将有序数组转换成二叉搜索树

正常递归，中序遍历
递归经常会把自己绕晕，还是得画图分析

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
class Solution {
private:
    //递归三部曲：确定递归的参数，返回的是依据数组下标从[l,r]构建的二叉树
    TreeNode* tree(vector<int>&nums,int l,int r)
    {
        //当下标不合适之后结束递归
        if(l>r)
            return NULL;
        //确定单层递归的逻辑
        //中间节点为头结点，左右子树分别为左右数组组成的二叉树
        int mid=(r-l)/2+l;
        TreeNode* root= new TreeNode(nums[mid]);
        root->left=tree(nums,l,mid-1);
        root->right=tree(nums,mid+1,r);
        return root;
    }
public:
    TreeNode* sortedArrayToBST(vector<int>& nums) {
        int l=0;
        int r=nums.size()-1;
        return tree(nums,l,r);
    }
};