[Leetcode] 230. Kth Smallest Element in a BST 解题报告

题目：

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?

思路：

BST的中序遍历结果一定是有序的，所以采用中序遍历二叉树，然后返回遍历到的第k个结果即可。如果要频繁查找的话，可以在数据结构中增加一个字段，记录左边节点有多少个。这样可以将查找的时间复杂度降低到O(h)，其中h是BST的高度。

代码：

/** * Definition for a binary tree node. * struct TreeNode { *     int val; *     TreeNode *left; *     TreeNode *right; *     TreeNode(int x) : val(x), left(NULL), right(NULL) {} * }; */class Solution {public:    int kthSmallest(TreeNode* root, int k) {        if(root == NULL) {            return -1;        }        int index = 0, val = 0;        inorderTraverse(root, index, k, val);        return val;    }private:    void inorderTraverse(TreeNode* node, int& index, int k, int& val) {        if(node->left) {            inorderTraverse(node->left, index, k, val);        }        if(++index == k) {            val = node->val;            return;        }        if(node->right) {            inorderTraverse(node->right, index, k, val);        }    }};