230. Kth Smallest Element in a BST

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?

Hint:

1. Try to utilize the property of a BST.
2. What if you could modify the BST node's structure?
3. The optimal runtime complexity is O(height of BST).

本题开始做的时候想到了还是用前序遍历的方法来做，但是对于follow up就做不出来了，看了答案才发现，可以把二分搜索树分成三部分，一部分是二分搜索树的左子树，一部分是当前节点，一部分是二分搜索树的右子树部分。只要数出左子树的个数就可以找到第k小的元素了，代码如下：

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    public int kthSmallest(TreeNode root, int k) {
        int count = countNodes(root.left);
        if(k<=count) return kthSmallest(root.left,k);
        else if(k>count+1) return kthSmallest(root.right,k-count-1);
        return root.val;
    }
    public int countNodes(TreeNode node){
        if(node==null) return 0;
        return 1+countNodes(node.left)+countNodes(node.right);
    }
}