814. Binary Tree Pruning

Problem

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.)

Note:

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

Example1

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.

Example2

Input: [1,0,1,0,0,0,1]
Output: [1,null,1,null,1]

Example3

Input: [1,1,0,1,1,0,1,0]
Output: [1,1,0,1,1,null,1]

Solution

后序遍历，递归。
在每个节点处，判断以它为根的树是否包含1。如果不包含，就将这棵树删掉。

/**
 * 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 root;
        root->left = pruneTree(root->left);
        root->right = pruneTree(root->right);

        if(!isTreeContainsOne(root))
        {
            deleteTreeFromCurrentNode(root);
            return NULL;
        }

        return root;

    }

    bool isTreeContainsOne(TreeNode *root)
    {
        if(!root)
            return false;
        if(root->val ==1)
            return true;
        return isTreeContainsOne(root->left) || isTreeContainsOne(root->right);
    }

    void deleteTreeFromCurrentNode(TreeNode *root)
    {
        if(!root)
            return;
        deleteTreeFromCurrentNode(root->left);
        deleteTreeFromCurrentNode(root->right);

        root->left = NULL;;
        root->right = NULL;
        delete root;
    }
};