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.

What is a BST：
若任意节点的左子树不空，则左子树上所有结点的值均小于它的根结点的值；
任意节点的右子树不空，则右子树上所有结点的值均大于它的根结点的值；
任意节点的左、右子树也分别为二叉查找树。

What is height-balanced:
An empty tree is height-balanced. A non-empty binary tree T is balanced if: 1) Left subtree of T is balanced 2) Right subtree of T is balanced 3) The difference between heights of left subtree and right subtree is not more than 1.

/**
 * 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 left, int right){
        if (left > right){
            return null;
        } 
        int mid = left + (right - left) / 2;
        TreeNode node = new TreeNode(nums[mid]);
        node.left = helper(nums, left, mid - 1);
        node.right = helper(nums, mid + 1, right);
        return node;
    }
}