173. Binary Search Tree Iterator - Explanation

Problem Link

Description

Implement the BSTIterator class that represents an iterator over the in-order traversal of a binary search tree (BST):

• BSTIterator(TreeNode root) Initializes an object of the BSTIterator class. The root of the BST is given as part of the constructor. The pointer should be initialized to a non-existent number smaller than any element in the BST.
• boolean hasNext() Returns true if there exists a number in the traversal to the right of the pointer, otherwise returns false.
• int next() Moves the pointer to the right, then returns the number at the pointer.

Notice that by initializing the pointer to a non-existent smallest number, the first call to next() will return the smallest element in the BST.

You may assume that next() calls will always be valid. That is, there will be at least a next number in the in-order traversal when next() is called.

Example 1:

Input: ["BSTIterator","next","next","hasNext","next","hasNext","next","hasNext","next","hasNext"], [[[7,3,15,null,null,9,20]],[],[],[],[],[],[],[],[],[]]

Output: [null,3,7,true,9,true,15,true,20,false]

Explanation:
BSTIterator bSTIterator = new BSTIterator([7, 3, 15, null, null, 9, 20]);
bSTIterator.next(); // return 3
bSTIterator.next(); // return 7
bSTIterator.hasNext(); // return True
bSTIterator.next(); // return 9
bSTIterator.hasNext(); // return True
bSTIterator.next(); // return 15
bSTIterator.hasNext(); // return True
bSTIterator.next(); // return 20
bSTIterator.hasNext(); // return False

Constraints:

• 1 <= The number of nodes in the tree <= 100,000.
• 0 <= Node.val <= 1,000,000
• At most 100,000 calls will be made to hasNext, and next.

Follow up:

• Could you implement next() and hasNext() to run in average O(1) time and use O(h) memory, where h is the height of the tree?

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1. Flattening the BST (DFS)

# 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 BSTIterator:

    def __init__(self, root: Optional[TreeNode]):
        self.arr = []
        self.itr = 0

        def dfs(node):
            if not node:
                return
            dfs(node.left)
            self.arr.append(node.val)
            dfs(node.right)

        dfs(root)

    def next(self) -> int:
        val = self.arr[self.itr]
        self.itr += 1
        return val

    def hasNext(self) -> bool:
        return self.itr < len(self.arr)

Time & Space Complexity

• Time complexity:
  • $O(n)$ time for initialization.
  • $O(n)$ time for each $next()$ and $hasNext()$ function calls.
• Space complexity: $O(n)$

---

2. Flatten the BST (Iterative DFS)

# 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 BSTIterator:

    def __init__(self, root: Optional[TreeNode]):
        self.arr = []
        self.itr = 0

        stack = []
        while root or stack:
            while root:
                stack.append(root)
                root = root.left
            root = stack.pop()
            self.arr.append(root.val)
            root = root.right

    def next(self) -> int:
        val = self.arr[self.itr]
        self.itr += 1
        return val

    def hasNext(self) -> bool:
        return self.itr < len(self.arr)

Time & Space Complexity

• Time complexity:
  • $O(n)$ time for initialization.
  • $O(n)$ time for each $next()$ and $hasNext()$ function calls.
• Space complexity: $O(n)$

---

3. Iterative DFS - I

# 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 BSTIterator:

    def __init__(self, root: Optional[TreeNode]):
        self.stack = []
        while root:
            self.stack.append(root)
            root = root.left

    def next(self) -> int:
        res = self.stack.pop()
        cur = res.right
        while cur:
            self.stack.append(cur)
            cur = cur.left
        return res.val

    def hasNext(self) -> bool:
        return bool(self.stack)

Time & Space Complexity

• Time complexity: $O(1)$ in average for each function call.
• Space complexity: $O(h)$

Where $n$ is the number of nodes and $h$ is the height of the given tree.

---

4. Iterative DFS - II

# 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 BSTIterator:

    def __init__(self, root: Optional[TreeNode]):
        self.cur = root
        self.stack = []

    def next(self) -> int:
        while self.cur:
            self.stack.append(self.cur)
            self.cur = self.cur.left

        node = self.stack.pop()
        self.cur = node.right
        return node.val

    def hasNext(self) -> bool:
        return bool(self.cur) or bool(self.stack)

Time & Space Complexity

• Time complexity: $O(1)$ in average for each function call.
• Space complexity: $O(h)$

Where $n$ is the number of nodes and $h$ is the height of the given tree.