Convert Sorted Array to Binary Search Tree

Given an array where elements are sorted in ascending order, convert it to a height balanced BST.
Solution: recursion
把一个数组看成一棵树（也就是以中点为根，左右为左右子树，依次下去）数组就等价于一个二分查找树。
把中间元素转化为根，然后递归构造左右子树。用二叉树递归的方法来实现，以根作为返回值，每层递归函数取中间元素，作为当前根和赋上结点值，然后左右结点接上左右区间的递归函数返回值。
时间复杂度还是一次树遍历O(n)，总的空间复杂度是栈空间O(logn), stack space.

/**
 * Definition for binary tree
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    public TreeNode sortedArrayToBST(int[] num) {
        if(num == null || num.length == 0)
            return null;
        
        return sortHelper(num, 0, num.length - 1);
    }
    
    public TreeNode sortHelper(int[] num, int start, int end) {
        if(start > end)
            return null;
        
        int mid = (start + end) / 2;
        TreeNode root = new TreeNode(num[mid]);
        root.left = sortHelper(num, start, mid - 1);
        root.right = sortHelper(num, mid + 1, end);
        return root;
    }
}