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.
My code:
# Definition for a binary tree node.
# class TreeNode(object):
# def __init__(self, x):
# self.val = x
# self.left = None
# self.right = None
class Solution(object):
def rob1(self, root):
"""
:type root: TreeNode
:rtype: int
"""
check_results1=self.check_rob(root,True)
check_results2=self.check_rob(root,False)
return max(check_results1,check_results2)
def check_rob(self,root,label):
if root==None:
return 0
#print('check root val %d'%root.val)
if label:
count1=self.check_rob(root.left,False)
count2=self.check_rob(root.right,False)
choose_result=root.val+count1+count2
else:
count1=self.check_rob(root.left,True)
count2=self.check_rob(root.right,True)
count3=self.check_rob(root.left,False)
count1=max(count1,count3)
count4=self.check_rob(root.right,False)
count2=max(count2,count4)
choose_result=count1+count2
return choose_result
def rob(self, root):
return max(self.f(root))
def f(self,node):
if node is None:return [0,0]
nl,nr = self.f(node.left),self.f(node.right)
return [node.val+ nl[1]+nr[1], max(nl)+max(nr)]Solution: https://leetcode.com/problems/house-robber-iii/submissions/