[LeetCode]Convert Sorted Array to Binary Search Tree

题目：

Convert Sorted Array to Binary Search Tree

 

Given an array where elements are sorted in ascending order, convert it to a height balanced BST.

来源：https://oj.leetcode.com/problems/convert-sorted-array-to-binary-search-tree/

思路：

将排序数组转换成二叉平衡树。

设数组的下标范围为[left, right]，它的中间元素的下标为middle = (left + right) / 2;
middle 将数组分为[left, middle - 1]和[middle + 1, right];
middle作为一个相对根节点，它的左右子子树分别为[left, middle - 1]和[middle + 1, right]的所有元素，依次递归

C++ AC代码：

/** * Definition for binary tree * struct TreeNode { *     int val; *     TreeNode *left; *     TreeNode *right; *     TreeNode(int x) : val(x), left(NULL), right(NULL) {} * }; */class Solution {public:    TreeNode *sortedArrayToBST(vector<int> &num) {        return sortedArrayToBST(num, 0, num.size()-1);    }TreeNode *sortedArrayToBST(vector<int> &num, int left, int right) {    if ( left > right ) return NULL;int middle = (left + right)/2;TreeNode *root = new TreeNode(num[middle]);root->left = sortedArrayToBST(num, left, middle-1);root->right = sortedArrayToBST(num, middle+1, right);return root;}};

运行时间 92ms