[LeetCode]House Robber III

最新推荐文章于 2020-05-10 21:00:27 发布

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最新推荐文章于 2020-05-10 21:00:27 发布

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本文探讨了一种特殊情况下窃贼如何在不触动警报的情况下从房屋组成的二叉树结构中获得最大金额的问题。通过递归算法，我们展示了两种不同的策略，并提供了一个具体的代码实现案例。

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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:

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.

题解：有2中方式如下图：

code:

public class Solution {
    public int rob(TreeNode root) {
        
       if(root == null)
            return 0;
        int first = root.val,second = 0;
        if(root.left != null)
        {
            first += rob(root.left.left);
            first += rob(root.left.right);
            second += rob(root.left);
        }
        if(root.right != null)
        {
            first += rob(root.right.left);
            first += rob(root.right.right);
            second += rob(root.right);
        }
        return first>second?first:second;
    }
   
}

参考： http://www.cnblogs.com/leesf456/p/5349176.html