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]
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]
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.