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题目：
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
Maximum amount of money the thief can rob = 3 + 3 + 1 = 7.
Example 2:
3
/ 4 5
/ \ \
1 3 1
Maximum amount of money the thief can rob = 4 + 5 = 9.
题目解答：
这个题目是在之前House Robber 和House Robber II之上的变体。它要求的是在树上的动归。需要用到深度优先搜索和动归的思想。一样的，相邻节点不能偷。
另外借助一个辅助结构，来帮助我们存动归中需要比较的两个值。
代码：
/**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
struct robMoney //辅助数据结构，用来记录动态规划过程中的两个整型值
{
int noRob; //不偷当前这一家，可以得到的金钱数目
int curRob; //在当前这个位置，可以获得到的金钱数目最大值
robMoney():noRob(0),curRob(0){}
};
class Solution {
public:
int rob(TreeNode* root) {
if(root == NULL)
return 0;
robMoney res = dfs(root);
return res.curRob;//在当前位置所能获得的金钱数目最大值
}
robMoney dfs(TreeNode *root) //深度优先搜索
{
robMoney money;
if(root == NULL)
{
return money;
}
robMoney left = dfs(root -> left);
robMoney right = dfs(root -> right);
money.noRob = left.curRob + right.curRob; //当前这家选择不偷
money.curRob = max(money.noRob, left.noRob + right.noRob + root -> val); //取二者的最大值
return money;
}
int max(int a, int b)
{
return a > b ? a : b;
}
};
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原文地址：http://www.cnblogs.com/CodingGirl121/p/5548992.html
