[C language brush leetcode] 814. Binary tree pruning (m)

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The root node of your binary tree  root , In addition, the value of each node of the tree is either 0 , Or 1 .

Return to remove all excluded 1 The original binary tree of the subtree .

node node The subtree of is node Plus all itself node The offspring of .

Example 1：

Input ：root = [1,null,0,0,1]
Output ：[1,null,0,null,1]
explain ：
Only the red nodes meet the conditions “ All do not contain 1 The subtree of ”. On the right is the answer .
Example 2：

Input ：root = [1,0,1,0,0,0,1]
Output ：[1,null,1,null,1]
Example 3：

Input ：root = [1,1,0,1,1,0,1,0]
Output ：[1,1,0,1,1,null,1]
 

Tips ：

The number of nodes in the tree is in the range [1, 200] Inside
Node.val by 0 or 1

source ： Power button （LeetCode）
link ：https://leetcode.cn/problems/binary-tree-pruning
Copyright belongs to the network . For commercial reprint, please contact the official authority , Non-commercial reprint please indicate the source .

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Say it's a medium question , It looks like a simple problem , It is estimated that we should pay attention to some details .

int dfs(struct TreeNode* root)
{
    if (root == NULL) { //  achieve NULL, Start back to 
        return 0;
    }

    int lf = dfs(root->left);
    int rf = dfs(root->right);

    if (lf == 0) {
        root->left = NULL;
    }

    if (rf == 0) {
        root->right = NULL;
    }

    if (lf == 0 && rf == 0 && root->val == 0) {//  If the left and right nodes are empty and they are equal to 0, return 0, It means it will be deleted 
        return 0;
    }

    return 1;
}

struct TreeNode* pruneTree(struct TreeNode* root){
    int rf = dfs(root); //  Start a deep search for , And return whether the current node is a leaf node 

    if (rf == 0 && root->val == 0) {
        return NULL;
    }

    return root;
}