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;    }};