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.

思路：BST的中序遍历就是一个升序，满足题目的迭代要求。考虑到要O(h)的空间复杂度，用栈可以保存当前最小node的父节点，每次最小值在栈顶，O(1)的时间复杂度。

public class BSTIterator {

Stack<TreeNode> stack = new Stack<>();

public BSTIterator(TreeNode root) {
    while (root != null) {
        stack.push(root);
        root = root.left;
    }
}

/** @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();
    TreeNode dummy = node.right;
    while (dummy != null) {
        stack.push(dummy);
        dummy = dummy.left;
    }
    return node.val;
}
}