337. House Robber III 【M】【vip】

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.

Credits:

Special thanks to @dietpepsi for adding this problem and creating all test cases.

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本质上是一个树的遍历 问题

这个刚开始想错了，为什么不能用广搜呢，因为并不是如果我不rob树根，就一定要rob他的两个子节点。也就是说，有可能我只rob他的一个子节点会更好。

f1(node)

be the value of maximum money we can rob from the subtree with

node

as
root ( we can rob

node

if necessary).

f2(node)

be the value of maximum money we can rob from the subtree with

node

as
root but without robbing

node

.

Then we have

f2(node) = f1(node.left) + f1(node.right)

and

f1(node) = max( f2(node.left)+f2(node.right)+node.value, f2(node) )

.

class Solution(object):
def do(self,root):
if root == None:
return (0,0)

l = self.do(root.left)
r = self.do(root.right)

return (l[1] + r[1] , max(l[1] + r[1] , l[0] + r[0] + root.val))

def rob(self, root):
return self.do(root)[1]