Leetcode 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.

使用DFS，比如对于根节点，分两种情况，如果选根节点（第0层），则不能选第一层结点，如果不选根节点，则第一层可选可不选（到底选不选，用max），这样递归的往下层进行。

/**
* 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 {
vector<int> getMoney(TreeNode* node){
vector<int> ret(2,0);
if(!node) return ret;
vector<int> lRet=getMoney(node->left);
vector<int> rRet=getMoney(node->right);
ret[0]=lRet[1]+rRet[1]+node->val;      //表示选这层结点
ret[1]=max(lRet[0],lRet[1])+max(rRet[0],rRet[1]); //表示不选这层结点
return ret;
}
public:
int rob(TreeNode* root) {
vector<int> ret=getMoney(root);
return max(ret[0],ret[1]);
}
};