Solutions

Prerequisites

Before attempting this problem, you should be comfortable with:

Binary Search Tree (BST) Properties - Understanding that inorder traversal of a valid BST produces a sorted sequence
Inorder Traversal - Both recursive and iterative implementations for visiting nodes in left-root-right order
Tree Validation - Checking if a binary tree satisfies BST constraints using min/max bounds
Morris Traversal (Optional) - Space-optimized tree traversal technique using threading for O(1) space solution

1. Brute Force

Intuition

The simplest approach is to try every possible pair of nodes and check if swapping their values produces a valid BST. Since exactly two nodes were swapped, one such pair must restore the tree to its correct state.

For each pair of nodes, we swap their values, validate whether the resulting tree is a valid BST, and either keep the swap (if valid) or revert it (if invalid). While inefficient, this approach guarantees finding the solution by exhaustively checking all possibilities.

Algorithm

Define a helper function isBST that validates whether a tree is a valid BST using BFS with min/max bounds for each node.
Use a nested DFS approach: the outer DFS (dfs) iterates through each node as a potential first swap candidate.
For each first candidate, the inner DFS (dfs1) iterates through every other node as the second swap candidate.
For each pair, swap their values, check if the tree is now a valid BST, and if so, return true. Otherwise, swap back and continue.
Once a valid swap is found, the tree is corrected in place.

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution:
    def recoverTree(self, root: Optional[TreeNode]) -> None:
        """
        Do not return anything, modify root in-place instead.
        """
        def isBST(node):
            if not node:
                return True

            q = deque([(node, float("-inf"), float("inf"))])
            while q:
                cur, left, right = q.popleft()
                if not (left < cur.val < right):
                    return False
                if cur.left:
                    q.append((cur.left, left, cur.val))
                if cur.right:
                    q.append((cur.right, cur.val, right))

            return True

        def dfs1(node1, node2):
            if not node2 or node1 == node2:
                return False
            
            node1.val, node2.val = node2.val, node1.val
            if isBST(root):
                return True
            
            node1.val, node2.val = node2.val, node1.val
            return dfs1(node1, node2.left) or dfs1(node1, node2.right)


        def dfs(node):
            if not node:
                return False
            
            if dfs1(node, root):
                return True
            
            return dfs(node.left) or dfs(node.right)
        
        dfs(root)
        return root

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
public class Solution {
    public void recoverTree(TreeNode root) {
        dfs(root, root);
    }

    private boolean dfs(TreeNode node1, TreeNode root) {
        if (node1 == null) return false;
        if (dfs1(node1, root, root)) return true;
        return dfs(node1.left, root) || dfs(node1.right, root);
    }

    private boolean dfs1(TreeNode node1, TreeNode node2, TreeNode root) {
        if (node2 == null || node1 == node2) return false;

        swap(node1, node2);
        if (isBST(root)) return true;
        swap(node1, node2);

        return dfs1(node1, node2.left, root) || dfs1(node1, node2.right, root);
    }

    private boolean isBST(TreeNode node) {
        Queue<Object[]> q = new LinkedList<>();
        q.offer(new Object[]{node, Long.MIN_VALUE, Long.MAX_VALUE});

        while (!q.isEmpty()) {
            Object[] curr = q.poll();
            TreeNode n = (TreeNode) curr[0];
            long left = (long) curr[1];
            long right = (long) curr[2];

            if (n == null) continue;
            if (n.val <= left || n.val >= right) return false;

            q.offer(new Object[]{n.left, left, (long) n.val});
            q.offer(new Object[]{n.right, (long) n.val, right});
        }

        return true;
    }

    private void swap(TreeNode a, TreeNode b) {
        int tmp = a.val;
        a.val = b.val;
        b.val = tmp;
    }
}

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
class Solution {
public:
    void recoverTree(TreeNode* root) {
        dfs(root, root, root);
    }

    bool dfs(TreeNode* node1, TreeNode* node2, TreeNode* root) {
        if (!node1) return false;
        if (dfs1(node1, node2, root)) return true;
        return dfs(node1->left, node2, root) || dfs(node1->right, node2, root);
    }

    bool dfs1(TreeNode* node1, TreeNode* node2, TreeNode* root) {
        if (!node2 || node1 == node2) return false;

        swap(node1->val, node2->val);
        if (isBST(root)) return true;
        swap(node1->val, node2->val);

        return dfs1(node1, node2->left, root) || dfs1(node1, node2->right, root);
    }

    bool isBST(TreeNode* node) {
        TreeNode* prev = nullptr;
        return inorder(node, prev);
    }

    bool inorder(TreeNode* node, TreeNode*& prev) {
        if (!node) return true;
        if (!inorder(node->left, prev)) return false;
        if (prev && prev->val >= node->val) return false;
        prev = node;
        return inorder(node->right, prev);
    }
};

