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:

    / \

    \   \
        

Maximum amount of money the thief can rob = 3 + 3 + 1 = 7.

Example 2:

    / \

  / \   \
       

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=,lcur=,rpre=,rcur=;
         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=,cur=;
         rob(root,pre,cur);
         return max(pre,cur);
     }
 };