -> true
            root2.`val` > target2 -> binarySearch(root2.left, target2)
            else -> binarySearch(root2.right, target2)
        }
    }

    private fun dfs(root1: TreeNode?, root2: TreeNode?, target: Int): Boolean {
        if (root1 == null) {
            return false
        }
        if (binarySearch(root2, target - root1.`val`)) {
            return true
        }
        return dfs(root1.left, root2, target) || dfs(root1.right, root2, target)
    }

    fun twoSumBSTs(root1: TreeNode?, root2: TreeNode?, target: Int): Boolean {
        return dfs(root1, root2, target)
    }
}

class Solution {
    func twoSumBSTs(_ root1: TreeNode?, _ root2: TreeNode?, _ target: Int) -> Bool {
        func binarySearch(_ root2: TreeNode?, _ target2: Int) -> Bool {
            guard let root2 = root2 else {
                return false
            }
            if root2.val == target2 {
                return true
            } else if root2.val > target2 {
                return binarySearch(root2.left, target2)
            } else {
                return binarySearch(root2.right, target2)
            }
        }

        func dfs(_ root1: TreeNode?, _ target: Int) -> Bool {
            guard let root1 = root1 else {
                return false
            }
            if binarySearch(root2, target - root1.val) {
                return true
            }
            return dfs(root1.left, target) || dfs(root1.right, target)
        }

        return dfs(root1, target)
    }
}

Time & Space Complexity

Time complexity: $O(m \cdot \log n)$
Space complexity: $O(\log m + \log n)$

Where $m$ and $n$ are the number of nodes in the two trees.

3. Hash Set

Intuition

We can trade space for time by storing all values from one tree in a hash set. Then for each value in the other tree, we check if the complement exists in the set in O(1) time.

Algorithm

Traverse the first tree and add all values to a hash set.
Traverse the second tree and add all values to another hash set.
For each value in the first set, check if target - value exists in the second set.
Return true if found, false otherwise.

class Solution:
    def twoSumBSTs(self, root1: Optional[TreeNode], root2: Optional[TreeNode], target: int) -> bool:
        def dfs(curr_node, node_set):
            if not curr_node:
                return
            dfs(curr_node.left, node_set)
            node_set.add(curr_node.val)
            dfs(curr_node.right, node_set)
        
        node_set1, node_set2 = set(), set()
        dfs(root1, node_set1)
        dfs(root2, node_set2)
        
        for value1 in node_set1:
            if target - value1 in node_set2:
                return True
        return False

class Solution {
    private void dfs(TreeNode currNode, Set<Integer> nodeSet) {
        if (currNode == null) {
            return;
        }
        dfs(currNode.left, nodeSet);
        nodeSet.add(currNode.val);
        dfs(currNode.right, nodeSet);
    }

    public boolean twoSumBSTs(TreeNode root1, TreeNode root2, int target) {
        Set<Integer> nodeSet1 = new HashSet<>();
        Set<Integer> nodeSet2 = new HashSet<>();
        dfs(root1, nodeSet1);
        dfs(root2, nodeSet2);

        for (int value1 : nodeSet1) {
            if (nodeSet2.contains(target - value1)) {
                return true;
            }
        }

        return false;
    }
}

class Solution {
public:
    void dfs(TreeNode* currNode, unordered_set<int>& nodeSet) {
        if (currNode == nullptr) {
            return;
        }
        dfs(currNode->left, nodeSet);
        nodeSet.insert(currNode->val);
        dfs(currNode->right, nodeSet);
    }

    bool twoSumBSTs(TreeNode* root1, TreeNode* root2, int target) {
        unordered_set<int> nodeSet1, nodeSet2;
        dfs(root1, nodeSet1);
        dfs(root2, nodeSet2);

        for (int value1 : nodeSet1) {
            if (nodeSet2.count(target - value1)) {
                return true;
            }
        }

        return false;
    }
};

class Solution {
    /**
     * @param {TreeNode} root1
     * @param {TreeNode} root2
     * @param {number} target
     * @return {boolean}
     */
    twoSumBSTs(root1, root2, target) {
        const dfs = (currNode, nodeSet) => {
            if (!currNode) {
                return;
            }
            dfs(currNode.left, nodeSet);
            nodeSet.add(currNode.val);
            dfs(currNode.right, nodeSet);
        };

