LeetCode 337. House Robber III

最新推荐文章于 2022-03-05 22:30:26 发布

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最新推荐文章于 2022-03-05 22:30:26 发布

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分类专栏： LeetCode 题解

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本文探讨了一个有趣的问题：如何在一个形如二叉树布局的社区中，窃贼可以不触动报警系统的情况下最大化抢劫金额。文章通过递归算法解决了这个问题，并给出了具体的实现代码。

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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.

This is a very interesting question!

/**
 * 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) {
        if(!root) return 0;
        if(!root->left && !root->right) return root->val;
        int rIn = root->val, rOut = 0;
        if(root->left) {
            rIn += rob(root->left->left) + rob(root->left->right);
            rOut = rob(root->left);
        }
        if(root->right) {
            rIn += rob(root->right->left) + rob(root->right->right);
            rOut += rob(root->right);
        }
        return rIn > rOut ? rIn : rOut;
    }
};