题目地址：https://leetcode.com/problems/house-robber-iii/description/
题目描述
Thethief 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.
题目大意
从一棵二叉树中取出一些数字，使得取得数字的和最大。取的规则是不能同时取直接相连的两个节点。
解题方法
这个是限定规则下的博弈过程。曾经看过左程云的视频教程，对这个过程印象比较深刻。
本题的做法，就是求本节点+孙子更深节点vs儿子节点+重孙更深的节点的比较。
道理能想明白，代码有点难写。用了dfs函数，虽然递归是自顶向下的，但是因为是不断的return，所以真正求值是从底向上的。用到了一个有两个元素的列表，分别保存了之前层的，不取节点和取节点的情况。然后遍历左右子树，求出当前节点取和不取能得到的值，再返回给上一层。注意这个里面的robcurr是当前节点能达到的最大值，所以最后返回结果的时候试试返回的root节点robcurr的值。
# 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 dfs(root):
from bottom to top
if not root: return [0, 0] before layer, no robcurr, robcurr
robleft = dfs(root.left)
robright = dfs(root.right)
norobcurr = robleft[1] + robright[1]
robcurr = max(root.val + robleft[0] + robright[0], norobcurr)
return [norobcurr, robcurr]
return dfs(root)[1]
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22
二刷的时候换了一种解法，使用的仍然是递归，不过不用返回两个值，而是直接一个值：无论用还是不用情况下，能得到的最好结果。必须使用记忆化递归，否则超时。
# 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
"""
memo = dict()
return self.helper(root, memo)
def helper(self, root, memo):
if not root:
return 0
if root in memo:
return memo[root]
res = 0
notused = self.helper(root.left, memo) + self.helper(root.right, memo)
used = 0
if root.left:
used += self.helper(root.left.left, memo) + self.helper(root.left.right, memo)
if root.right:
used += self.helper(root.right.left, memo) + self.helper(root.right.right, memo)
res = max(notused, used + root.val)
memo[root] = res
return res
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
三刷的时候，代码思路更简洁明了。递归函数增加一个变量，表示当前节点的父亲节点是否用过。根节点没有父亲节点，所以其父亲节点肯定没用过。然后我们判断在某个节点的父亲用过和没用过的情况下，当前节点能不能用，最优的结果分别是多少。
# 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
"""
self.d = dict()
return self.helper(root, False)
def helper(self, root, parentUsed):
if not root: return 0
if (root, parentUsed) in self.d:
return self.d[(root, parentUsed)]
res = 0
if parentUsed:
res = self.helper(root.left, False) + self.helper(root.right, False)
else:
res = max(root.val + self.helper(root.left, True) + self.helper(root.right, True), self.helper(root.left, False) + self.helper(root.right, False))
self.d[(root, parentUsed)] = res
return res
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
DDKK.COM 弟弟快看-教程，程序员编程资料站，版权归原作者所有
本文经作者：负雪明烛 授权发布，任何组织或个人未经作者授权不得转发