        const nodeSet1 = new Set(), nodeSet2 = new Set();
        dfs(root1, nodeSet1);
        dfs(root2, nodeSet2);

        for (const value1 of nodeSet1) {
            if (nodeSet2.has(target - value1)) {
                return true;
            }
        }

        return false;
    }
}

public class Solution {
    private void Dfs(TreeNode currNode, HashSet<int> nodeSet) {
        if (currNode == null) {
            return;
        }
        Dfs(currNode.left, nodeSet);
        nodeSet.Add(currNode.val);
        Dfs(currNode.right, nodeSet);
    }

    public bool TwoSumBSTs(TreeNode root1, TreeNode root2, int target) {
        HashSet<int> nodeSet1 = new HashSet<int>();
        HashSet<int> nodeSet2 = new HashSet<int>();
        Dfs(root1, nodeSet1);
        Dfs(root2, nodeSet2);

        foreach (int value1 in nodeSet1) {
            if (nodeSet2.Contains(target - value1)) {
                return true;
            }
        }

        return false;
    }
}

func twoSumBSTs(root1 *TreeNode, root2 *TreeNode, target int) bool {
    var dfs func(currNode *TreeNode, nodeSet map[int]bool)
    dfs = func(currNode *TreeNode, nodeSet map[int]bool) {
        if currNode == nil {
            return
        }
        dfs(currNode.Left, nodeSet)
        nodeSet[currNode.Val] = true
        dfs(currNode.Right, nodeSet)
    }

    nodeSet1 := make(map[int]bool)
    nodeSet2 := make(map[int]bool)
    dfs(root1, nodeSet1)
    dfs(root2, nodeSet2)

    for value1 := range nodeSet1 {
        if nodeSet2[target-value1] {
            return true
        }
    }

    return false
}

class Solution {
    private fun dfs(currNode: TreeNode?, nodeSet: MutableSet<Int>) {
        if (currNode == null) {
            return
        }
        dfs(currNode.left, nodeSet)
        nodeSet.add(currNode.`val`)
        dfs(currNode.right, nodeSet)
    }

    fun twoSumBSTs(root1: TreeNode?, root2: TreeNode?, target: Int): Boolean {
        val nodeSet1 = mutableSetOf<Int>()
        val nodeSet2 = mutableSetOf<Int>()
        dfs(root1, nodeSet1)
        dfs(root2, nodeSet2)

        for (value1 in nodeSet1) {
            if (nodeSet2.contains(target - value1)) {
                return true
            }
        }

        return false
    }
}

class Solution {
    func twoSumBSTs(_ root1: TreeNode?, _ root2: TreeNode?, _ target: Int) -> Bool {
        func dfs(_ currNode: TreeNode?, _ nodeSet: inout Set<Int>) {
            guard let currNode = currNode else {
                return
            }
            dfs(currNode.left, &nodeSet)
            nodeSet.insert(currNode.val)
            dfs(currNode.right, &nodeSet)
        }

        var nodeSet1 = Set<Int>(), nodeSet2 = Set<Int>()
        dfs(root1, &nodeSet1)
        dfs(root2, &nodeSet2)

        for value1 in nodeSet1 {
            if nodeSet2.contains(target - value1) {
                return true
            }
        }

        return false
    }
}

Time & Space Complexity

Time complexity: $O(m + n)$
Space complexity: $O(m + n)$

Where $m$ and $n$ are the number of nodes in the two trees.

4. Two Pointers

Intuition

An inorder traversal of a BST produces a sorted list. If we have sorted lists from both trees, we can use the classic two-pointer technique: one pointer starts at the beginning of the first list (smallest), and another starts at the end of the second list (largest). We adjust pointers based on whether the current sum is too small or too large.

