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

Example 2:

     3
    /    4   5
  / \   \ 
 1   3   1

分析：

以下的答案有錯，不知道錯在哪裏！

！

！難道不是統計偶數層的和與奇數層的和，然後比較大小就可得出結果？

/**
 * 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;//用來總是保存當層的節點  
        que.push(root);  
        int oddsum =root->val;//用於統計奇數層的和
        int evensum=0;  //用於統偶數層的和
        //獲取每一層的節點
        int curlevel=2;
        while(!que.empty())  
        {  
            int levelSize = que.size();//通過size來推斷當層的結束  
            for(int i=0; i<levelSize; i++)   
            {  
                if(que.front()->left != NULL) //先獲取該節點下一層的左右子，再獲取該節點的元素。由於一旦壓入必彈出。所以先處理左右子  
                    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;//奇數層的和與偶數層的和誰更大誰就是結果
    }
};

學習別人的代碼：

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);
}

註：本博文為EbowTang原創，興許可能繼續更新本文。假設轉載，請務必復制本條信息！

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題解索引：http://blog.csdn.net/ebowtang/article/details/50668895

<LeetCode OJ> 337. House Robber III