leetcode--230 Kth Smallest Element in a BST(二分搜索树第K小的值)

Given a binary search tree, write a function kthSmallest to find thekth 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?

在二叉搜索树种，找到第K个元素。

算法如下：

1、计算左子树元素个数left。

2、 left+1 = K，则根节点即为第K个元素

      left >=k, 则第K个元素在左子树中，

     left +1 <k, 则转换为在右子树中，寻找第K-left-1元素。

代码如下：

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