177 - 337 打家劫舍3

题目

在上次打劫完一条街道之后和一圈房屋后，小偷又发现了一个新的可行窃的地区。这个地区只有一个入口，我们称之为“根”。 除了“根”之外，每栋房子有且只有一个“父“房子与之相连。一番侦察之后，聪明的小偷意识到“这个地方的所有房屋的排列类似于一棵二叉树”。 如果两个直接相连的房子在同一天晚上被打劫，房屋将自动报警。

计算在不触动警报的情况下，小偷一晚能够盗取的最高金额。

示例 1:

输入: [3,2,3,null,3,null,1]

 3
/ \

2 3 3 1

输出: 7 解释: 小偷一晚能够盗取的最高金额 = 3 + 3 + 1 = 7.

示例 2:

输入: [3,4,5,1,3,null,1]

 3
/ \

4 5 / 1 3 1

输出: 9 解释: 小偷一晚能够盗取的最高金额 = 4 + 5 = 9.

解答

递归

思路就是比较爷+4孙与2子之间，哪个值更大。

class Solution:
    def rob(self, root: TreeNode) -> int:
        if not root:
            return 0

        money = root.val
        if root.left != None:
            money += Solution.rob(self, root.left.left)+Solution.rob(self,
                                                                     root.left.right)
        if root.right != None:
            money += Solution.rob(self, root.right.left)+Solution.rob(self,
                                                                      root.right.right)
        return max(money, Solution.rob(self, root.left)+Solution.rob(self, root.right))

超时了。。

小本本记一下

class Solution:
    def rob(self, root: TreeNode) -> int:
        memo = dict()

        def helper(root):
            if not root:
                return 0
            if root in memo:
                return memo.get(root)

            money = root.val
            if root.left != None:
                money += helper(root.left.left)+helper(root.left.right)
            if root.right != None:
                money += helper(root.right.left)+helper(root.right.right)

            result = max(money, helper(root.left)+helper(root.right))
            memo[root] = result
            return result

        return helper(root)

Runtime: 52 ms, faster than 70.88% of Python3 online submissions for House Robber III.

Memory Usage: 15.2 MB, less than 86.67% of Python3 online submissions for House Robber III.

另一种写法：

class Solution:
    def rob(self, root: TreeNode) -> int:
        memo = dict()

        def helper(root):
            if not root:
                return 0
            if root in memo:
                return memo.get(root)
            do_it = root.val + (helper(root.left.left)+helper(root.left.right)
                                if root.left else
                                0)+(helper(root.right.left)+helper(root.right.right)
                                    if root.right else 0)
            not_do = helper(root.left)+helper(root.right)
            res = max(do_it, not_do)
            memo[root] = res
            return res

        return helper(root)

Runtime: 72 ms, faster than 11.95% of Python3 online submissions for House Robber III.

Memory Usage: 15.3 MB, less than 80.00% of Python3 online submissions for House Robber III.

用数组

存[不抢， 抢]

class Solution:
    def rob(self, root: TreeNode) -> int:
        def dp(root):
            if not root:
                return [0, 0]
            left = dp(root.left)
            right = dp(root.right)
            rob = root.val+left[0]+right[0]
            not_rob = max(left[0], left[1])+max(right[0], right[1])
            return [not_rob, rob]

        res = dp(root)
        return max(res[0], res[1])

Runtime: 52 ms, faster than 70.88% of Python3 online submissions for House Robber III.

Memory Usage: 14.7 MB, less than 100.00% of Python3 online submissions for House Robber III.