/**
 * Definition for a binary tree node.
 * class TreeNode {
 *     constructor(val = 0, left = null, right = null) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
 class Solution {
    /**
     * @param {TreeNode} root
     * @return {void} Do not return anything, modify root in-place instead.
     */
    recoverTree(root) {
        const isBST = (node) => {
            const q = [[node, -Infinity, Infinity]];
            while (q.length) {
                const [cur, left, right] = q.shift();
                if (!cur) continue;
                if (!(left < cur.val && cur.val < right)) return false;
                q.push([cur.left, left, cur.val]);
                q.push([cur.right, cur.val, right]);
            }
            return true;
        };

        const dfs1 = (node1, node2) => {
            if (!node2 || node1 === node2) return false;
            [node1.val, node2.val] = [node2.val, node1.val];
            if (isBST(root)) return true;
            [node1.val, node2.val] = [node2.val, node1.val];
            return dfs1(node1, node2.left) || dfs1(node1, node2.right);
        };

        const dfs = (node) => {
            if (!node) return false;
            if (dfs1(node, root)) return true;
            return dfs(node.left) || dfs(node.right);
        };

        dfs(root);
    }
 }

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     public int val;
 *     public TreeNode left;
 *     public TreeNode right;
 *     public TreeNode(int val=0, TreeNode left=null, TreeNode right=null) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
public class Solution {
    public void RecoverTree(TreeNode root) {
        Dfs(root, root);
    }

    private bool Dfs(TreeNode node1, TreeNode root) {
        if (node1 == null) return false;
        if (Dfs1(node1, root, root)) return true;
        return Dfs(node1.left, root) || Dfs(node1.right, root);
    }

    private bool Dfs1(TreeNode node1, TreeNode node2, TreeNode root) {
        if (node2 == null || node1 == node2) return false;

        Swap(node1, node2);
        if (IsBST(root)) return true;
        Swap(node1, node2);

        return Dfs1(node1, node2.left, root) || Dfs1(node1, node2.right, root);
    }

    private bool IsBST(TreeNode node) {
        var q = new Queue<(TreeNode node, long min, long max)>();
        q.Enqueue((node, long.MinValue, long.MaxValue));

        while (q.Count > 0) {
            var (cur, min, max) = q.Dequeue();
            if (cur == null) continue;
            long val = cur.val;
            if (val <= min || val >= max) return false;

            q.Enqueue((cur.left, min, val));
            q.Enqueue((cur.right, val, max));
        }
        return true;
    }

    private void Swap(TreeNode a, TreeNode b) {
        int temp = a.val;
        a.val = b.val;
        b.val = temp;
    }
}

/**
 * Definition for a binary tree node.
 * type TreeNode struct {
 *     Val int
 *     Left *TreeNode
 *     Right *TreeNode
 * }
 */
func recoverTree(root *TreeNode) {
    var dfs func(node1 *TreeNode) bool
    var dfs1 func(node1, node2 *TreeNode) bool

    isBST := func(node *TreeNode) bool {
        type item struct {
            n          *TreeNode
            min, max   int64
        }
        q := []item{{node, math.MinInt64, math.MaxInt64}}
        for len(q) > 0 {
            cur := q[0]
            q = q[1:]
            if cur.n == nil {
                continue
            }
            val := int64(cur.n.Val)
            if val <= cur.min || val >= cur.max {
                return false
            }
            q = append(q, item{cur.n.Left, cur.min, val})
            q = append(q, item{cur.n.Right, val, cur.max})
        }
        return true
    }

    dfs1 = func(node1, node2 *TreeNode) bool {
        if node2 == nil || node1 == node2 {
            return false
        }
        node1.Val, node2.Val = node2.Val, node1.Val
        if isBST(root) {
            return true
        }
        node1.Val, node2.Val = node2.Val, node1.Val
        return dfs1(node1, node2.Left) || dfs1(node1, node2.Right)
    }

    dfs = func(node1 *TreeNode) bool {
        if node1 == nil {
            return false
        }
        if dfs1(node1, root) {
            return true
        }
        return dfs(node1.Left) || dfs(node1.Right)
    }

