337 leetcode House Robber III

337 leetcode House Robber III

The thief has found himself a new place for his thievery again. There is only one entrance to this area, called the “root.” Besides the root, each house has one and only one parent house. After a tour, the smart thief realized that “all houses in this place forms a binary tree”. It will automatically contact the police if two directly-linked houses were broken into on the same night.

Determine the maximum amount of money the thief can rob tonight without alerting the police.

Example 1:

 3
/ \

2 3
\ \
3 1

 3
/ \

2 3
\ \
3 1
Maximum amount of money the thief can rob = 3 + 3 + 1 = 7.

Example 2:

 3
/ \

4 5
/ \ \
1 3 1

 3
/ \

4 5
/ \ \
1 3 1
Maximum amount of money the thief can rob = 4 + 5 = 9.

爲啥我的代碼通不過

#include <iostream>

    struct TreeNode {
        int val;
        TreeNode *left;
        TreeNode *right;
        TreeNode(int x) : val(x), left(NULL), right(NULL) {}
  };

class Solution {
public:
    int rob(TreeNode* root) {
        if(!root)
            return 0;
        std::queue<TreeNode*> level[2];
        int current = 0;
        int next =1;
        int sum[2] = {0, 0};

        level[0].push(root);
        while(!level[0].empty() || !level[1].empty())
        {
            TreeNode* pNode = level[current].top();
            level[0].pop();
            sum[current] += pNode->val;
            if(current == 0){
                if(pNode->left != NULL)
                    level[next].push(pNode->left);
                if(pNode->right != NULL)
                    level[next].push(pNOde->right);
            }
            else{
                if(pNode->left != NULL)
                    level[next].push(pNode->left);
                if(pNode->right != NULL)
                    level[next].push(pNode->right);
            }
            if(level[current].empty()){
                current = 1-current;
                next = 1- next;
            }

        }
        return sum[0]>sum[1]?sum[0]: sum[1];

    }
};

不就是計算奇偶層的和麼

參考別人的代碼：

C++

class Solution {
public:
    int rob(TreeNode* root) {
        return dfs(root).first;
    }

    pair<int, int> dfs(TreeNode* root){
        pair<int, int> dp = make_pair(0,0);
        if(root){
            pair<int, int> dp_L = dfs(root->left);
            pair<int, int> dp_R = dfs(root->right);
            dp.second = dp_L.first + dp_R.first;
            dp.first = max(dp.second ,dp_L.second + dp_R.second + root->val);
        }
        return dp;
    }
};