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.

这是一道很经典的题目，考的非递归的中序遍历。其实这道题等价于写一个二叉树中序遍历的迭代器。需要内置一个栈，一开始先存储到最左叶子节点的路径。在遍历的过程中，只要当前节点存在右孩子，则进入右孩子，存除从此处开始到当前子树里最左叶子节点的路径。

public class BSTIterator {
	Stack<TreeNode> stack;

	public BSTIterator(TreeNode root) {
		stack = new Stack<TreeNode>();
		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();
		int ret = node.val;
		if (node.right != null) {
			node = node.right;
			while (node != null) {
				stack.push(node);
				node = node.left;
			}
		}
		return ret;
	}
}

[LeetCode] Binary Search Tree Iterator