    dfs(root)
}

/**
 * Example:
 * var ti = TreeNode(5)
 * var v = ti.`val`
 * Definition for a binary tree node.
 * class TreeNode(var `val`: Int) {
 *     var left: TreeNode? = null
 *     var right: TreeNode? = null
 * }
 */
class Solution {
    fun recoverTree(root: TreeNode?): Unit {
        fun isBST(node: TreeNode?): Boolean {
            val q = ArrayDeque<Triple<TreeNode?, Long, Long>>()
            q.add(Triple(node, Long.MIN_VALUE, Long.MAX_VALUE))
            while (q.isNotEmpty()) {
                val (cur, min, max) = q.removeFirst()
                if (cur == null) continue
                val v = cur.`val`.toLong()
                if (v <= min || v >= max) return false
                q.add(Triple(cur.left, min, v))
                q.add(Triple(cur.right, v, max))
            }
            return true
        }

        fun dfs1(node1: TreeNode, node2: TreeNode?): Boolean {
            if (node2 == null || node1 === node2) return false
            val tmp = node1.`val`
            node1.`val` = node2.`val`
            node2.`val` = tmp
            if (isBST(root)) return true
            node2.`val` = node1.`val`
            node1.`val` = tmp
            return dfs1(node1, node2.left) || dfs1(node1, node2.right)
        }

        fun dfs(node1: TreeNode?): Boolean {
            if (node1 == null) return false
            if (dfs1(node1, root)) return true
            return dfs(node1.left) || dfs(node1.right)
        }

        dfs(root)
    }
}

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     public var val: Int
 *     public var left: TreeNode?
 *     public var right: TreeNode?
 *     public init() { self.val = 0; self.left = nil; self.right = nil; }
 *     public init(_ val: Int) { self.val = val; self.left = nil; self.right = nil; }
 *     public init(_ val: Int, _ left: TreeNode?, _ right: TreeNode?) {
 *         self.val = val
 *         self.left = left
 *         self.right = right
 *     }
 * }
 */
class Solution {
    func recoverTree(_ root: TreeNode?) {
        func isBST(_ node: TreeNode?) -> Bool {
            var q: [(TreeNode?, Int64, Int64)] = [(node, Int64.min, Int64.max)]
            while !q.isEmpty {
                let (cur, minVal, maxVal) = q.removeFirst()
                guard let cur = cur else { continue }
                let v = Int64(cur.val)
                if v <= minVal || v >= maxVal { return false }
                q.append((cur.left, minVal, v))
                q.append((cur.right, v, maxVal))
            }
            return true
        }

        func dfs1(_ node1: TreeNode, _ node2: TreeNode?) -> Bool {
            guard let node2 = node2 else { return false }
            if node1 === node2 { return false }
            let tmp = node1.val
            node1.val = node2.val
            node2.val = tmp
            if isBST(root) { return true }
            node2.val = node1.val
            node1.val = tmp
            return dfs1(node1, node2.left) || dfs1(node1, node2.right)
        }

        func dfs(_ node1: TreeNode?) -> Bool {
            guard let node1 = node1 else { return false }
            if dfs1(node1, root) { return true }
            return dfs(node1.left) || dfs(node1.right)
        }

        _ = dfs(root)
    }
}

Time & Space Complexity

Time complexity: $O(n ^ 2)$
Space complexity: $O(n)$

2. Inorder Traversal

Intuition

A key property of BSTs is that an inorder traversal produces values in sorted order. When two nodes are swapped incorrectly, this sorted sequence will have one or two "inversions" where a value is greater than the next value.

If the swapped nodes are adjacent in the inorder sequence, there will be exactly one inversion. If they are not adjacent, there will be two inversions. In the first inversion, the larger (out-of-place) node is the first swapped node. In the second inversion (or the same one if only one exists), the smaller node is the second swapped node.

By collecting all nodes during inorder traversal and then scanning for these inversions, we can identify the two swapped nodes and swap their values back.

Algorithm

Perform an inorder traversal of the tree and store all nodes in a list.
Scan the list to find inversions (where arr[i].val > arr[i+1].val).
On the first inversion, mark node1 = arr[i] and node2 = arr[i+1].
If a second inversion is found, update node2 = arr[i+1] (the first swap target is already correct).
Swap the values of node1 and node2 to restore the BST.

