108. Convert Sorted Array to Binary Search Tree

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


用一个已排序的数组构建一个平衡二叉搜索树

C++（16ms）：

/**
 * 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) {
        int len = nums.size() ;
        if (len == 0)
            return NULL ;
        if (len == 1)
            return new TreeNode(nums[0]) ;
        int mid = len/2 ;
        TreeNode* root = new TreeNode(nums[mid]) ;
        vector<int> leftArr(nums.begin() , nums.begin()+mid ) ;
        vector<int> rightArr(nums.begin()+mid+1 , nums.end() ) ;
        root->left = sortedArrayToBST(leftArr) ;
        root->right = sortedArrayToBST(rightArr) ;
        return root ;
    }
};