[leetcode]814. Binary Tree Pruning

[leetcode]814. Binary Tree Pruning

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Analysis

周四不知道快不快乐—— [有点想回家过暑假的冲动！]

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.)
删除一棵树中所有不含1的子树，递归解决，要注意的是判断某节点要不要删除要在处理完其左右子树之后（i.e. example 2）~

Implement

/**
 * 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->left && !root->right && root->val == 0)
            return NULL;
        return root;
    }
};