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  • [leetcode-337-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:

    Basically you want to compare which one is bigger between 1) you + sum of your grandchildren and 2) sum of your children. 

    用l表示下一层左子树的值,r表示下一层右子树的值。那么对于root的值只需比较

    max(l + r, root->val + ll + lr + rl + rr)即可。
    int rob3(TreeNode* root, int& l, int& r)
        {
            if (root == NULL)return 0;
            int ll = 0, lr = 0, rl = 0, rr = 0;
            l = rob3(root->left, ll, lr);
            r = rob3(root->right, rl, rr);
            return max(l + r, root->val + ll + lr + rl + rr);
        }
        int rob3(TreeNode* root)
        {
            int l, r;
            return rob3(root, l, r);
        }

    参考:

    https://discuss.leetcode.com/topic/40847/simple-c-solution

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  • 原文地址:https://www.cnblogs.com/hellowooorld/p/7235447.html
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