将有序数组转换为二叉搜索树

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。

提示：

• 1 <= nums.length <= 104
• -104 <= nums[i] <= 104
• nums 按 严格递增 顺序排列

/**
 * 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 == null && nums.length == 0) {
            return null;
        }

        int middle = nums.length / 2;
        int[] numsLeft = Arrays.copyOf(nums, middle);
        int[] numsRight = Arrays.copyOfRange(nums, middle + 1, nums.length);
        TreeNode treeNode = new TreeNode(nums[middle]);
        sortedArrayToBST(numsLeft, numsRight, treeNode);
        return treeNode;
    }

    public void sortedArrayToBST(int[] numsLeft, int[] numsRight, TreeNode treeNode) {
        if (numsLeft != null && numsLeft.length > 0) {
            int middleLeft = numsLeft.length / 2;
            int[] numsLeftTemp = Arrays.copyOf(numsLeft, middleLeft);
            int[] numsRightTemp = Arrays.copyOfRange(numsLeft, middleLeft + 1, numsLeft.length);
            treeNode.left = new TreeNode(numsLeft[middleLeft]);
            sortedArrayToBST(numsLeftTemp, numsRightTemp, treeNode.left);
        }
        if (numsRight != null && numsRight.length > 0) {
            int middleRight = numsRight.length / 2;
            int[] numsLeftTemp = Arrays.copyOf(numsRight, middleRight);
            int[] numsRightTemp = Arrays.copyOfRange(numsRight, middleRight + 1, numsRight.length);
            treeNode.right = new TreeNode(numsRight[middleRight]);
            sortedArrayToBST(numsLeftTemp, numsRightTemp, treeNode.right);
        }
    }
}