337. House Robber III

最新推荐文章于 2025-12-22 20:10:38 发布

转载最新推荐文章于 2025-12-22 20:10:38 发布 · 92 阅读

0 ·

CC 4.0 BY-SA版权

原文链接：http://www.cnblogs.com/Zzz-y/p/8001246.html

文章标签：

#数据结构与算法

本文探讨了一种基于二叉树结构的算法问题，即在不触动报警机制的情况下，如何计算从二叉树中能获得的最大收益。通过递归地考虑每个节点作为根的子树，算法实现了对最大和的选择，确保了相邻节点最多只选其一，从而避免了同时选择直接相连的两个节点。文章详细解析了算法的实现过程，并提供了具体的代码示例。

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.

给定一棵二叉树，求最大和，相邻节点只能最多选其一。

对于每个子节点，以该节点为根root，则可以得到的最大和为：加上该节点的值res[1]或不加上该节点的值res[0].

当加上该节点，则相邻的左右子节点都不能加。要是不加上该节点，则相邻的左右子节点可加可不加。

故而res[0] = max(left.res[0], left.res[1]) + max(right.res[0], right.res[1]);

　　res[1] = root->val + left.res[0] + right.res[0];

class Solution {
public:
    vector<int> helper(TreeNode* root) {
        vector<int> res(2, 0);
        if (root) {
            vector<int> l = helper(root->left);
            vector<int> r = helper(root->right);
            res[0] = max(l[0],  l[1]) + max(r[0],  r[1]);
            res[1] = root->val + l[0] + r[0];
        }
        return res;
    }
    int rob(TreeNode* root) {
        vector<int> v = helper(root);
        return max(v[0],  v[1]);
    }
};