【LeetCode】814. Binary Tree Pruning 解题报告（Python & C++）

原创

负雪明烛 2021-07-14 10:52:34 ©著作权

文章标签 算法 LeetCode 文章分类 代码人生

©著作权归作者所有：来自51CTO博客作者负雪明烛的原创作品，请联系作者获取转载授权，否则将追究法律责任

作者：负雪明烛
id： fuxuemingzhu
个人博客： http://fuxuemingzhu.cn/

题目描述

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]

【LeetCode】814. Binary Tree Pruning 解题报告（Python & C++）_LeetCode

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]

【LeetCode】814. Binary Tree Pruning 解题报告（Python & C++）_算法_02

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

【LeetCode】814. Binary Tree Pruning 解题报告（Python & C++）_算法_03

Note:

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

题目大意

把一棵树的所有不含1的子树都删除。子树的定义是自身节点和所有子节点。

解题方法

后序遍历

这个题一看还是dfs啊～习惯了新定义一个函数dfs了，但这次不需要了。我们直接把节点的左孩子和右孩子重新设置就好了。这个题是后序遍历！

一定要注意的是，我们判断这个节点是叶子节点并且节点值是1的这个步骤要放在左右子树处理之后。可以从Example2中看出来，如果0节点的左右子节点都是0，那么把左右节点都减去了之后，还要判断自身是不是0，然后把自己也剪了。也就是说这一步相当于后序遍历，把孩子都处理结束之后，然后再处理自身。

# Definition for a binary tree node.
# class TreeNode(object):
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution(object):
    def pruneTree(self, root):
        """
        :type root: TreeNode
        :rtype: TreeNode
        """
        if not root: return
        root.left = self.pruneTree(root.left)
        root.right = self.pruneTree(root.right)
        if not root.left and not root.right and root.val == 0:
            return None
        return root

C++版本代码如下：

/**
 * 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) return nullptr;
        root->left = pruneTree(root->left);
        root->right = pruneTree(root->right);
        if (!root->left && !root->right)
            return root->val == 1 ? root : nullptr;
        return root;
    }
};