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?

題意：從一個二叉搜索樹上找第k小元素。
followup：如果這個二叉樹經常改變（增刪），然后又會經常查第k小元素，該怎么優化？

1、暴力的思路：把二叉搜索樹轉為一個有序數組，返回數組第k-1個元素。要遍歷所有樹節點，時間復雜度O(n)；要用一個數組存儲，空間復雜度O(n)。

2、先找到最小元素，通過棧存儲父節點，利用前序遍歷找到第k個。

    public int kthSmallest(TreeNode root, int k) {

        //1 bst轉為有序數組，返回第k個，時間 O(n), 空間 O(n)

        //2 先找到bst中最小的葉子節點，然后遞歸向上搜索第k個，父節點存儲在棧中
        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;
    }