[LeetCode] 337. House Robber III

最新推荐文章于 2025-04-15 17:56:05 发布

原创最新推荐文章于 2025-04-15 17:56:05 发布 · 191 阅读

0 ·

CC 4.0 BY-SA版权

算法 Algorithm 同时被 2 个专栏收录

38 篇文章

订阅专栏

深度优先算法 DFS

14 篇文章

订阅专栏

解决一个关于二叉树中如何最大化抢劫金额的问题，确保不会触发报警机制。通过递归算法来确定最佳抢劫策略。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

Problem:

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.

Solution:

/**
 * 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:
    int rob(TreeNode* root) {
        int l = 0, r = 0;
        int max = dfsRob(root, l, r);
        return max;
    }
    
    /* 对每颗子树：
      计算其根节点 + 四个孙子节点的值（若不存在即为0） root->val + ll + lr + rl + rr
      计算其左子节点值 + 右子节点的值 l + r
      比较并返回较大者*/
    int dfsRob(TreeNode* root, int& l, int& r) {
        if (root == NULL) return 0;
        int ll = 0, lr = 0, rl = 0, rr = 0;
        l = dfsRob(root->left, ll, lr);
        r = dfsRob(root->right, rl, rr);
        return max(root->val + ll + lr + rl + rr, l + r);
    }
};

关键点说明：对每颗子树来说，有两个各不相邻的节点的集合（假设节点均不为空）：

{根节点，四个孙子节点} 和 {左子节点，右子节点}

所以，只需要比较这两个集合的节点值的和即可得到每颗子树的要求的最大值，通过递归可以得到最终的最大值。