[Swift]LeetCode337. 打家劫舍 III | House Robber III

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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   1

Output: 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   1

Output: 9
Explanation: 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

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     public var val: Int
 *     public var left: TreeNode?
 *     public var right: TreeNode?
 *     public init(_ val: Int) {
 *         self.val = val
 *         self.left = nil
 *         self.right = nil
 *     }
 * }
 */
class Solution {
    func rob(_ root: TreeNode?) -> Int {
        let maxs = robMaxs(root)

        return max(maxs.0, maxs.1)
    }

    func robMaxs(_ root : TreeNode?) -> (Int, Int) {
        if root == nil {
            return (0,0)
        }

        let leftMaxs = robMaxs(root?.left)
        let rightMaxs = robMaxs(root?.right)
        
        
        return (leftMaxs.1 + rightMaxs.1 + root!.val, max(leftMaxs.1 + rightMaxs.1,leftMaxs.0 + rightMaxs.0, leftMaxs.0 + rightMaxs.1, leftMaxs.1 + rightMaxs.0))
    }
}

---

56ms

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     public var val: Int
 *     public var left: TreeNode?
 *     public var right: TreeNode?
 *     public init(_ val: Int) {
 *         self.val = val
 *         self.left = nil
 *         self.right = nil
 *     }
 * }
 */
class Solution {
    func rob(_ root: TreeNode?) -> Int {
        guard let node = root else {
            return 0
        }
        return max(helper(node)[0], helper(node)[1])
    }
    private func helper(_ root: TreeNode?) -> [Int] {
        guard let node = root else {
            return [0, 0]
        }
        var res = [Int](repeating: 0, count: 2)
        let left = helper(node.left)
        let right = helper(node.right)
        res[0] = node.val + left[1] + right[1]
        res[1] = max(left[0], left[1]) + max(right[0], right[1])
        return res
    }
}

---

60ms

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     public var val: Int
 *     public var left: TreeNode?
 *     public var right: TreeNode?
 *     public init(_ val: Int) {
 *         self.val = val
 *         self.left = nil
 *         self.right = nil
 *     }
 * }
 */
class Solution {
    
    func rob(_ root: TreeNode?) -> Int {
        let res = findMax(root)
        return max(res.0, res.1)
    }
    
    func findMax(_ root:TreeNode?) -> (Int, Int){  //do rob root, do not rob
        guard let root = root else {
            return (0,0)
        }
        let left = findMax(root.left)
        let right = findMax(root.right)
        let robCur = left.1 + right.1 + root.val
        let noRobC = max(left.1,left.0) + max(right.1,right.0)
                
        return (robCur, noRobC)
    }
}