814. Binary Tree Pruning

问题描述：

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:

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

解题思路：

　　问题是给树剪枝，当其所有的子树为0时，就可以剪掉它。

　　一般遇到树我首先就会考虑递归。如果不能递归，那就用栈来辅助遍历。

　　递归的basic state是root为NULL

　　其他情况我们要对左右子树分别调用方法，并且最后根据节点的值来判断是否将该节点剪掉。

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;
        root->left = pruneTree(root->left);
        root->right = pruneTree(root->right);
        if(root->val == 0 && !root->left && !root->right)
            return NULL;
        return root;
    }
};