814. Binary Tree Pruning

Link: https://leetcode.com/problems/binary-tree-pruning/

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.

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題目翻譯:

檢查每個節點的左子樹和右子樹有無包含 1 的點，如果沒有就將其節點砍掉。

程式思路:

遞迴基本練習。

/**
 * 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 == nullptr)
            return nullptr;
        root->left = pruneTree (root->left);
        root->right = pruneTree (root->right);
        if(!root->left  && !root->right && root->val == 0)
        {
            delete root;
            return nullptr;
        }
        else
           return root;
    }
};