树

1、最小高度二叉搜索树（面试题 04.02）

首先确定递归 DFS

递归出口：l > r

创建根节点

1 填值，注意mid = (r - l)/2 + l

2 创建左子树和右子树 DFS(nums, l, mid-1)和DFS(nums, mid+1, r)

class Solution {
public:
    TreeNode* sortedArrayToBST(vector<int>& nums) {

        return DFS(nums, 0, nums.size()-1);

    }

    TreeNode* DFS(vector<int>& nums, int l, int r){

        if(l>r){return nullptr;}

        TreeNode* root = new TreeNode();

        int mid = (r-l)/2 + l;
        root->val = nums[mid];
        root->left = DFS(nums, l, mid-1);
        root->right = DFS(nums, mid+1, r);
        return root;
    }
};