标签：form val efi object mat bin mount after example
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.
# 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 3 => 3 3 => 3 / 2 3 3 =>3=max(3,3) 3 / \ => 3+2=5=max(3, 2+3) 2 3 3 / 2 3 \ 3 => 3+3=6=max(3+3,2+3)=max(3+node2,3 not choose, node2,3 choosed) 1 / 4 1 / \ \ 1 1 5 => 4+5=9=max(3+3+1+1,4+5)=9 """ return max(self.rob_helper(root)) def rob_helper(self, root): if root is None: return [0, 0] ans = [0]*2 ans_left = self.rob_helper(root.left) ans_right = self.rob_helper(root.right) ans[0] = max(ans_left[0], ans_left[1]) + max(ans_right[0], ans_right[1]) ans[1] = ans_left[0] + ans_right[0] + root.val return ans
337. House Robber III——树的题目几乎都是BFS、DFS，要么递归要么循环
标签：form val efi object mat bin mount after example
原文地址：http://www.cnblogs.com/bonelee/p/6220463.html