Algorithm

Perform inorder traversal on both trees to get two sorted lists.
Initialize pointer1 = 0 (start of first list) and pointer2 = len(list2) - 1 (end of second list).
While both pointers are valid:
- If sum equals target, return true.
- If sum is less than target, increment pointer1.
- If sum is greater than target, decrement pointer2.
Return false if no pair is found.

class Solution:
    def twoSumBSTs(self, root1: Optional[TreeNode], root2: Optional[TreeNode], target: int) -> bool:
        def dfs(curr_node, node_list):
            if not curr_node:
                return
            dfs(curr_node.left, node_list)
            node_list.append(curr_node.val)
            dfs(curr_node.right, node_list)
        
        node_list1, node_list2 = [], []
        dfs(root1, node_list1)
        dfs(root2, node_list2)
        
        pointer1 = 0
        pointer2 = len(node_list2) - 1
        while pointer1 < len(node_list1) and pointer2 >= 0:
            if node_list1[pointer1] + node_list2[pointer2] == target:
                return True
            elif node_list1[pointer1] + node_list2[pointer2] < target:
                pointer1 += 1
            else:
                pointer2 -= 1
        return False

class Solution {
    private void dfs(TreeNode currNode, List<Integer> nodeList) {
        if (currNode == null) {
            return;
        }
        dfs(currNode.left, nodeList);
        nodeList.add(currNode.val);
        dfs(currNode.right, nodeList);
    }

    public boolean twoSumBSTs(TreeNode root1, TreeNode root2, int target) {
        List<Integer> nodeList1 = new ArrayList<>();
        List<Integer> nodeList2 = new ArrayList<>();
        dfs(root1, nodeList1);
        dfs(root2, nodeList2);

        int pointer1 = 0, pointer2 = nodeList2.size() - 1;
        while (pointer1 < nodeList1.size() && pointer2 >= 0) {
            if (nodeList1.get(pointer1) + nodeList2.get(pointer2) == target) {
                return true;
            } else if (nodeList1.get(pointer1) + nodeList2.get(pointer2) < target) {
                pointer1++;
            } else {
                pointer2--;
            }
        }

        return false;
    }
}

class Solution {
public:
    void dfs(TreeNode* currNode, vector<int>& nodeList) {
        if (currNode == nullptr) {
            return;
        }
        dfs(currNode->left, nodeList);
        nodeList.push_back(currNode->val);
        dfs(currNode->right, nodeList);
    }

    bool twoSumBSTs(TreeNode* root1, TreeNode* root2, int target) {
        vector<int> nodeList1, nodeList2;
        dfs(root1, nodeList1);
        dfs(root2, nodeList2);

        int pointer1 = 0, pointer2 = nodeList2.size() - 1;
        while (pointer1 < (int)nodeList1.size() && pointer2 >= 0) {
            if (nodeList1[pointer1] + nodeList2[pointer2] == target) {
                return true;
            } else if (nodeList1[pointer1] + nodeList2[pointer2] < target) {
                pointer1++;
            } else {
                pointer2--;
            }
        }

        return false;
    }
};

class Solution {
    /**
     * @param {TreeNode} root1
     * @param {TreeNode} root2
     * @param {number} target
     * @return {boolean}
     */
    twoSumBSTs(root1, root2, target) {
        const dfs = (currNode, nodeList) => {
            if (!currNode) {
                return;
            }
            dfs(currNode.left, nodeList);
            nodeList.push(currNode.val);
            dfs(currNode.right, nodeList);
        };

        const nodeList1 = [], nodeList2 = [];
        dfs(root1, nodeList1);
        dfs(root2, nodeList2);

        let pointer1 = 0, pointer2 = nodeList2.length - 1;
        while (pointer1 < nodeList1.length && pointer2 >= 0) {
            if (nodeList1[pointer1] + nodeList2[pointer2] === target) {
                return true;
            } else if (nodeList1[pointer1] + nodeList2[pointer2] < target) {
                pointer1++;
            } else {
                pointer2--;
            }
        }

        return false;
    }
}

public class Solution {
    private void Dfs(TreeNode currNode, List<int> nodeList) {
        if (currNode == null) {
            return;
        }
        Dfs(currNode.left, nodeList);
        nodeList.Add(currNode.val);
        Dfs(currNode.right, nodeList);
    }

    public bool TwoSumBSTs(TreeNode root1, TreeNode root2, int target) {
        List<int> nodeList1 = new List<int>();
        List<int> nodeList2 = new List<int>();
        Dfs(root1, nodeList1);
        Dfs(root2, nodeList2);

