Given a binary search tree, write a function kthSmallest to find the kth smallest element in it.

Note:
You may assume k is always valid, 1 ≤ k ≤ BST’s total elements.

Follow up:
What if the BST is modified (insert/delete operations) often and you need to find the kth smallest frequently? How would you optimize the kthSmallest routine?

题意：从一个二叉搜索树上找第k小元素。
followup：如果这个二叉树经常改变（增删），然后又会经常查第k小元素，该怎么优化？

1、暴力的思路：把二叉搜索树转为一个有序数组，返回数组第k-1个元素。要遍历所有树节点，时间复杂度O(n)；要用一个数组存储，空间复杂度O(n)。

2、先找到最小元素，通过栈存储父节点，利用前序遍历找到第k个。

    public int kthSmallest(TreeNode root, int k) {

        //1 bst转为有序数组，返回第k个，时间 O(n), 空间 O(n)

        //2 先找到bst中最小的叶子节点，然后递归向上搜索第k个，父节点存储在栈中
        Stack<TreeNode> stack = new Stack<>();
        TreeNode dummy = root;
        while (dummy != null) {
            stack.push(dummy);
            dummy = dummy.left;
        }

        while (k > 1) {
            TreeNode top = stack.pop();
            dummy = top.right;
            while (dummy != null) {
                stack.push(dummy);
                dummy = dummy.left;
            }
            k--;
        }

        return stack.peek().val;
    }