Convert Sorted Array to Binary Search Tree

Problem:

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

Analysis:

Use binary search, nums[mid] is root, left and right are sub trees' mid. But condition has to be low < high, instead of regular template condition low + 1 < high. Because low == high is valid. 

Solution:

 /**  
  * Definition for a binary tree node.  
  * public class TreeNode {  
  *   int val;  
  *   TreeNode left;  
  *   TreeNode right;  
  *   TreeNode(int x) { val = x; }  
  * }  
  */  
 class Solution {  
   public TreeNode sortedArrayToBST(int[] nums) {  
     if (nums == null || nums.length ==0) return null;  
           return helper(nums, 0, nums.length - 1);  
   }  
      private TreeNode helper(int[] nums, int low, int high) {  
           if (low > high) return null;  
           int mid = low + (high - low) / 2;  
           TreeNode root = new TreeNode(nums[mid]);  
           root.left = helper(nums, low, mid - 1);  
           root.right = helper(nums, mid + 1, high);  
           return root;  
      }  
 }