Problem description
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.
Solution
We can use top down dynamic programming with memoization to solve the problem. There are two cases, the root is robbed or the root is not robbed. We compare these two cases and return the larger one between these two.
Below is my python implementation
# Definition for a binary tree node.
# class TreeNode:
# def __init__(self, x):
# self.val = x
# self.left = None
# self.right = None
class Solution:
import functools
@functools.lru_cache(maxsize=200)
def rob(self, root: TreeNode) -> int:
if not root:
return 0
#case 1
res1 = root.val
if root.left and root.left.left:
res1 += self.rob(root.left.left)
if root.left and root.left.right:
res1 += self.rob(root.left.right)
if root.right and root.right.left:
res1 += self.rob(root.right.left)
if root.right and root.right.right:
res1 += self.rob(root.right.right)
#case 2
res2 = self.rob(root.left) + self.rob(root.right)
return max(res1, res2)