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

  /   

 13   1

Maximum amount of money the thief can rob = 4 + 5 = 9.

Credits:
Special thanks to @dietpepsi for adding this problem and creating all test cases.

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    思路，dfs。递归函数返回两个值，分别是窃取root节点和不窃取root节点时窃取的资金总额。当根节点为空时，直接返回(0,0).根节点不为空的时候，窃取根节点的时候,窃取的资金为left.second+right.second+root->val(此时，不能窃取根节点左右子结点).而不窃取根节点的时候，窃取的资金为max(left.first,left.second)+max(right.first,right.second),即左右子节点能够窃取的最大值之和。

    最终，在实际调用完成之后，我们返回窃取root节点和不窃取root节点偷得的资金的最大值。

/**
 * 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 {
pair<int,int> help(TreeNode *root){
    if(!root){
        return make_pair(0,0);
    }
    auto left = help(root->left);
    auto right = help(root->right);
    //窃取root节点
    int rob_root = left.second+right.second +root->val;
    //不窃取root节点
    int not_rob_root = max(left.first,left.second)+max(right.first,right.second);
    return make_pair(rob_root,not_rob_root);
}
public:
    int rob(TreeNode* root) {
        auto result =help(root);
        return max(result.first,result.second);
    }
};