# 173 Binary Search Tree Iterator – Medium

### Problem:

Implement an iterator over a binary search tree (BST). Your iterator will be initialized with the root node of a BST.

Calling next() will return the next smallest number in the BST.

Note: next() and hasNext() should run in average O(1) time and uses O(h) memory, where h is the height of the tree.

### Thoughts:

Using a stack if helpful to implement the iterator.

Implementing iterator is like a inorder walk. which is left, root, right.

### Solutions:

``````/**
* Definition for binary tree
* public class TreeNode {
*     int val;
*     TreeNode left;
*     TreeNode right;
*     TreeNode(int x) { val = x; }
* }
*/

public class BSTIterator {
Stack<TreeNode> helper = new Stack<TreeNode>();
public BSTIterator(TreeNode root) {
while (root != null) {
helper.push(root);
root = root.left;
}
}

/** @return whether we have a next smallest number */
public boolean hasNext() {
return helper.size() != 0;
}

/** @return the next smallest number */
public int next() {
TreeNode result = helper.pop();
TreeNode node = result.right;
while (node != null) {
helper.push(node);
node = node.left;
}
return result.val;
}
}

/**
* Your BSTIterator will be called like this:
* BSTIterator i = new BSTIterator(root);
* while (i.hasNext()) v[f()] = i.next();
*/
``````