[LeetCode] 173. Binary Search Tree Iterator

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.

Thought process:
The idea is to design a class that implements the iterative version of in-order traversal, like in "94. Binary Tree Inorder Traversal". 
I will create a stack as an instance variable. In the constructor, I will push the root and its left branch onto the stack (go down root.left until bottom). hasNext() method can return whether the stack is empty. next() will pop a node off the stack, push the left branch of its right child onto stack if it has one, and return the node's value.

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

public class BSTIterator {
    private Stack<TreeNode> stack;

    public BSTIterator(TreeNode root) {
        stack = new Stack<>();
        pushLeftPath(root);
    }

    /** @return whether we have a next smallest number */
    public boolean hasNext() {
        return !stack.isEmpty();
    }

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

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

Time complexity: amortized O(1) for both next() and hasNext().
Location: Philadelphia, PA, USA

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