108. Convert Sorted Array to Bin

题目链接
tag:

• Easy；

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

For this problem, a height-balanced binary tree is defined as a binary tree in which the depth of the two subtrees of every node never differ by more than 1.

Example:

Given the sorted array: [-10,-3,0,5,9],
One possible answer is: [0,-3,9,-10,null,5], which represents the following height balanced BST:
0
/ \
-3 9
/ /
-10 5

思路：
  所谓二叉搜索树，是一种始终满足左<根<右的特性，如果将二叉搜索树按中序遍历的话，得到的就是一个有序数组了。那么反过来，根节点应该是有序数组的中间点，从中间点分开为左右两个有序数组，在分别找出其中间点作为原中间点的左右两个子节点，这就是二分查找法的核心思想。代码如下：

/**
 * 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:
    TreeNode* sortedArrayToBST(vector<int>& nums) {
        if (nums.empty()) 
            return NULL;
        int mid = nums.size() / 2;
        TreeNode *cur = new TreeNode(nums[mid]);
        vector<int> left(nums.begin(), nums.begin()+mid), right(nums.begin()+mid+1, nums.end());
        cur->left = sortedArrayToBST(left);
        cur->right = sortedArrayToBST(right);
        return cur;
    }
};