# 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:

`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 superrob(node):
# returns tuple of size two (now, later)
# now: max money earned if input node is robbed
# later: max money earned if input node is not robbed

# base case
if not node: return (0, 0)

# get values
left, right = superrob(node.left), superrob(node.right)

# rob now
now = node.val + left[1] + right[1]

# rob later
later = max(left) + max(right)

return (now, later)

return max(superrob(root))```