        int pointer1 = 0, pointer2 = nodeList2.Count - 1;
        while (pointer1 < nodeList1.Count && pointer2 >= 0) {
            if (nodeList1[pointer1] + nodeList2[pointer2] == target) {
                return true;
            } else if (nodeList1[pointer1] + nodeList2[pointer2] < target) {
                pointer1++;
            } else {
                pointer2--;
            }
        }

        return false;
    }
}

func twoSumBSTs(root1 *TreeNode, root2 *TreeNode, target int) bool {
    var dfs func(currNode *TreeNode, nodeList *[]int)
    dfs = func(currNode *TreeNode, nodeList *[]int) {
        if currNode == nil {
            return
        }
        dfs(currNode.Left, nodeList)
        *nodeList = append(*nodeList, currNode.Val)
        dfs(currNode.Right, nodeList)
    }

    var nodeList1, nodeList2 []int
    dfs(root1, &nodeList1)
    dfs(root2, &nodeList2)

    pointer1, pointer2 := 0, len(nodeList2)-1
    for pointer1 < len(nodeList1) && pointer2 >= 0 {
        if nodeList1[pointer1]+nodeList2[pointer2] == target {
            return true
        } else if nodeList1[pointer1]+nodeList2[pointer2] < target {
            pointer1++
        } else {
            pointer2--
        }
    }

    return false
}

class Solution {
    private fun dfs(currNode: TreeNode?, nodeList: MutableList<Int>) {
        if (currNode == null) {
            return
        }
        dfs(currNode.left, nodeList)
        nodeList.add(currNode.`val`)
        dfs(currNode.right, nodeList)
    }

    fun twoSumBSTs(root1: TreeNode?, root2: TreeNode?, target: Int): Boolean {
        val nodeList1 = mutableListOf<Int>()
        val nodeList2 = mutableListOf<Int>()
        dfs(root1, nodeList1)
        dfs(root2, nodeList2)

        var pointer1 = 0
        var pointer2 = nodeList2.size - 1
        while (pointer1 < nodeList1.size && pointer2 >= 0) {
            when {
                nodeList1[pointer1] + nodeList2[pointer2] == target -> return true
                nodeList1[pointer1] + nodeList2[pointer2] < target -> pointer1++
                else -> pointer2--
            }
        }

        return false
    }
}

class Solution {
    func twoSumBSTs(_ root1: TreeNode?, _ root2: TreeNode?, _ target: Int) -> Bool {
        func dfs(_ currNode: TreeNode?, _ nodeList: inout [Int]) {
            guard let currNode = currNode else {
                return
            }
            dfs(currNode.left, &nodeList)
            nodeList.append(currNode.val)
            dfs(currNode.right, &nodeList)
        }

        var nodeList1 = [Int](), nodeList2 = [Int]()
        dfs(root1, &nodeList1)
        dfs(root2, &nodeList2)

        var pointer1 = 0, pointer2 = nodeList2.count - 1
        while pointer1 < nodeList1.count && pointer2 >= 0 {
            if nodeList1[pointer1] + nodeList2[pointer2] == target {
                return true
            } else if nodeList1[pointer1] + nodeList2[pointer2] < target {
                pointer1 += 1
            } else {
                pointer2 -= 1
            }
        }

        return false
    }
}

Time & Space Complexity

Time complexity: $O(m + n)$
Space complexity: $O(m + n)$

Where $m$ and $n$ are the number of nodes in the two trees.

5. Morris Traversal

Intuition

The two-pointer approach requires O(m + n) space to store the sorted lists. Morris traversal lets us iterate through a BST in sorted order using O(1) extra space by temporarily modifying tree pointers. We use forward Morris traversal on one tree and reverse Morris traversal on the other to simulate the two-pointer technique without extra space.

