leetcode笔记:House Robber III

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:precurr

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。

时间: 2024-10-12 14:42:54

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