# 814.Binary Tree Pruning

**814.Binary Tree Pruning**

难度:Medium

> 给定二叉树根结点 root ，此外树的每个结点的值要么是 0，要么是 1。

返回移除了所有不包含 1 的子树的原二叉树。

( 节点 X 的子树为 X 本身，以及所有 X 的后代。)

示例1: 输入: \[1,null,0,0,1] 输出: \[1,null,0,null,1] ![1028\_2.png](https://i.loli.net/2019/10/16/xk2N9PuaV7r6lch.png)

解释: 只有红色节点满足条件“所有不包含 1 的子树”。 右图为返回的答案。

示例2: 输入: \[1,0,1,0,0,0,1] 输出: \[1,null,1,null,1] ![1028\_1.png](https://i.loli.net/2019/10/16/gphcfuUWX3CS8ea.png)

示例3: 输入: \[1,1,0,1,1,0,1,0] 输出: \[1,1,0,1,1,null,1] ![1028.png](https://i.loli.net/2019/10/16/rZIwfhVbT6iyUYF.png)

说明: 给定的二叉树最多有 100 个节点。 每个节点的值只会为 0 或 1 。

递归剪枝，如果左子树或者右子树都为空，而且此时节点值为0（除去根节点），则将该节点剪掉。

```
/**
 * 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 {
bool isRoot = true;
public:
    TreeNode* pruneTree(TreeNode* root) {
        if(!root) return root;
        if(!root->left && !root->right) return  (!isRoot && root->val == 0) ? nullptr: root;
        isRoot=false;
      // cout<<root->val<<endl;
        if(root->left   ) root->left = pruneTree(root->left);
        if(root->right) root->right = pruneTree(root->right);
        if( root->val ==0 && !root->left && !root->right) return nullptr;
        return root; 
    }
};
```

> 执行用时 :8 ms, 在所有 C++ 提交中击败了48.30%的用户\
> 内存消耗 :9.7 MB, 在所有 C++ 提交中击败了74.11%的用户


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