<LeetCode OJ> 337. House Robber III

全栈程序员-用户IM • 2022年3月5日下午2:00 • 未分类

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Total Accepted: 1341
Total Submissions: 3744
Difficulty: Medium

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:

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

Example 2:

Maximum amount of money the thief can rob =
4 +
5 =
9.

分析：

以下的答案有错，不知道错在哪里！

！

！难道不是统计偶数层的和与奇数层的和，然后比較大小就可得出结果？

/**
 * 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:
    int rob(TreeNode* root) {
        if(root==NULL)  
            return 0;  
        queue<TreeNode*> que;//用来总是保存当层的节点  
        que.push(root);  
        int oddsum =root->val;//用于统计奇数层的和
        int evensum=0;  //用于统偶数层的和
        //获取每一层的节点
        int curlevel=2;
        while(!que.empty())  
        {  
            int levelSize = que.size();//通过size来推断当层的结束  
            for(int i=0; i<levelSize; i++)   
            {  
                if(que.front()->left != NULL) //先获取该节点下一层的左右子，再获取该节点的元素。由于一旦压入必弹出。所以先处理左右子  
                    que.push(que.front()->left);  
                if(que.front()->right != NULL)   
                    que.push(que.front()->right);  
                if(curlevel % 2 ==1)
                    oddsum  += que.front()->val;
                else
                    evensum += que.front()->val;
                que.pop();  
            }  
            curlevel++;
        }
        return oddsum > evensum ? oddsum : evensum;//奇数层的和与偶数层的和谁更大谁就是结果
    }
};

学习别人的代码：

int rob(TreeNode* root) {
    int child = 0, childchild = 0;
    rob(root, child, childchild);
    return max(child, childchild);
}

void rob(TreeNode* root, int &child, int &childchild) {
    if(!root) return;

    int l1 = 0, l2 = 0, r1 = 0, r2 = 0;
    rob(root->left, l1, l2);
    rob(root->right, r1, r2);

    child = l2 + r2 + root->val;
    childchild = max(l1, l2) + max(r1, r2);
}

注：本博文为EbowTang原创，兴许可能继续更新本文。假设转载，请务必复制本条信息！

原文地址：http://blog.csdn.net/ebowtang/article/details/50890931

原作者博客：http://blog.csdn.net/ebowtang

本博客LeetCode题解索引：http://blog.csdn.net/ebowtang/article/details/50668895

转载于:https://www.cnblogs.com/yutingliuyl/p/7225828.html

发布者：全栈程序员-用户IM，转载请注明出处：https://javaforall.cn/108303.html原文链接：https://javaforall.cn

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