Leetcode 337. House Robber III

337. House Robber III

• Total Accepted: 18475
• Total Submissions: 47725
• Difficulty: Medium

 

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：基本思路和Leetcode 198. House Robber相同，只是将传递公式植入到二叉树的DFS过程中。

 

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:
    void rob(TreeNode* root,int& pre,int& cur){
        if(!root) return;
        int lpre=0,lcur=0,rpre=0,rcur=0;
        rob(root->left,lpre,lcur);
        rob(root->right,rpre,rcur);
        pre=lcur+rcur;
        cur=max(lpre+rpre+root->val,pre);
    }
    int rob(TreeNode* root) {
        int pre=0,cur=0;
        rob(root,pre,cur);
        return max(pre,cur);
    }
};