题目描述：

给你一个整数数组 nums ，其中元素已经按 升序 排列，请你将其转换为一棵 

平衡 二叉搜索树。

示例 1：

输入：nums = [-10,-3,0,5,9]
输出：[0,-3,9,-10,null,5]
解释：[0,-10,5,null,-3,null,9] 也将被视为正确答案：

示例 2：

输入：nums = [1,3]
输出：[3,1]
解释：[1,null,3] 和 [3,1] 都是高度平衡二叉搜索树。

代码思路：

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
class Solution {
    public TreeNode sortedArrayToBST(int[] nums) {
        if (nums.length==0){
            return null;
        }
        TreeNode root = new TreeNode();
        locat(root , 0, nums.length-1,nums);
       return root;
    }
    public void locat(TreeNode root,int low,int high,int[] nums){
        int mid = (low+high)/2;
        root.val = nums[mid];
        if(low<mid){
            TreeNode lx = new TreeNode();
            root.left = lx;
            locat(lx,low,mid-1,nums);
        }
        if(mid<high){
            TreeNode rx = new TreeNode();
            root.right = rx;
            locat(rx,mid+1,high,nums);
        }
    }
}