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:
3
/
2 3
3 1
- Maximum amount of money the thief can rob = 3 + 3 + 1 = 7.
Example 2:
3
/
4 5
/
1 3 1
- Maximum amount of money the thief can rob = 4 + 5 = 9.
Python Solution:
# 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 rob(self, root):
"""
:type root: TreeNode
:rtype: int
"""
def postorder(root):
if not root: return 0, 0
l_yes, l_no = postorder(root.left)
r_yes, r_no = postorder(root.right)
yes = l_no + r_no + root.val
no = max(l_yes, l_no) + max(r_yes, r_no)
return yes, no
yes, no = postorder(root)
return max(yes, no)
Summary:
- postorder
LeetCode: 337. House Robber III
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