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
.
分析
dfs返回的是【不选，选】
# 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):
"""
:type root: TreeNode
:rtype: int
"""
res = self.helper(root)
return max(res)
def helper(self,root):
# every house you have two option, rob or not rob.
# [value1, value2] value1: not rob current house, rob current house
if not root:
return [0,0]
if not root.left and not root.right:
return [0,root.val]
left = self.helper(root.left)
right = self.helper(root.right)
return [max(left[0],left[1])+max(right[0],right[1]), root.val+left[0]+right[0]]