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：
1 /** 2 * Definition for a binary tree node. 3 * struct TreeNode { 4 * int val; 5 * TreeNode *left; 6 * TreeNode *right; 7 * TreeNode(int x) : val(x), left(NULL), right(NULL) {} 8 * }; 9 */ 10 class Solution { 11 public: 12 void rob(TreeNode* root,int& pre,int& cur){ 13 if(!root) return; 14 int lpre=0,lcur=0,rpre=0,rcur=0; 15 rob(root->left,lpre,lcur); 16 rob(root->right,rpre,rcur); 17 pre=lcur+rcur; 18 cur=max(lpre+rpre+root->val,pre); 19 } 20 int rob(TreeNode* root) { 21 int pre=0,cur=0; 22 rob(root,pre,cur); 23 return max(pre,cur); 24 } 25 };