LeetCode 337. House Robber III 题解

337. House Robber III

 

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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:

     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.

解题思路：

注意到若选取父节点，则其左右子节点均不可选。如果不选取父节点，则左右子节点可以选也可以不选。

代码展示：

/** * 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 {private:     vector<int> help(TreeNode* root){        if(!root) return vector<int> (2,0);        vector<int> left = help(root->left);        vector<int> right = help(root->right);        vector<int> res(2,0);        //res[0]保存不添加root结点的情况        //res[1]保存添加root结点的情况        // res[0] 不包括root结点，所以其子节点可以选择取或者不取        res[0]=max(left[0],left[1])+max(right[0],right[1]);        // res[1] 包括root结点， 所以其子节点不可选取        res[1]=root->val+left[0]+right[0];        return res;     } public:    int rob(TreeNode* root){        vector<int> result = help(root);        return max(result[0], result[1]);    }};