Algorithm

Create a forward Morris iterator for the first tree (yields values in ascending order).
Create a reverse Morris iterator for the second tree (yields values in descending order).
Get the first value from each iterator.
While both values are valid:
- If sum equals target, return true.
- If sum is less than target, advance the forward iterator.
- If sum is greater than target, advance the reverse iterator.
Return false if no pair is found.

class Solution:
    def twoSumBSTs(self, root1: Optional[TreeNode], root2: Optional[TreeNode], target: int) -> bool:
        def morris_traversal(root):
            current = root
            while current:
                if not current.left:
                    yield current.val
                    current = current.right
                else:
                    pre = current.left
                    while pre.right and pre.right != current:
                        pre = pre.right
                    if not pre.right:
                        pre.right = current
                        current = current.left
                    else:
                        pre.right = None
                        yield current.val
                        current = current.right

        def reversed_morris_traversal(root):
            current = root
            while current:
                if not current.right:
                    yield current.val
                    current = current.left
                else:
                    pre = current.right
                    while pre.left and pre.left != current:
                        pre = pre.left
                    if not pre.left:
                        pre.left = current
                        current = current.right
                    else:
                        pre.left = None
                        yield current.val
                        current = current.left
                        
        iterater1 = morris_traversal(root1)
        iterater2 = reversed_morris_traversal(root2)
        value1 = next(iterater1, None)
        value2 = next(iterater2, None)

        while value1 is not None and value2 is not None:
            if value1 + value2 == target:
                return True
            elif value1 + value2 < target:
                value1 = next(iterater1, None)
            else:
                value2 = next(iterater2, None)
        return False

class MorrisIterator implements Iterator<Integer> {
    private TreeNode current;
    private TreeNode pre;

    public MorrisIterator(TreeNode root) {
        current = root;
        pre = null;
    }

    public boolean hasNext() {
        return current != null;
    }

    public Integer next() {
        Integer val = null;
        while (current != null) {
            if (current.left == null) {
                val = current.val;
                current = current.right;
                break;
            } else {
                pre = current.left;
                while (pre.right != null && pre.right != current) {
                    pre = pre.right;
                }
                if (pre.right == null) {
                    pre.right = current;
                    current = current.left;
                } else {
                    pre.right = null;
                    val = current.val;
                    current = current.right;
                    break;
                }
            }
        }
        return val;
    }
}

class ReversedMorrisIterator implements Iterator<Integer> {
    private TreeNode current;
    private TreeNode pre;

    public ReversedMorrisIterator(TreeNode root) {
        current = root;
        pre = null;
    }

    public boolean hasNext() {
        return current != null;
    }

    public Integer next() {
        Integer val = null;
        while (current != null) {
            if (current.right == null) {
                val = current.val;
                current = current.left;
                break;
            } else {
                pre = current.right;
                while (pre.left != null && pre.left != current) {
                    pre = pre.left;
                }
                if (pre.left == null) {
                    pre.left = current;
                    current = current.right;
                } else {
                    pre.left = null;
                    val = current.val;
                    current = current.left;
                    break;
                }
            }
        }
        return val;
    }
}

class Solution {
    public boolean twoSumBSTs(TreeNode root1, TreeNode root2, int target) {
        MorrisIterator iterator1 = new MorrisIterator(root1);
        ReversedMorrisIterator iterator2 = new ReversedMorrisIterator(root2);
        
        int value1 = iterator1.next(), value2 = iterator2.next();
        
        while (value1 != Integer.MIN_VALUE && value2 != Integer.MIN_VALUE) {
            if (value1 + value2 == target) {
                return true;
            } else if (value1 + value2 < target) {
                if (iterator1.hasNext()) {
                    value1 = iterator1.next();
                } else {
                    value1 = Integer.MIN_VALUE;
                }
            } else {
                if (iterator2.hasNext()) {
                    value2 = iterator2.next();
                } else {
                    value2 = Integer.MIN_VALUE;
                }
            }
        }
        
        return false;
    }
}

class MorrisIterator {
private:
    TreeNode* current;
    TreeNode* pre;
    
public:
    MorrisIterator(TreeNode* root) : current(root), pre(nullptr) {}
    
    bool hasNext() {
        return current != nullptr;
    }
    
    int next() {
        int val = INT_MIN;
        while (current != nullptr) {
            if (current->left == nullptr) {
                val = current->val;
                current = current->right;
                break;
            } else {
                pre = current->left;
                while (pre->right != nullptr && pre->right != current) {
                    pre = pre->right;
                }
                if (pre->right == nullptr) {
                    pre->right = current;
                    current = current->left;
                } else {
                    pre->right = nullptr;
                    val = current->val;
                    current = current->right;
                    break;
                }
            }
        }
        return val;
    }
};

