House Robber III | Mingzhe Du

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))

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