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[学习笔记] Leetcode 337. House Robber III

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发表于 2020-8-1 00:52:37 | 显示全部楼层 |阅读模式

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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, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution:
    def rob(self, root: TreeNode) -> int:
        result = self.helper(root)
        return max(result)
    
    def helper(self, root: TreeNode) -> int:
        if root == None:
            return [0, 0]
        result = [0, 0]
        left = self.helper(root.left)
        right = self.helper(root.right)
        result[0] = max(left) + max(right)
        result[1] = left[0] + right[0] + root.val
        return result

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