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))
