leetcode/Python/house-robber-iii.py at master · OnlyChristmas/leetcode · GitHub

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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:

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.
'''


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

class Solution:
    def rob(self, root: TreeNode) -> int:
        def robHelper(root):
            if not root:
                return (0, 0)
            left, right = robHelper(root.left), robHelper(root.right)
            return (root.val + left[1] + right[1], max(left) + max(right))

        return max(robHelper(root))