class ReversedMorrisIterator {
private:
    TreeNode* current;
    TreeNode* pre;
    
public:
    ReversedMorrisIterator(TreeNode* root) : current(root), pre(nullptr) {}
    
    bool hasNext() {
        return current != nullptr;
    }
    
    int next() {
        int val = INT_MIN;
        while (current != nullptr) {
            if (current->right == nullptr) {
                val = current->val;
                current = current->left;
                break;
            } else {
                pre = current->right;
                while (pre->left != nullptr && pre->left != current) {
                    pre = pre->left;
                }
                if (pre->left == nullptr) {
                    pre->left = current;
                    current = current->right;
                } else {
                    pre->left = nullptr;
                    val = current->val;
                    current = current->left;
                    break;
                }
            }
        }
        return val;
    }
};

class Solution {
public:
    bool twoSumBSTs(TreeNode* root1, TreeNode* root2, int target) {
        MorrisIterator iterator1(root1);
        ReversedMorrisIterator iterator2(root2);
        
        int value1 = iterator1.next();
        int value2 = iterator2.next();
        
        while (value1 != INT_MIN && value2 != INT_MIN) {
            if (value1 + value2 == target) {
                return true;
            } else if (value1 + value2 < target) {
                if (iterator1.hasNext()) {
                    value1 = iterator1.next();
                } else {
                    value1 = INT_MIN;
                }
            } else {
                if (iterator2.hasNext()) {
                    value2 = iterator2.next();
                } else {
                    value2 = INT_MIN;
                }
            }
        }
        
        return false;
    }
};

class Solution {
    /**
     * @param {TreeNode} root1
     * @param {TreeNode} root2
     * @param {number} target
     * @return {boolean}
     */
    twoSumBSTs(root1, root2, target) {
        function* morrisTraversal(root) {
            let current = root;
            while (current) {
                if (!current.left) {
                    yield current.val;
                    current = current.right;
                } else {
                    let pre = current.left;
                    while (pre.right && pre.right !== current) {
                        pre = pre.right;
                    }
                    if (!pre.right) {
                        pre.right = current;
                        current = current.left;
                    } else {
                        pre.right = null;
                        yield current.val;
                        current = current.right;
                    }
                }
            }
        }

        function* reversedMorrisTraversal(root) {
            let current = root;
            while (current) {
                if (!current.right) {
                    yield current.val;
                    current = current.left;
                } else {
                    let pre = current.right;
                    while (pre.left && pre.left !== current) {
                        pre = pre.left;
                    }
                    if (!pre.left) {
                        pre.left = current;
                        current = current.right;
                    } else {
                        pre.left = null;
                        yield current.val;
                        current = current.left;
                    }
                }
            }
        }

        const iterator1 = morrisTraversal(root1);
        const iterator2 = reversedMorrisTraversal(root2);

        let result1 = iterator1.next();
        let result2 = iterator2.next();

        let value1 = result1.done ? null : result1.value;
        let value2 = result2.done ? null : result2.value;

        while (value1 !== null && value2 !== null) {
            if (value1 + value2 === target) {
                return true;
            } else if (value1 + value2 < target) {
                result1 = iterator1.next();
                value1 = result1.done ? null : result1.value;
            } else {
                result2 = iterator2.next();
                value2 = result2.done ? null : result2.value;
            }
        }

        return false;
    }
}

public class Solution {
    private class MorrisIterator {
        private TreeNode current;
        private TreeNode pre;

        public MorrisIterator(TreeNode root) {
            current = root;
            pre = null;
        }

        public bool HasNext() {
            return current != null;
        }

        public int? Next() {
            int? val = null;
            while (current != null) {
                if (current.left == null) {
                    val = current.val;
                    current = current.right;
                    break;
                } else {
                    pre = current.left;
                    while (pre.right != null && pre.right != current) {
                        pre = pre.right;
                    }
                    if (pre.right == null) {
                        pre.right = current;
                        current = current.left;
                    } else {
                        pre.right = null;
                        val = current.val;
                        current = current.right;
                        break;
                    }
                }
            }
            return val;
        }
    }

    private class ReversedMorrisIterator {
        private TreeNode current;
        private TreeNode pre;

