【LeetCode】337. House Robber III

版权声明：本文为博主原创文章，未经博主允许不得转载。 https://blog.csdn.net/zpalyq110/article/details/86651809

337. House Robber III

Description：
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.
Difficulty：Medium

Example:

Input: [3,2,3,null,3,null,1]

     3
    / \
   2   3
    \   \ 
     3   1

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

方法：Deep First Search

• Time complexity : $O\left ( n\right )$
• Space complexity : $O\left ( n \right )$
  思路：
  针对一个节点记录两个值，第一个值是包含此节点的最大数，第二个值是不包含此节点的最大数，这样能极大减少递归次数

class Solution {
public:
    int rob(TreeNode* root) {
        return helper(root).first;
    }
    
    pair<int, int> helper(TreeNode* root){
        if(!root) return make_pair(0, 0);
        pair<int, int> left = helper(root->left);
        pair<int, int> right = helper(root->right);
        int max_without_self = left.first + right.first;
        int max_may_with_self = max((root->val + left.second + right.second),
                                   max_without_self);
        return make_pair(max_may_with_self, max_without_self);
    }
};

第一版，很傻的代码，时间复杂度: $O\left ( 2^n\right )$

class Solution {
public:
    int rob(TreeNode* root) {
        if(!root) return 0;
        if(!root->left && !root->right)
            return  root->val;
        if(root->left && root->right)
            return max((root->val + 
                       rob(root->left->left) + 
                       rob(root->left->right) + 
                       rob(root->right->left) + 
                       rob(root->right->right)), 
                       rob(root->left) + rob(root->right) 
                      );
        if(!root->left)
            return  max((root->val +             
                       rob(root->right->left) + 
                       rob(root->right->right)), 
                        rob(root->right) 
                      );
        if(!root->right)
            return  max((root->val +             
                       rob(root->left->left) + 
                       rob(root->left->right)), 
                        rob(root->left) 
                      );
        return 0;
    }
};

第二版，简洁，但是还是慢

class Solution {
public:
    int rob(TreeNode* root) {
        if(!root) return 0;
        int t1 = 0;
        int t2 = root->val;
        
        t1 = rob(root->left) + rob(root->right);
        
        if(root->left)
            t2 += rob(root->left->left) + rob(root->left->right);
        if(root->right)
            t2 += rob(root->right->left) + rob(root->right->right);
        
        return max(t1, t2);
    }
};