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution:
    def recoverTree(self, root: Optional[TreeNode]) -> None:
        """
        Do not return anything, modify root in-place instead.
        """
        arr = []
        def inorder(node):
            if not node:
                return
            
            inorder(node.left)
            arr.append(node)
            inorder(node.right)

        inorder(root)
        node1, node2 = None, None

        for i in range(len(arr) - 1):
            if arr[i].val > arr[i + 1].val:
                node2 = arr[i + 1]
                if node1 is None:
                    node1 = arr[i]
                else:
                    break
        node1.val, node2.val = node2.val, node1.val

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
public class Solution {
    public void recoverTree(TreeNode root) {
        List<TreeNode> arr = new ArrayList<>();
        inorder(root, arr);

        TreeNode node1 = null, node2 = null;
        for (int i = 0; i < arr.size() - 1; i++) {
            if (arr.get(i).val > arr.get(i + 1).val) {
                node2 = arr.get(i + 1);
                if (node1 == null) node1 = arr.get(i);
                else break;
            }
        }

        int temp = node1.val;
        node1.val = node2.val;
        node2.val = temp;
    }

    private void inorder(TreeNode node, List<TreeNode> arr) {
        if (node == null) return;
        inorder(node.left, arr);
        arr.add(node);
        inorder(node.right, arr);
    }
}

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
class Solution {
public:
    void recoverTree(TreeNode* root) {
        vector<TreeNode*> arr;
        inorder(root, arr);

        TreeNode* node1 = nullptr;
        TreeNode* node2 = nullptr;

        for (int i = 0; i < arr.size() - 1; i++) {
            if (arr[i]->val > arr[i + 1]->val) {
                node2 = arr[i + 1];
                if (!node1) node1 = arr[i];
                else break;
            }
        }

        swap(node1->val, node2->val);
    }

    void inorder(TreeNode* node, vector<TreeNode*>& arr) {
        if (!node) return;
        inorder(node->left, arr);
        arr.push_back(node);
        inorder(node->right, arr);
    }
};

/**
 * Definition for a binary tree node.
 * class TreeNode {
 *     constructor(val = 0, left = null, right = null) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
 class Solution {
    /**
     * @param {TreeNode} root
     * @return {void} Do not return anything, modify root in-place instead.
     */
    recoverTree(root) {
        const arr = [];

        const inorder = node => {
            if (!node) return;
            inorder(node.left);
            arr.push(node);
            inorder(node.right);
        };

        inorder(root);

        let node1 = null, node2 = null;
        for (let i = 0; i < arr.length - 1; i++) {
            if (arr[i].val > arr[i + 1].val) {
                node2 = arr[i + 1];
                if (node1 === null) node1 = arr[i];
                else break;
            }
        }

        [node1.val, node2.val] = [node2.val, node1.val];
    }
 }

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     public int val;
 *     public TreeNode left;
 *     public TreeNode right;
 *     public TreeNode(int val=0, TreeNode left=null, TreeNode right=null) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
public class Solution {
    public void RecoverTree(TreeNode root) {
        var arr = new List<TreeNode>();
        Inorder(root, arr);

        TreeNode node1 = null, node2 = null;
        for (int i = 0; i < arr.Count - 1; i++) {
            if (arr[i].val > arr[i + 1].val) {
                node2 = arr[i + 1];
                if (node1 == null) node1 = arr[i];
                else break;
            }
        }

        int temp = node1.val;
        node1.val = node2.val;
        node2.val = temp;
    }

    private void Inorder(TreeNode node, List<TreeNode> arr) {
        if (node == null) return;
        Inorder(node.left, arr);
        arr.Add(node);
        Inorder(node.right, arr);
    }
}

/**
 * Definition for a binary tree node.
 * type TreeNode struct {
 *     Val int
 *     Left *TreeNode
 *     Right *TreeNode
 * }
 */
func recoverTree(root *TreeNode) {
    var arr []*TreeNode
    var inorder func(node *TreeNode)
    inorder = func(node *TreeNode) {
        if node == nil {
            return
        }
        inorder(node.Left)
        arr = append(arr, node)
        inorder(node.Right)
    }

    inorder(root)

    var node1, node2 *TreeNode
    for i := 0; i < len(arr)-1; i++ {
        if arr[i].Val > arr[i+1].Val {
            node2 = arr[i+1]
            if node1 == nil {
                node1 = arr[i]
            } else {
                break
            }
        }
    }

    node1.Val, node2.Val = node2.Val, node1.Val
}

/**
 * Example:
 * var ti = TreeNode(5)
 * var v = ti.`val`
 * Definition for a binary tree node.
 * class TreeNode(var `val`: Int) {
 *     var left: TreeNode? = null
 *     var right: TreeNode? = null
 * }
 */
class Solution {
    fun recoverTree(root: TreeNode?): Unit {
        val arr = mutableListOf<TreeNode>()

        fun inorder(node: TreeNode?) {
            if (node == null) return
            inorder(node.left)
            arr.add(node)
            inorder(node.right)
        }

        inorder(root)

        var node1: TreeNode? = null
        var node2: TreeNode? = null
        for (i in 0 until arr.size - 1) {
            if (arr[i].`val` > arr[i + 1].`val`) {
                node2 = arr[i + 1]
                if (node1 == null) node1 = arr[i]
                else break
            }
        }

        val temp = node1!!.`val`
        node1.`val` = node2!!.`val`
        node2.`val` = temp
    }
}

