337. House Robber III

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.
A:
首先：就是遍历一遍。然后考虑2种情况。 取或者不取 node。
要防止多次（重复）遍历。 有个简单的方法，就是用hash. 这里，我们用hashMap来保存结果

/**
 * 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(not root)
            return 0;
        if(M.find(root) != M.end())
            return M[root];
        
        int withRoot = root->val + (root->left? rob(root->left->left) + rob(root->left->right) :0 )+ \
            (root->right ? rob(root->right->left) + rob(root->right->right) : 0);
        int noRoot = rob(root->left) + rob(root->right);
        M[root] = max(withRoot, noRoot);
        return M[root];
    }
private:
    unordered_map<TreeNode*, int> M;
};

Mistakes:
1：  withRoot 计算的时候， 三目运算符， 要括起来。