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);
}
};
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