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.

URL: https://leetcode.com/problems/house-robber-iii/

# Definition for a binary tree node.
# class TreeNode(object):
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution(object):
    def rob(self, root):
        """
        :type root: TreeNode
        :rtype: int
        """
        if root == None:
            return 0
        else:
            result = self.rob_max(root)
            return max(result[0], result[1])

    def rob_max(self, root):
        if root == None:
            return [0, 0]
        else:
            left_res = self.rob_max(root.left)
            right_res = self.rob_max(root.right)
            result = [0]*2
            result[0] = root.val + left_res[1] + right_res[1]
            result[1] = max(left_res[0], left_res[1]) + max(right_res[0], right_res[1])
            return result