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     public var val: Int
 *     public var left: TreeNode?
 *     public var right: TreeNode?
 *     public init() { self.val = 0; self.left = nil; self.right = nil; }
 *     public init(_ val: Int) { self.val = val; self.left = nil; self.right = nil; }
 *     public init(_ val: Int, _ left: TreeNode?, _ right: TreeNode?) {
 *         self.val = val
 *         self.left = left
 *         self.right = right
 *     }
 * }
 */
class Solution {
    func recoverTree(_ root: TreeNode?) {
        var arr = [TreeNode]()

        func inorder(_ node: TreeNode?) {
            guard let node = node else { return }
            inorder(node.left)
            arr.append(node)
            inorder(node.right)
        }

        inorder(root)

        var node1: TreeNode? = nil
        var node2: TreeNode? = nil
        for i in 0..<(arr.count - 1) {
            if arr[i].val > arr[i + 1].val {
                node2 = arr[i + 1]
                if node1 == nil { node1 = arr[i] }
                else { break }
            }
        }

        let temp = node1!.val
        node1!.val = node2!.val
        node2!.val = temp
    }
}

Time & Space Complexity

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

3. Iterative Inorder Traversal

Intuition

This approach uses the same logic as the recursive inorder traversal but implements it iteratively using an explicit stack. The advantage is that we can detect inversions on the fly during traversal rather than collecting all nodes first.

By keeping track of the previously visited node, we can immediately detect when the current node's value is less than the previous node's value, signaling an inversion. This allows us to identify the swapped nodes in a single pass through the tree.

Algorithm

Initialize an empty stack and pointers for node1, node2, prev, and curr.
Traverse the tree iteratively using the stack for inorder traversal.
For each visited node, compare it with prev. If prev.val > curr.val, an inversion is found.
On the first inversion, set node1 = prev and node2 = curr.
On the second inversion, update node2 = curr and break early.
After traversal, swap the values of node1 and node2.

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution:
    def recoverTree(self, root: Optional[TreeNode]) -> None:
        stack = []
        node1 = node2 = prev = None
        curr = root

        while stack or curr:
            while curr:
                stack.append(curr)
                curr = curr.left

            curr = stack.pop()
            if prev and prev.val > curr.val:
                node2 = curr
                if not node1:
                    node1 = prev
                else:
                    break
            prev = curr
            curr = curr.right

        node1.val, node2.val = node2.val, node1.val

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
public class Solution {
    public void recoverTree(TreeNode root) {
        Stack<TreeNode> stack = new Stack<>();
        TreeNode node1 = null, node2 = null, prev = null, curr = root;

        while (!stack.isEmpty() || curr != null) {
            while (curr != null) {
                stack.push(curr);
                curr = curr.left;
            }

            curr = stack.pop();
            if (prev != null && prev.val > curr.val) {
                node2 = curr;
                if (node1 == null) node1 = prev;
                else break;
            }
            prev = curr;
            curr = curr.right;
        }

        int temp = node1.val;
        node1.val = node2.val;
        node2.val = temp;
    }
}

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
class Solution {
public:
    void recoverTree(TreeNode* root) {
        stack<TreeNode*> stack;
        TreeNode *node1 = nullptr, *node2 = nullptr, *prev = nullptr, *curr = root;

        while (!stack.empty() || curr) {
            while (curr) {
                stack.push(curr);
                curr = curr->left;
            }

            curr = stack.top(); stack.pop();
            if (prev && prev->val > curr->val) {
                node2 = curr;
                if (!node1) node1 = prev;
                else break;
            }
            prev = curr;
            curr = curr->right;
        }

        swap(node1->val, node2->val);
    }
};

/**
 * Definition for a binary tree node.
 * class TreeNode {
 *     constructor(val = 0, left = null, right = null) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
 class Solution {
    /**
     * @param {TreeNode} root
     * @return {void} Do not return anything, modify root in-place instead.
     */
    recoverTree(root) {
        let stack = [];
        let node1 = null, node2 = null, prev = null, curr = root;

        while (stack.length > 0 || curr) {
            while (curr) {
                stack.push(curr);
                curr = curr.left;
            }

            curr = stack.pop();
            if (prev && prev.val > curr.val) {
                node2 = curr;
                if (!node1) node1 = prev;
                else break;
            }
            prev = curr;
            curr = curr.right;
        }

