House Robber III

最新推荐文章于 2025-09-10 22:03:14 发布

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最新推荐文章于 2025-09-10 22:03:14 发布

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分类专栏： leetcode 文章标签： leetcode House Robber III

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解决一个关于二叉树中如何最大化抢劫金额同时避免触动报警机制的问题。通过递归算法实现，考虑每个节点抢或不抢两种状态，最终得出最大收益。

题目：

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.

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:
    int rob(TreeNode* root) {
        if(root==NULL) return 0;
        vector<int> res(2,0);
        res=subrot(root);
        return max(res[0],res[1]);
    }
    vector<int> subrot(TreeNode* root){
        vector<int> res(2,0);
        if(root==NULL) return res;
        
        vector<int> left(2,0);
        left=subrot(root->left);
        vector<int> right(2,0);
        right=subrot(root->right);
        
        res[0]=(root->val)+left[1]+right[1];
        res[1]=max(left[0],left[1])+max(right[0],right[1]);
        return res;
    }
    
    
};

分析：

首先想到迭代的方法，rob(root)=max(rob(root->left)+rob(root->right), root->val+rob(root->left->left)+rob(root->left->right)+rob(root->right->left)+rob(root->right->right)),由于重复计算，在计算rob(root->left)时，需要计算rob(root->left->left),rob(root->left->right),存在重复计算，超时。需要一种方法能够记录算出的结果，其中left[0]为左子树计算根节点情况的值，left[1]为左子树不计算根节点的值。