一. 题目描述
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
.
二. 题目分析
该题是House Robber II和House Robber的后续。
题目的意思是,小偷找到了一个新的偷盗场所。这片区域只有一个入口,叫做“根”。除了根以外,每个屋子有且仅有一个父屋子。在踩点之后盗贼发现,所有的房间构造形成了一棵二叉树。如果两个直接相连的屋子在同时被盗窃,就会惊动警察。
编写函数判断盗贼在不惊动警察的情况下最多可以偷到的金钱数目。测试用例如题目描述。
针对该题,可使用深度优先搜索来解决。对于当前节点,只有盗窃和不盗窃两种操作,这取决于当前节点的子节点的状态,可用两个变量表示:pre
和curr
:
pre:当前节点(屋子)不偷盗时,所能获得的收益;取决于在该节点的两个子节点被偷盗时的收益之和。
curr:当前节点(屋子)偷盗时,所能获得的收益;由于相邻节点不能同时偷盗否则会引来警察,所以两个子节点不能偷盗,此时收益等于:父节点->val + 左子节点->pre + 右子节点->pre。
比较两种收益pre和curr,取较大者作为当前节点的最大收益。
三. 示例代码
/**
* 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:
struct Money {
int pre; // 该节点不偷,收益为子节点偷的情况下的收益和
int curr; // 该节点处偷,加上两子节点不偷时的收益
Money():pre(0), curr(0){}
};
int rob(TreeNode* root) {
Money sum = dfs(root);
return sum.curr;
}
Money dfs(TreeNode* root)
{
if (root == NULL) return Money();
Money leftMoney = dfs(root->left); // 当前节点的左子节点收益情况
Money rightMoney = dfs(root->right); // 当前节点的右子节点收益情况
Money sumMoney;
sumMoney.pre = leftMoney.curr + rightMoney.curr; // 当前节点不偷
sumMoney.curr = max(sumMoney.pre, root->val + leftMoney.pre + rightMoney.pre);
return sumMoney;
}
};
四. 小结
该题属于DFS的经典题目。