leetcode-rob house

最新推荐文章于 2023-02-25 22:41:55 发布

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探讨在一棵二叉树形式的社区中，窃贼如何在不触动报警的前提下最大化抢劫金额的问题。通过递归和动态规划的方法，实现高效计算。

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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.

思路:递归,从上到下要么偷要么不偷。缺点：效率不高，容易超时
DP:加个hashmap，防止相同节点重复计算

/**
 * 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 {
    //思路：
public:
    int rob(TreeNode* root) {
        if(root == NULL)
        {
            return 0;
        }


        //偷
        int stolenRootSum = root->val;
        if(root->left)
        {
            stolenRootSum+= rob(root->left->left) + rob(root->left->right);
        }
        if(root->right)
        {
            stolenRootSum+= rob(root->right->left) + rob(root->right->right);
        }
        return max(stolenRootSum,rob(root->left)+rob(root->right));

    }
};