ps:练习来自力扣

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。

// 定义二叉树节点类
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) {
        return buildBST(nums, 0, nums.length - 1);
    }

    private TreeNode buildBST(int[] nums, int left, int right) {
        // 如果左边界大于右边界，说明子数组为空，返回 null
        if (left > right) {
            return null;
        }
        // 找到中间位置
        int mid = left + (right - left) / 2;
        // 创建根节点，值为中间元素的值
        TreeNode root = new TreeNode(nums[mid]);
        // 递归构建左子树
        root.left = buildBST(nums, left, mid - 1);
        // 递归构建右子树
        root.right = buildBST(nums, mid + 1, right);
        return root;
    }
}