Problem discription:

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

Accepted Code:

1 /** 2  * Definition for binary tree 3  * struct TreeNode { 4  *     int val; 5  *     TreeNode *left; 6  *     TreeNode *right; 7  *     TreeNode(int x) : val(x), left(NULL), right(NULL) {} 8  * }; 9  */10 class Solution {11 public:12     TreeNode *sortedArrayToBST(vector
     
       &num, int start, int end) {13         if (start > end) {14             return NULL;15         } 16         // if you write " int middle = start + (end - start) >> 1"17         // instead of the following sentence, you should keep in 18         // mind that the priority of ">>" is weaker than "+" operator19         int middle = start + (end - start) / 2;20         TreeNode *p = new TreeNode(num[middle]);21         // elements at the left of num[middle] construct22         // the left child tree of "p" and its right child23         // tree consists of elements located at the right24         // of num[middle]...25         // because the given array is sorted26         if (start != end) {27             p->left = sortedArrayToBST(num, start, middle - 1);28             p->right = sortedArrayToBST(num, middle + 1, end);29         }30         return p;31     }32     TreeNode *sortedArrayToBST(vector
      
        &num) {33         // call the overload function34         return sortedArrayToBST(num, 0, num.size() - 1);35     }36 };