337. House Robber III
這一題是 198. House Robber 和 213. House Robber II 的延伸題組，題目的情境相同，但是做法完全不同，是一題後序遍歷的問題。
# Definition for a binary tree node.
# class TreeNode:
# def __init__(self, val=0, left=None, right=None):
# self.val = val
# self.left = left
# self.right = right
class Solution:
def rob(self, root: Optional[TreeNode]) -> int:
def helper(node):
if not node:
return (0, 0)
left = helper(node.left)
right = helper(node.right)
rob = node.val + left[1] + right[1]
not_rob = max(left) + max(right)
return rob, not_rob
return max(helper(root))