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➤微信公众号：山青咏芝（shanqingyongzhi）➤博客园地址：山青咏芝（）➤GitHub地址：➤原文地址： ➤如果链接不是山青咏芝的博客园地址，则可能是爬取作者的文章。➤原文已修改更新！强烈建议点击原文地址阅读！支持作者！支持原创！★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★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:
Input: [3,2,3,null,3,null,1] 3 / \ 2 3 \ \ 3 1Output: 7 Explanation: Maximum amount of money the thief can rob = 3 + 3 + 1 = 7.
Example 2:
Input: [3,4,5,1,3,null,1] 3 / \ 4 5 / \ \ 1 3 1Output: 9Explanation: Maximum amount of money the thief can rob = 4 + 5 = 9.
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.
52ms
1 /** 2 * Definition for a binary tree node. 3 * public class TreeNode { 4 * public var val: Int 5 * public var left: TreeNode? 6 * public var right: TreeNode? 7 * public init(_ val: Int) { 8 * self.val = val 9 * self.left = nil10 * self.right = nil11 * }12 * }13 */14 class Solution {15 func rob(_ root: TreeNode?) -> Int {16 let maxs = robMaxs(root)17 18 return max(maxs.0, maxs.1)19 }20 21 func robMaxs(_ root : TreeNode?) -> (Int, Int) {22 if root == nil {23 return (0,0)24 }25 26 let leftMaxs = robMaxs(root?.left)27 let rightMaxs = robMaxs(root?.right)28 29 30 return (leftMaxs.1 + rightMaxs.1 + root!.val, max(leftMaxs.1 + rightMaxs.1,leftMaxs.0 + rightMaxs.0, leftMaxs.0 + rightMaxs.1, leftMaxs.1 + rightMaxs.0))31 }32 } 56ms
1 /** 2 * Definition for a binary tree node. 3 * public class TreeNode { 4 * public var val: Int 5 * public var left: TreeNode? 6 * public var right: TreeNode? 7 * public init(_ val: Int) { 8 * self.val = val 9 * self.left = nil10 * self.right = nil11 * }12 * }13 */14 class Solution {15 func rob(_ root: TreeNode?) -> Int {16 guard let node = root else {17 return 018 }19 return max(helper(node)[0], helper(node)[1])20 }21 private func helper(_ root: TreeNode?) -> [Int] {22 guard let node = root else {23 return [0, 0]24 }25 var res = [Int](repeating: 0, count: 2)26 let left = helper(node.left)27 let right = helper(node.right)28 res[0] = node.val + left[1] + right[1]29 res[1] = max(left[0], left[1]) + max(right[0], right[1])30 return res31 }32 } 60ms
1 /** 2 * Definition for a binary tree node. 3 * public class TreeNode { 4 * public var val: Int 5 * public var left: TreeNode? 6 * public var right: TreeNode? 7 * public init(_ val: Int) { 8 * self.val = val 9 * self.left = nil10 * self.right = nil11 * }12 * }13 */14 class Solution {15 16 func rob(_ root: TreeNode?) -> Int {17 let res = findMax(root)18 return max(res.0, res.1)19 }20 21 func findMax(_ root:TreeNode?) -> (Int, Int){ //do rob root, do not rob22 guard let root = root else {23 return (0,0)24 }25 let left = findMax(root.left)26 let right = findMax(root.right)27 let robCur = left.1 + right.1 + root.val28 let noRobC = max(left.1,left.0) + max(right.1,right.0)29 30 return (robCur, noRobC)31 }32 }