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.

Solution

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

public class BSTIterator {

    private Stack<TreeNode> ancestorPath;
    private TreeNode curNode;

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

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

    /** @return the next smallest number */
    public int next() {
        if(ancestorPath.size() == 0) return 0;

        TreeNode curNode = ancestorPath.pop();
        TreeNode iter = curNode.right;
        while(iter != null) {
            ancestorPath.push(iter);
            iter = iter.left;
        }
        return curNode.val;
    }
}

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