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:

Try to utilize the property of a BST.
What if you could modify the BST node's structure?
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：

 1 /**
 2  * Definition for a binary tree node.
 3  * public class TreeNode {
 4  *     int val;
 5  *     TreeNode left;
 6  *     TreeNode right;
 7  *     TreeNode(int x) { val = x; }
 8  * }
 9  */
10 public class Solution {
11     public int kthSmallest(TreeNode root, int k) {
12         int count = countNodes(root.left);
13         if(k<=count) return kthSmallest(root.left,k);
14         else if(k>count+1) return kthSmallest(root.right,k-count-1);
15         return root.val;
16     }
17     public int countNodes(TreeNode node){
18         if(node==null) return 0;
19         return 1+countNodes(node.left)+countNodes(node.right);
20     }
21 }