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);}}
}