Binary Search Tree Iterator

Binary Search Tree Iterator - Problem

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.

Input & Output

Example 1 — Basic BST Iterator

$ Input: commands = ["BSTIterator", "next", "next", "hasNext", "next", "hasNext", "next", "hasNext", "next", "hasNext"], values = [[[7, 3, 15, null, null, 9, 20]], [], [], [], [], [], [], [], [], []]

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

💡 Note: Initialize iterator with BST root 7. In-order traversal gives 3,7,9,15,20. Each next() returns the next smallest value, hasNext() checks if more values exist.

Example 2 — Single Node Tree

$ Input: commands = ["BSTIterator", "next", "hasNext"], values = [[[5]], [], []]

› Output: [null, 5, false]

💡 Note: Tree with single node 5. First next() returns 5, then hasNext() returns false as no more values exist.

Example 3 — Left Skewed Tree

$ Input: commands = ["BSTIterator", "next", "next", "hasNext"], values = [[[3, 1, null, null, 2]], [], [], []]

› Output: [null, 1, 2, true]

💡 Note: Left-skewed BST. In-order traversal: 1,2,3. After getting 1 and 2, hasNext() is true because 3 remains.

Constraints

The number of nodes in the tree is in the range [1, 10⁵]
0 ≤ Node.val ≤ 10⁶
At most 10⁵ calls will be made to hasNext, and next

Visualization

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Asked in

F Facebook 35 G Google 28 a Amazon 22 M Microsoft 18

The key insight is to use a stack to simulate the call stack of recursive in-order traversal, allowing us to generate values on-demand. The optimal approach uses controlled recursion with a stack, achieving O(1) amortized time per operation and O(h) space where h is tree height. Time: O(n) total, Space: O(h)

Common Approaches

✓ Optimized

⏱️ Time: N/A Space: N/A

Controlled Recursion with Stack

⏱️ Time: O(n) Space: O(h)

Maintain a stack to keep track of nodes to visit next. Push all left children onto the stack, then pop for next value and process right subtree. This simulates the call stack of recursive in-order traversal.

Pre-compute All Values

⏱️ Time: O(n) Space: O(n)

Perform a complete in-order traversal during construction and store all values in an array. Then use an index pointer to iterate through the pre-computed values.

Algorithm Steps — Algorithm Steps

Code -

solution.c — C

Time & Space Complexity

Time Complexity

⏱️

✓ Linear Growth

Space Complexity

⚡ Linearithmic Space

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