！力扣 108. 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

方法：中序遍历，总是选择中间位置左边的数字作为根节点

/*** Definition for a binary tree node.* struct TreeNode {*     int val;*     struct TreeNode *left;*     struct TreeNode *right;* };*/
//自己定义的build函数要放在前面
struct TreeNode* build(int* nums,int left,int right){if(left>right){return NULL;}int mid=(left+right)/2;
//创建新节点struct TreeNode* root =(struct TreeNode*)malloc(sizeof(struct TreeNode));root->val=nums[mid];root->left=build(nums,left,mid-1);root->right=build(nums,mid+1,right);return root;
}struct TreeNode* sortedArrayToBST(int* nums, int numsSize) {return build(nums,0,numsSize-1);
}