LeetCode 814. Binary Tree Pruning

题目链接：https://leetcode.com/problems/binary-tree-pruning/description/

We are given the head node root of a binary tree, where additionally every node's value is either a 0 or a 1.

Return the same tree where every subtree (of the given tree) not containing a 1 has been removed.

(Recall that the subtree of a node X is X, plus every node that is a descendant of X.)

Example 1:
Input: [1,null,0,0,1]
Output: [1,null,0,null,1]
 
Explanation: 
Only the red nodes satisfy the property "every subtree not containing a 1".
The diagram on the right represents the answer.

Example 2:
Input: [1,0,1,0,0,0,1]
Output: [1,null,1,null,1]

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

Note:

• The binary tree will have at most 100 nodes.
• The value of each node will only be 0 or 1.

题目解析：大意就是删掉值为0且以此节点为根的子树中无1的节点。很容易想到用递归来处理，自下而上删除节点，处理好判断条件即可。

代码如下：0ms Accepted beating 100%

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    TreeNode* pruneTree(TreeNode* root) {
        if (root == NULL)
            return NULL;
        root->left = pruneTree(root->left);
        root->right = pruneTree(root->right);
        if (root->val == 0 && root->left == NULL && root->right == NULL)
        {
            free(root);
            return NULL;
        }
        return root;
    }
};