house-robber-iii.py

# Time:  O(n)
# Space: O(h)

# 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.
#
# 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
        """
        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))