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 .

Credits:

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

这个题目是动态规划的思想，就是第i层的左右结点选或者不选，分两种情况考虑。

代码是参考别人的写的。。自己没做出来。。- -

/**
 * 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:
    void dfs(TreeNode* root,int &sum1,int &sum2){
        if(!root)return;
        int l1=0,l2=0,r1=0,r2=0;
        dfs(root->left,l1,l2);
        dfs(root->right,r1,r2);
        sum1=root->val+l2+r2;
        sum2=max(l1,l2)+max(r1,r2);
    }
    int rob(TreeNode* root) {
        int sum1=0,sum2=0;
        dfs(root,sum1,sum2);
        return max(sum1,sum2);
    }

};