$dfs$，记录当前节点，以及当前节点是否加入结果

如果$root$删除，那么子节点有可能加入结果

如果$root$不删除，那么子节点不可能加入结果

/**
 * 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:
    vector<TreeNode*> res;
    set<int>d;
    vector<TreeNode*> delNodes(TreeNode* root, vector<int>& to_delete) {
        for(auto x:to_delete)
            d.insert(x);
        dfs(root,true);
        return res;
    }

    void dfs(TreeNode *&root,bool add)
    {
        if(!root) return ;

        if(d.count(root->val))
        {
            dfs(root->left,true);
            dfs(root->right,true);
            root=NULL;
        }
        else
        {
            dfs(root->left,false);
            dfs(root->right,false);
            if(add)
                res.push_back(root);
        }

    }
};