        [node1.val, node2.val] = [node2.val, node1.val];
    }
 }

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     public int val;
 *     public TreeNode left;
 *     public TreeNode right;
 *     public TreeNode(int val=0, TreeNode left=null, TreeNode right=null) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
public class Solution {
    public void RecoverTree(TreeNode root) {
        var stack = new Stack<TreeNode>();
        TreeNode node1 = null, node2 = null, prev = null, curr = root;

        while (stack.Count > 0 || curr != null) {
            while (curr != null) {
                stack.Push(curr);
                curr = curr.left;
            }

            curr = stack.Pop();
            if (prev != null && prev.val > curr.val) {
                node2 = curr;
                if (node1 == null) node1 = prev;
                else break;
            }
            prev = curr;
            curr = curr.right;
        }

        int temp = node1.val;
        node1.val = node2.val;
        node2.val = temp;
    }
}

/**
 * Definition for a binary tree node.
 * type TreeNode struct {
 *     Val int
 *     Left *TreeNode
 *     Right *TreeNode
 * }
 */
func recoverTree(root *TreeNode) {
    var stack []*TreeNode
    var node1, node2, prev *TreeNode
    curr := root

    for len(stack) > 0 || curr != nil {
        for curr != nil {
            stack = append(stack, curr)
            curr = curr.Left
        }

        curr = stack[len(stack)-1]
        stack = stack[:len(stack)-1]

        if prev != nil && prev.Val > curr.Val {
            node2 = curr
            if node1 == nil {
                node1 = prev
            } else {
                break
            }
        }
        prev = curr
        curr = curr.Right
    }

    node1.Val, node2.Val = node2.Val, node1.Val
}

/**
 * Example:
 * var ti = TreeNode(5)
 * var v = ti.`val`
 * Definition for a binary tree node.
 * class TreeNode(var `val`: Int) {
 *     var left: TreeNode? = null
 *     var right: TreeNode? = null
 * }
 */
class Solution {
    fun recoverTree(root: TreeNode?): Unit {
        val stack = ArrayDeque<TreeNode>()
        var node1: TreeNode? = null
        var node2: TreeNode? = null
        var prev: TreeNode? = null
        var curr = root

        while (stack.isNotEmpty() || curr != null) {
            while (curr != null) {
                stack.addLast(curr)
                curr = curr.left
            }

            curr = stack.removeLast()
            if (prev != null && prev.`val` > curr!!.`val`) {
                node2 = curr
                if (node1 == null) node1 = prev
                else break
            }
            prev = curr
            curr = curr?.right
        }

        val temp = node1!!.`val`
        node1.`val` = node2!!.`val`
        node2.`val` = temp
    }
}

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     public var val: Int
 *     public var left: TreeNode?
 *     public var right: TreeNode?
 *     public init() { self.val = 0; self.left = nil; self.right = nil; }
 *     public init(_ val: Int) { self.val = val; self.left = nil; self.right = nil; }
 *     public init(_ val: Int, _ left: TreeNode?, _ right: TreeNode?) {
 *         self.val = val
 *         self.left = left
 *         self.right = right
 *     }
 * }
 */
class Solution {
    func recoverTree(_ root: TreeNode?) {
        var stack = [TreeNode]()
        var node1: TreeNode? = nil
        var node2: TreeNode? = nil
        var prev: TreeNode? = nil
        var curr = root

        while !stack.isEmpty || curr != nil {
            while curr != nil {
                stack.append(curr!)
                curr = curr!.left
            }

            curr = stack.removeLast()
            if prev != nil && prev!.val > curr!.val {
                node2 = curr
                if node1 == nil { node1 = prev }
                else { break }
            }
            prev = curr
            curr = curr?.right
        }

        let temp = node1!.val
        node1!.val = node2!.val
        node2!.val = temp
    }
}

Time & Space Complexity

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

4. Morris Traversal

Intuition

Morris traversal allows us to perform inorder traversal without using a stack or recursion, achieving O(1) space complexity. The technique works by temporarily modifying the tree structure: for each node with a left child, we find its inorder predecessor and create a temporary link back to the current node.

This temporary threading allows us to return to ancestor nodes after processing the left subtree without needing a stack. We can detect inversions during this traversal just like in the iterative approach, but without the extra space for a stack.

Algorithm

Initialize pointers for node1, node2, prev, and curr (starting at root).
While curr is not null:
- If curr has no left child, process it (check for inversion with prev), update prev, and move to the right child.
- If curr has a left child, find its inorder predecessor (rightmost node in left subtree).
  - If the predecessor's right pointer is null, create a thread to curr and move left.
  - If the predecessor's right pointer points to curr, remove the thread, process curr, update prev, and move right.
After traversal, swap the values of node1 and node2.

