[LeetCode] Binary Tree Pruning

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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.

对二叉树进行剪枝。剪枝条件：结点的所有子节点值为0。

利用递归后序遍历进行判断。如果节点值为0并且其左右孩子为空，则令其为空。

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 == NULL && root->right == NULL)
            return NULL;
        return root;
    }
};