337. House Robber III

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LeetCode

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Problem

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:

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.

Example 2:

Input: [3,4,5,1,3,null,1]

     3
    / \
   4   5
  / \   \
 1   3   1

Output: 9
Explanation: Maximum amount of money the thief can rob = 4 + 5 = 9.

Code

class Solution {
    public int[] helper(TreeNode node) {
        if (node == null) {
            return new int[] { 0, 0 };
        }

        int left[] = helper(node.left);
        int right[] = helper(node.right);

        int rob = node.val + left[1] + right[1];
        int notRob = Math.max(left[0], left[1]) + Math.max(right[0], right[1]);

        return new int[] { rob, notRob };
    }

    public int rob(TreeNode root) {
        int[] res = helper(root);
        return Math.max(res[0], res[1]);
    }
}