        public ReversedMorrisIterator(TreeNode root) {
            current = root;
            pre = null;
        }

        public bool HasNext() {
            return current != null;
        }

        public int? Next() {
            int? val = null;
            while (current != null) {
                if (current.right == null) {
                    val = current.val;
                    current = current.left;
                    break;
                } else {
                    pre = current.right;
                    while (pre.left != null && pre.left != current) {
                        pre = pre.left;
                    }
                    if (pre.left == null) {
                        pre.left = current;
                        current = current.right;
                    } else {
                        pre.left = null;
                        val = current.val;
                        current = current.left;
                        break;
                    }
                }
            }
            return val;
        }
    }

    public bool TwoSumBSTs(TreeNode root1, TreeNode root2, int target) {
        var iterator1 = new MorrisIterator(root1);
        var iterator2 = new ReversedMorrisIterator(root2);

        int? value1 = iterator1.Next();
        int? value2 = iterator2.Next();

        while (value1 != null && value2 != null) {
            if (value1 + value2 == target) {
                return true;
            } else if (value1 + value2 < target) {
                value1 = iterator1.HasNext() ? iterator1.Next() : null;
            } else {
                value2 = iterator2.HasNext() ? iterator2.Next() : null;
            }
        }

        return false;
    }
}

func twoSumBSTs(root1 *TreeNode, root2 *TreeNode, target int) bool {
    morrisChan := func(root *TreeNode) <-chan int {
        ch := make(chan int)
        go func() {
            defer close(ch)
            current := root
            for current != nil {
                if current.Left == nil {
                    ch <- current.Val
                    current = current.Right
                } else {
                    pre := current.Left
                    for pre.Right != nil && pre.Right != current {
                        pre = pre.Right
                    }
                    if pre.Right == nil {
                        pre.Right = current
                        current = current.Left
                    } else {
                        pre.Right = nil
                        ch <- current.Val
                        current = current.Right
                    }
                }
            }
        }()
        return ch
    }

    reversedMorrisChan := func(root *TreeNode) <-chan int {
        ch := make(chan int)
        go func() {
            defer close(ch)
            current := root
            for current != nil {
                if current.Right == nil {
                    ch <- current.Val
                    current = current.Left
                } else {
                    pre := current.Right
                    for pre.Left != nil && pre.Left != current {
                        pre = pre.Left
                    }
                    if pre.Left == nil {
                        pre.Left = current
                        current = current.Right
                    } else {
                        pre.Left = nil
                        ch <- current.Val
                        current = current.Left
                    }
                }
            }
        }()
        return ch
    }

    ch1 := morrisChan(root1)
    ch2 := reversedMorrisChan(root2)

    value1, ok1 := <-ch1
    value2, ok2 := <-ch2

    for ok1 && ok2 {
        if value1+value2 == target {
            return true
        } else if value1+value2 < target {
            value1, ok1 = <-ch1
        } else {
            value2, ok2 = <-ch2
        }
    }

    return false
}

class Solution {
    private class MorrisIterator(root: TreeNode?) : Iterator<Int> {
        private var current: TreeNode? = root
        private var pre: TreeNode? = null
        private var nextVal: Int? = null

        init {
            advance()
        }

        private fun advance() {
            nextVal = null
            while (current != null) {
                if (current!!.left == null) {
                    nextVal = current!!.`val`
                    current = current!!.right
                    break
                } else {
                    pre = current!!.left
                    while (pre!!.right != null && pre!!.right != current) {
                        pre = pre!!.right
                    }
                    if (pre!!.right == null) {
                        pre!!.right = current
                        current = current!!.left
                    } else {
                        pre!!.right = null
                        nextVal = current!!.`val`
                        current = current!!.right
                        break
                    }
                }
            }
        }

        override fun hasNext(): Boolean = nextVal != null

        override fun next(): Int {
            val result = nextVal!!
            advance()
            return result
        }
    }

    private class ReversedMorrisIterator(root: TreeNode?) : Iterator<Int> {
        private var current: TreeNode? = root
        private var pre: TreeNode? = null
        private var nextVal: Int? = null

