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:

1
2
3
4
5
  3
 / \
2   3
 \   \ 
  3   1

Maximum amount of money the thief can rob = 3 + 3 + 1 = 7.

Example 2:

1
2
3
4
5
    3
   / \
  4   5
 / \   \ 
1   3   1

Maximum amount of money the thief can rob = 4 + 5 = 9.

思路

就是照着参考资料的思路做的……自己并没有想出来TAT……

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
# 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
        """
        nodemap = {}
        if (root):
            return self.DFS(root, nodemap)
        else:
            return 0
        
    def DFS(self, cur, nodemap):
        if not cur:
            return 0
        if nodemap.has_key(cur.val):
            return nodemap[cur.val]
        
        leftchild = 0
        rightchild = 0
        if cur.left:
            
            leftchild += self.DFS(cur.left.left, nodemap)
            leftchild += self.DFS(cur.left.right, nodemap)
        
            
        if cur.right:
            
            rightchild += self.DFS(cur.right.left, nodemap)
            rightchild += self.DFS(cur.right.right, nodemap)
            
                 
        nodemap[cur.val] = max(cur.val + rightchild + leftchild, self.DFS(cur.right, nodemap) + self.DFS(cur.left, nodemap))
        return nodemap[cur.val]

参考资料：https://discuss.leetcode.com/topic/39834/step-by-step-tackling-of-the-problem