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.

Solution

/**
 * 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:
    vector<int> help(TreeNode* root) {
        vector<int> res(2, 0);
        if ( root == NULL ) return res;
        if ( root->left == NULL and root->right == NULL ) { res[0] = root->val; return res;}
        vector<int> left = help(root->left), right = help(root->right);
        res[0] = root->val + left[1] + right[1];
        // Note we take the maximum!
        res[1] = max(left[0], left[1]) + max(right[0], right[1]);
        return res;
    }
    int rob(TreeNode* root) {
        if ( root == NULL ) return 0;
        vector<int> res = help(root);
        return max(res[0], res[1]);
    }
};

Notes
Note the following sentence in the above solution:

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

If we do not take the maximum as:

res[1] = left[0] + right[0];

we will get wrong answer for the following situation:

       4
      /
     1
    /
   2
  /
 3