814. Binary Tree Pruning（python+cpp）

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题目：

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.

解释：
把不包括1的子树移除。
递归写法：
python代码：

# 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 None
        if root.left:
            root.left=self.pruneTree(root.left)
        if root.right:
            root.right=self.pruneTree(root.right)
        if not root.left and not root.right and not root.val:
            return None 
        return root

c++代码：

/**
 * 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 NULL;
        if (root->left)
            root->left=pruneTree(root->left);
        if(root->right)
            root->right=pruneTree(root->right);
        if (!root->left &&!root->right && !root->val)
            return NULL;
        return root;
    }
};

总结：
可以用函数本身递归的，就不需要在写一个helper()函数了。