Leetcode 814 binary tree pruning [dfs] the leetcode road of heroding

Their thinking ：
A very standard dfs subject , Keep going down to the bottom of the original node , When encountering an empty node, it returns null , Leaf node is encountered and the value is 0 Return null of , Logically speaking, in the process of pruning , If the current node is 0, If the left and right child nodes do not exist or the values are 0 Just pruning , But in dfs In the process , The value is 0 The leaf node of has returned nullptr 了 , That is, every time you prune, it is the leaf node （ It is not a leaf, but also becomes a leaf because of the pruning of the left and right child nodes ）, The code is as follows ：

/** * 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* pruneTree(TreeNode* root) {
    
        if(root == nullptr) return nullptr;
        root->left = pruneTree(root->left);
        root->right = pruneTree(root->right);
        if(root->left == nullptr && root->right == nullptr && root->val == 0) {
    
            return nullptr;
        }
        return root;
    }
};