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.

給定一棵二叉樹，求最大和，相鄰節點只能最多選其一。

對於每個子節點，以該節點為根root，則可以得到的最大和為：加上該節點的值res[1]或不加上該節點的值res[0].

當加上該節點，則相鄰的左右子節點都不能加。要是不加上該節點，則相鄰的左右子節點可加可不加。

故而res[0] = max(left.res[0], left.res[1]) + max(right.res[0], right.res[1]);

　　res[1] = root->val + left.res[0] + right.res[0];

class Solution {
public:
    vector<int> helper(TreeNode* root) {
        vector<int> res(2, 0);
        if (root) {
            vector<int> l = helper(root->left);
            vector
 
<int> r = helper(root->right);
            res[0] = max(l[0],  l[1]) + max(r[0],  r[1]);
            res[1] = root->val + l[0] + r[0];
        }
        return res;
    }
    int rob(TreeNode* root) {
        vector<int> v = helper(root);
        return max(v[0],  v[1]);
    }
};

337. House Robber III