LeetCode 337 House Robber III

最新推荐文章于 2024-01-18 14:02:27 发布

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最新推荐文章于 2024-01-18 14:02:27 发布

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分类专栏： LeetCode 文章标签： leetcode ac

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本文探讨了一个有趣的问题：窃贼如何在不触动警报的情况下从一个形如二叉树的住宅区中偷取最多的财物。通过对每个节点进行递归判断，决定是否抢劫当前房屋，并计算能获得的最大金额。

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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.

public int rob(TreeNode root) {
         if (root == null) return 0;
         if (root.left == null && root.right == null) return root.val;
         int left = rob(root.left);
         int right = rob(root.right);
         int leftSon = 0, rightSon = 0;
         if (root.left != null)
            leftSon = rob(root.left.left) + rob(root.left.right);
         if (root.right != null)
            rightSon = rob(root.right.left) + rob(root.right.right);
         return Math.max(left + right, root.val + leftSon + rightSon);
}