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution:
    def recoverTree(self, root: Optional[TreeNode]) -> None:
        node1 = node2 = prev = None
        curr = root

        while curr:
            if not curr.left:
                if prev and prev.val > curr.val:
                    node2 = curr
                    if not node1:
                        node1 = prev
                prev = curr
                curr = curr.right
            else:
                pred = curr.left
                while pred.right and pred.right != curr:
                    pred = pred.right

                if not pred.right:
                    pred.right = curr
                    curr = curr.left
                else:
                    pred.right = None
                    if prev and prev.val > curr.val:
                        node2 = curr
                        if not node1:
                            node1 = prev
                    prev = curr
                    curr = curr.right

        node1.val, node2.val = node2.val, node1.val

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
public class Solution {
    public void recoverTree(TreeNode root) {
        TreeNode node1 = null, node2 = null, prev = null, curr = root;

        while (curr != null) {
            if (curr.left == null) {
                if (prev != null && prev.val > curr.val) {
                    node2 = curr;
                    if (node1 == null) node1 = prev;
                }
                prev = curr;
                curr = curr.right;
            } else {
                TreeNode pred = curr.left;
                while (pred.right != null && pred.right != curr) {
                    pred = pred.right;
                }

                if (pred.right == null) {
                    pred.right = curr;
                    curr = curr.left;
                } else {
                    pred.right = null;
                    if (prev != null && prev.val > curr.val) {
                        node2 = curr;
                        if (node1 == null) node1 = prev;
                    }
                    prev = curr;
                    curr = curr.right;
                }
            }
        }

        int temp = node1.val;
        node1.val = node2.val;
        node2.val = temp;
    }
}

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
class Solution {
public:
    void recoverTree(TreeNode* root) {
        TreeNode* node1 = nullptr;
        TreeNode* node2 = nullptr;
        TreeNode* prev = nullptr;
        TreeNode* curr = root;

        while (curr) {
            if (!curr->left) {
                if (prev && prev->val > curr->val) {
                    node2 = curr;
                    if (!node1) node1 = prev;
                }
                prev = curr;
                curr = curr->right;
            } else {
                TreeNode* pred = curr->left;
                while (pred->right && pred->right != curr) {
                    pred = pred->right;
                }

                if (!pred->right) {
                    pred->right = curr;
                    curr = curr->left;
                } else {
                    pred->right = nullptr;
                    if (prev && prev->val > curr->val) {
                        node2 = curr;
                        if (!node1) node1 = prev;
                    }
                    prev = curr;
                    curr = curr->right;
                }
            }
        }

        swap(node1->val, node2->val);
    }
};

/**
 * Definition for a binary tree node.
 * class TreeNode {
 *     constructor(val = 0, left = null, right = null) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
 class Solution {
    /**
     * @param {TreeNode} root
     * @return {void} Do not return anything, modify root in-place instead.
     */
    recoverTree(root) {
        let node1 = null, node2 = null, prev = null, curr = root;

        while (curr !== null) {
            if (curr.left === null) {
                if (prev !== null && prev.val > curr.val) {
                    node2 = curr;
                    if (node1 === null) node1 = prev;
                }
                prev = curr;
                curr = curr.right;
            } else {
                let pred = curr.left;
                while (pred.right !== null && pred.right !== curr) {
                    pred = pred.right;
                }

                if (pred.right === null) {
                    pred.right = curr;
                    curr = curr.left;
                } else {
                    pred.right = null;
                    if (prev !== null && prev.val > curr.val) {
                        node2 = curr;
                        if (node1 === null) node1 = prev;
                    }
                    prev = curr;
                    curr = curr.right;
                }
            }
        }

        let temp = node1.val;
        node1.val = node2.val;
        node2.val = temp;
    }
 }

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     public int val;
 *     public TreeNode left;
 *     public TreeNode right;
 *     public TreeNode(int val=0, TreeNode left=null, TreeNode right=null) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
public class Solution {
    public void RecoverTree(TreeNode root) {
        TreeNode node1 = null, node2 = null, prev = null, curr = root;

        while (curr != null) {
            if (curr.left == null) {
                if (prev != null && prev.val > curr.val) {
                    node2 = curr;
                    if (node1 == null) node1 = prev;
                }
                prev = curr;
                curr = curr.right;
            } else {
                TreeNode pred = curr.left;
                while (pred.right != null && pred.right != curr) {
                    pred = pred.right;
                }

                if (pred.right == null) {
                    pred.right = curr;
                    curr = curr.left;
                } else {
                    pred.right = null;
                    if (prev != null && prev.val > curr.val) {
                        node2 = curr;
                        if (node1 == null) node1 = prev;
                    }
                    prev = curr;
                    curr = curr.right;
                }
            }
        }

        int temp = node1.val;
        node1.val = node2.val;
        node2.val = temp;
    }
}

