House Robber III——Difficulty：Medium

Problem :

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:

Example 1:     3    / \   2   3    \   \      3   1Maximum amount of money the thief can rob = 3 + 3 + 1 = 7.Example 2:     3    / \   4   5  / \   \  1   3   1Maximum amount of money the thief can rob = 4 + 5 = 9.

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Algorithm:

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对于一个节点作为根节点，要找到它符合题意的最大值，有两种可能：
1. 节点本身的值+以节点的左节点的左节点和右节点为根节点的符合题意的最大值+以节点右节点的左节点和右节点为根节点的符合题意的最大值
2. 以节点的左节点为根节点的符合题意的最大值+以节点右节点为根节点符合题意的最大值

如果直接这样写会超时，因为在计算的时候重复的计算一些值，比如这个节点是4个节点的子节点，那么这个节点就会被计算四次，因此，可以来一个哈希表来记录已经算过的节点。

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Code：

/** * 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) {        unordered_map<TreeNode*,int> hash;        return al(root,hash);    }    int al(TreeNode* root,unordered_map<TreeNode*,int>& hash)    {        if(root==NULL)            return 0;        int count=0;        if(hash.count(root)) return hash[root];        if(root->left!=NULL)        {            count+=al(root->left->left,hash)+al(root->left->right,hash);        }        if(root->right!=NULL)        {            count+=al(root->right->left,hash)+al(root->right->right,hash);        }        int x=max(count+root->val,al(root->right,hash)+al(root->left,hash));        hash[root]=x;        return x;    }};