LeetCode题解：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).

题意：给定二叉搜索树，找出第K小的数。假如二叉树经常会被修改，而且频繁需要查找第K小的数呢？

解题思路：DFS

代码：

/** * 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) {        if(root == null){            return 0;        }        int leftNodeNums = countNodeNums(root.left);        if(leftNodeNums == k - 1){            return root.val;        }else if(leftNodeNums < k){            return kthSmallest(root.right, k - leftNodeNums - 1);        }else{            return kthSmallest(root.left, k);        }    }    private int countNodeNums(TreeNode root){        if(root == null){            return 0;        }        return 1 + countNodeNums(root.left) + countNodeNums(root.right);    }}