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(gè)二叉搜索樹上找第k小元素。
followup：如果這個(gè)二叉樹經(jīng)常改變（增刪），然后又會經(jīng)常查第k小元素，該怎么優(yōu)化？

1、暴力的思路：把二叉搜索樹轉(zhuǎn)為一個(gè)有序數(shù)組，返回?cái)?shù)組第k-1個(gè)元素。要遍歷所有樹節(jié)點(diǎn)，時(shí)間復(fù)雜度O(n)；要用一個(gè)數(shù)組存儲，空間復(fù)雜度O(n)。

2、先找到最小元素，通過棧存儲父節(jié)點(diǎn)，利用前序遍歷找到第k個(gè)。

    public int kthSmallest(TreeNode root, int k) {

        //1 bst轉(zhuǎn)為有序數(shù)組，返回第k個(gè)，時(shí)間 O(n), 空間 O(n)

        //2 先找到bst中最小的葉子節(jié)點(diǎn)，然后遞歸向上搜索第k個(gè)，父節(jié)點(diǎn)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;
    }