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]

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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

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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

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