        init {
            advance()
        }

        private fun advance() {
            nextVal = null
            while (current != null) {
                if (current!!.right == null) {
                    nextVal = current!!.`val`
                    current = current!!.left
                    break
                } else {
                    pre = current!!.right
                    while (pre!!.left != null && pre!!.left != current) {
                        pre = pre!!.left
                    }
                    if (pre!!.left == null) {
                        pre!!.left = current
                        current = current!!.right
                    } else {
                        pre!!.left = null
                        nextVal = current!!.`val`
                        current = current!!.left
                        break
                    }
                }
            }
        }

        override fun hasNext(): Boolean = nextVal != null

        override fun next(): Int {
            val result = nextVal!!
            advance()
            return result
        }
    }

    fun twoSumBSTs(root1: TreeNode?, root2: TreeNode?, target: Int): Boolean {
        val iterator1 = MorrisIterator(root1)
        val iterator2 = ReversedMorrisIterator(root2)

        var value1: Int? = if (iterator1.hasNext()) iterator1.next() else null
        var value2: Int? = if (iterator2.hasNext()) iterator2.next() else null

        while (value1 != null && value2 != null) {
            when {
                value1 + value2 == target -> return true
                value1 + value2 < target -> value1 = if (iterator1.hasNext()) iterator1.next() else null
                else -> value2 = if (iterator2.hasNext()) iterator2.next() else null
            }
        }

        return false
    }
}

class Solution {
    func twoSumBSTs(_ root1: TreeNode?, _ root2: TreeNode?, _ target: Int) -> Bool {
        class MorrisIterator: IteratorProtocol {
            private var current: TreeNode?
            private var pre: TreeNode?

            init(_ root: TreeNode?) {
                current = root
                pre = nil
            }

            func next() -> Int? {
                var val: Int? = nil
                while current != nil {
                    if current!.left == nil {
                        val = current!.val
                        current = current!.right
                        break
                    } else {
                        pre = current!.left
                        while pre!.right != nil && pre!.right !== current {
                            pre = pre!.right
                        }
                        if pre!.right == nil {
                            pre!.right = current
                            current = current!.left
                        } else {
                            pre!.right = nil
                            val = current!.val
                            current = current!.right
                            break
                        }
                    }
                }
                return val
            }
        }

        class ReversedMorrisIterator: IteratorProtocol {
            private var current: TreeNode?
            private var pre: TreeNode?

            init(_ root: TreeNode?) {
                current = root
                pre = nil
            }

            func next() -> Int? {
                var val: Int? = nil
                while current != nil {
                    if current!.right == nil {
                        val = current!.val
                        current = current!.left
                        break
                    } else {
                        pre = current!.right
                        while pre!.left != nil && pre!.left !== current {
                            pre = pre!.left
                        }
                        if pre!.left == nil {
                            pre!.left = current
                            current = current!.right
                        } else {
                            pre!.left = nil
                            val = current!.val
                            current = current!.left
                            break
                        }
                    }
                }
                return val
            }
        }

        var iterator1 = MorrisIterator(root1)
        var iterator2 = ReversedMorrisIterator(root2)

        var value1 = iterator1.next()
        var value2 = iterator2.next()

        while value1 != nil && value2 != nil {
            let sum = value1! + value2!
            if sum == target {
                return true
            } else if sum < target {
                value1 = iterator1.next()
            } else {
                value2 = iterator2.next()
            }
        }

        return false
    }
}

Time & Space Complexity

Time complexity: $O(m + n)$
Space complexity: $O(1)$

Where $m$ and $n$ are the number of nodes in the two trees.

Common Pitfalls

Treating the Trees as Regular Binary Trees

A common mistake is ignoring the BST property and treating both trees as generic binary trees. This leads to inefficient solutions that don't leverage the sorted structure for binary search or two-pointer approaches. Always consider whether the BST ordering can help optimize your search.

Computing the Complement Incorrectly

When searching for a pair that sums to the target, some forget to compute target - value correctly or accidentally search for the wrong value. For example, if target = 9 and the current node has value 4, you need to search for 5 in the other tree, not 9 - 4 = 5 with any other formula.

Modifying Tree Structure with Morris Traversal

Morris traversal temporarily modifies tree pointers to achieve O(1) space. A pitfall is forgetting to restore the original tree structure after traversal or exiting early without cleanup. This can corrupt the tree for subsequent operations or cause infinite loops.