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

Solution： recursion

1. 根据BST的特性，left < current < right，取数组的middle index数，作为当前root。
2. 以 nums: start, middle -1构建找子树
3. 以nums: middle + 1, end构建右子树
4. base case：start > end 退出。

/**
 * 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 sortedArrayToBSTHelper (nums, 0, nums.length - 1);
    }
    
    private TreeNode sortedArrayToBSTHelper (int[] nums, int start, int end) {
        if (start > end) {
            return null;
        }
        
        int mid = start + (end - start) / 2;
        TreeNode currentNode = new TreeNode (nums[mid]);
        TreeNode leftNode = sortedArrayToBSTHelper (nums, start, mid - 1);
        TreeNode rightNode = sortedArrayToBSTHelper (nums, mid + 1, end);
        
        currentNode.left = leftNode;
        currentNode.right = rightNode;
        
        return currentNode;
    }
}