One Code 814. Binary Tree Pruning Every Day

Code ideas
The first reaction should be to solve it recursively.
There are three conditions for the current node to be deleted:

1. The current node value is 0
2. Left subtree does not have 1
3. The right subtree does not have 1

code

/**
* 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:
   bool one(TreeNode *root){
    
    
       if(!root) return false;
       if(!one(root->left)) root->left = NULL;     //左子树没有1，则删除左子树
       if(!one(root->right)) root->right = NULL;   //右子树没有1，则删除右子树
       if(root->val) return true;                  //节点值为1，返回1
       else{
    
                                           
           if(!root->left && !root->right) return false;  //节点值为0,且左右子树都被删除，则返回0
           else return true;                              //有子树没被删除则返回1
       }
   }
   TreeNode* pruneTree(TreeNode* root) {
    
    
       if(!root) return root;
       if(one(root)) return root;
       else return NULL;
   }
};