LeetCode337. House Robber III

题目：

https://leetcode.com/problems/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.

思路：

题意：不能连续抢直接相连的两个节点。即例2中，抢了3就不能抢4，5。问最多能抢好多。

给一个二叉树，求它不直接相连的节点的val和最大为多少。

如果抢了当前节点，那么它的左右孩子就肯定不能抢了。

如果没有抢当前节点，左右孩子抢不抢取决于左右孩子的孩子的val大小。

QAQ，思路是这样。代码是参考别人的，自己没写出来……

代码：

/**
* 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> dfs(TreeNode* root){
vector<int> res(2,0);//res[0]代表抢，res[1]不抢
if(root==NULL){
return res;
}
vector<int> lRes=dfs(root->left);
vector<int> rRes=dfs(root->right);
res[0]=lRes[1]+rRes[1]+root->val;//抢当前节点，那么左右孩子就一定不能抢
res[1]=max(lRes[0],lRes[1])+max(rRes[0],rRes[1]);//不抢当前节点，左右孩子可抢可不抢
return res;
}
int rob(TreeNode* root) {
vector<int> res=dfs(root);
return max(res[0],res[1]);
}
};