题目链接/文章讲解： https://programmercarl.com/0669.%E4%BF%AE%E5%89%AA%E4%BA%8C%E5%8F%89%E6%90%9C%E7%B4%A2%E6%A0%91.html

视频讲解： https://www.bilibili.com/video/BV17P41177ud

1.这个题和上一个题删除二叉搜索树的节点的题有异曲同工之妙，这个题的思路其实做过昨天的题之后似乎就没有什么难点和坑了，要是有坑就在会直接将删除的节点置为NULL或者是直接return root->right,这样的话就相当于砍树了，因为在子树里还会有符合的节点呢。

CPP

/**
 * 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 {
public:
    TreeNode* traversal(TreeNode* node, int low, int high){
        if(node == NULL)return NULL;
        if(node->val < low){
            TreeNode* right = traversal(node->right, low, high);
            return right;
        }
        if(node->val > high){
            TreeNode* left = traversal(node->left, low, high);
            return left;
        }
        node->left = traversal(node->left, low, high);
        node->right = traversal(node->right, low, high);
        return node;
    }
    TreeNode* trimBST(TreeNode* root, int low, int high) {
        return traversal(root, low, high);
    }
};