# 814. Binary Tree Pruning 二叉树剪枝

@TOC

## # 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]
``````

``````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:

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

## # 解题方法

### # 后序遍历

``````# 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;
}
};
``````

## # 日期

2018 年 4 月 8 日 —— 网吧通宵了，然后睡了一天。。 2018 年 11 月 5 日 —— 打了羽毛球，有点累 2018 年 12 月 2 日 —— 又到了周日