[Medium]House Robber III

问题：
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.
解法：

源码：

/** * Definition for a binary tree node. * struct TreeNode { *     int val; *     TreeNode *left; *     TreeNode *right; *     TreeNode(int x) : val(x), left(NULL), right(NULL) {} * }; */#include <cmath>#include <pair>class Solution {public:    pair<int, int> find(TreeNode * root) {        if (root->left == NULL && root->right == NULL) {            return (make_pair(root->val, 0));        }        pair<int, int> pl = make_pair(0,0), pr = make_pair(0,0);        if (root->left != NULL) pl = find(root->left);        if (root->right != NULL) pr = find(root->right);        int a = root->val + pl.second + pr.second;        int b = max(pl.first, pl.second) + max(pr.first, pr.second);        return (make_pair(a, b));    }    int rob(TreeNode* root) {        pair<int, int> ans = make_pair(0,0);        if (root != NULL) ans = find(root);        return max(ans.first, ans.second);    }};