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.

Credits:
Special thanks to @ts for adding this problem and creating all test cases.

/**
 * Definition for binary tree
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class BSTIterator {
    private Stack<TreeNode> stk;
    public BSTIterator(TreeNode root) {
        stk = new Stack<TreeNode>();
        TreeNode p = root;
        while(p!=null) {
            stk.push(p);
            p = p.left;
        }
    }
    /** @return whether we have a next smallest number */
    public boolean hasNext() {
        return !stk.isEmpty();
    }
    /** @return the next smallest number */
    public int next() {
        if(!hasNext()) {
            // error here
        }
        TreeNode node = stk.pop();
        if(node.right!=null) {
            TreeNode tmp = node.right;
            while(tmp!=null) {
                stk.push(tmp);
                tmp = tmp.left;
            }
        }
        return node.val;
    }
}
/**
 * Your BSTIterator will be called like this:
 * BSTIterator i = new BSTIterator(root);
 * while (i.hasNext()) v[f()] = i.next();
 */