[leetcode] 337. House Robber III 解题报告

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题目链接: https://leetcode.com/problems/house-robber-iii/

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:

     3    / \   2   3    \   \      3   1

Maximum amount of money the thief can rob = 3 + 3 + 1 = 7.

Example 2:

     3    / \   4   5  / \   \  1   3   1

Maximum amount of money the thief can rob = 4 + 5 = 9.

思路: 一个基本的思路就是每一个结点做两种判断, 偷或者不偷, 所以在做DFS搜索的时候我们可以在每一个结点判断哪一个情况利益更大. 我们需要一个标记, 就是如果上一个结点被偷了, 这个结点就不可以再偷了, 所以一个基本的代码如下, 但是无法通过所有数据:

class Solution {public:    int DFS(TreeNode* root, bool flag)    {        if(!root) return 0;        int sum = DFS(root->left, true) + DFS(root->right, true);        if(flag == true)//如果当前结点可以偷, 那么下一个结点则不可以            sum = max(DFS(root->left, false)+ DFS(root->right, false)+root->val, sum);        return sum;    }    int rob(TreeNode* root) {        return max(DFS(root, false), DFS(root, true));    }};

我们可以看到这种方法包含了很多重复的计算, 那么我们可以将计算过的结点保存起来, 如果下次再遍历到这里的时候可以直接得到其值, 改进的方法如下:

/** * 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 DFS(TreeNode* root, int flag)    {        if(!root) return 0;        if(hash.count(root) && hash[root].count(flag)) return hash[root][flag];        int ans = DFS(root->left, 0) + DFS(root->right, 0);        if(!flag) ans=max(ans, root->val+DFS(root->left, 1)+DFS(root->right, 1));        hash[root][flag] = ans;        return ans;    }        int rob(TreeNode* root) {        if(!root) return 0;        return DFS(root, 0);    }private:    unordered_map<TreeNode*, unordered_map<int, int>> hash;};

这种方法已经足以通过所有数据, 并且也相当高效, 不过代码比较复杂, 参考别人的代码发现一种更好的方式, 并且不需要特殊的数据结构, 因此效率比我的代码更高.

代码如下:

class Solution {public:    int DFS(TreeNode* root, int& sonLeft, int& sonRight)    {        if(!root) return 0;        int leftSonLeft=0, leftSonRight=0, rightSonLeft=0, rightSonRight=0;        sonLeft = DFS(root->left, leftSonLeft, leftSonRight);        sonRight = DFS(root->right, rightSonLeft, rightSonRight);        int fromGrandson = leftSonLeft+leftSonRight+ rightSonLeft+ rightSonRight;        return max(root->val + fromGrandson, sonLeft + sonRight);    }    int rob(TreeNode* root) {        int sonLeft=0, sonRight=0;        return DFS(root, sonLeft, sonRight);    }};

参考链接:https://leetcode.com/discuss/93998/simple-c-solution

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