/**
 * Definition for a binary tree node.
 * type TreeNode struct {
 *     Val int
 *     Left *TreeNode
 *     Right *TreeNode
 * }
 */
func recoverTree(root *TreeNode) {
    var node1, node2, prev *TreeNode
    curr := root

    for curr != nil {
        if curr.Left == nil {
            if prev != nil && prev.Val > curr.Val {
                node2 = curr
                if node1 == nil {
                    node1 = prev
                }
            }
            prev = curr
            curr = curr.Right
        } else {
            pred := curr.Left
            for pred.Right != nil && pred.Right != curr {
                pred = pred.Right
            }

            if pred.Right == nil {
                pred.Right = curr
                curr = curr.Left
            } else {
                pred.Right = nil
                if prev != nil && prev.Val > curr.Val {
                    node2 = curr
                    if node1 == nil {
                        node1 = prev
                    }
                }
                prev = curr
                curr = curr.Right
            }
        }
    }

    node1.Val, node2.Val = node2.Val, node1.Val
}

/**
 * Example:
 * var ti = TreeNode(5)
 * var v = ti.`val`
 * Definition for a binary tree node.
 * class TreeNode(var `val`: Int) {
 *     var left: TreeNode? = null
 *     var right: TreeNode? = null
 * }
 */
class Solution {
    fun recoverTree(root: TreeNode?): Unit {
        var node1: TreeNode? = null
        var node2: TreeNode? = null
        var prev: TreeNode? = null
        var curr = root

        while (curr != null) {
            if (curr.left == null) {
                if (prev != null && prev.`val` > curr.`val`) {
                    node2 = curr
                    if (node1 == null) node1 = prev
                }
                prev = curr
                curr = curr.right
            } else {
                var pred = curr.left
                while (pred?.right != null && pred.right != curr) {
                    pred = pred.right
                }

                if (pred?.right == null) {
                    pred?.right = curr
                    curr = curr.left
                } else {
                    pred.right = null
                    if (prev != null && prev.`val` > curr.`val`) {
                        node2 = curr
                        if (node1 == null) node1 = prev
                    }
                    prev = curr
                    curr = curr.right
                }
            }
        }

        val temp = node1!!.`val`
        node1.`val` = node2!!.`val`
        node2.`val` = temp
    }
}

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     public var val: Int
 *     public var left: TreeNode?
 *     public var right: TreeNode?
 *     public init() { self.val = 0; self.left = nil; self.right = nil; }
 *     public init(_ val: Int) { self.val = val; self.left = nil; self.right = nil; }
 *     public init(_ val: Int, _ left: TreeNode?, _ right: TreeNode?) {
 *         self.val = val
 *         self.left = left
 *         self.right = right
 *     }
 * }
 */
class Solution {
    func recoverTree(_ root: TreeNode?) {
        var node1: TreeNode? = nil
        var node2: TreeNode? = nil
        var prev: TreeNode? = nil
        var curr = root

        while curr != nil {
            if curr!.left == nil {
                if prev != nil && prev!.val > curr!.val {
                    node2 = curr
                    if node1 == nil { node1 = prev }
                }
                prev = curr
                curr = curr!.right
            } else {
                var pred = curr!.left
                while pred?.right != nil && pred?.right !== curr {
                    pred = pred?.right
                }

                if pred?.right == nil {
                    pred?.right = curr
                    curr = curr!.left
                } else {
                    pred?.right = nil
                    if prev != nil && prev!.val > curr!.val {
                        node2 = curr
                        if node1 == nil { node1 = prev }
                    }
                    prev = curr
                    curr = curr!.right
                }
            }
        }

        let temp = node1!.val
        node1!.val = node2!.val
        node2!.val = temp
    }
}

Time & Space Complexity

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

Common Pitfalls

Assuming Only One Inversion Exists

When two nodes are swapped in a BST, there can be either one or two inversions in the inorder sequence depending on whether the swapped nodes are adjacent. Assuming only one inversion and not checking for a second one causes incorrect identification of the swapped nodes.

Swapping Node References Instead of Values

The problem asks to recover the tree by swapping values, not by restructuring node pointers. Attempting to swap the actual node positions in the tree is unnecessarily complex and error-prone. Simply swap the val fields of the two identified nodes.

Incorrectly Identifying First and Second Nodes

In the first inversion, node1 is the larger (out-of-place) element, and node2 is the smaller one. If a second inversion is found, node2 should be updated to the smaller element of that inversion, while node1 remains unchanged. Mixing up which node to update at each inversion leads to swapping the wrong pair.