[Swift]LeetCode508. 出现次数最多的子树元素和 | Most Frequent Subtree Sum

★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★
➤微信公众号：山青咏芝（shanqingyongzhi）
➤博客园地址：山青咏芝（https://www.cnblogs.com/strengthen/）
➤GitHub地址：https://github.com/strengthen/LeetCode
➤原文地址：https://www.cnblogs.com/strengthen/p/10392404.html
➤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！
★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★

Given the root of a tree, you are asked to find the most frequent subtree sum. The subtree sum of a node is defined as the sum of all the node values formed by the subtree rooted at that node (including the node itself). So what is the most frequent subtree sum value? If there is a tie, return all the values with the highest frequency in any order.

Examples 1
Input:

  5
 /  
2   -3

return [2, -3, 4], since all the values happen only once, return all of them in any order.

Examples 2
Input:

  5
 /  
2   -5

return [2], since 2 happens twice, however -5 only occur once.

Note: You may assume the sum of values in any subtree is in the range of 32-bit signed integer.

给出二叉树的根，找出出现次数最多的子树元素和。一个结点的子树元素和定义为以该结点为根的二叉树上所有结点的元素之和（包括结点本身）。然后求出出现次数最多的子树元素和。如果有多个元素出现的次数相同，返回所有出现次数最多的元素（不限顺序）。

示例 1
输入:

  5
 /  
2   -3

返回 [2, -3, 4]，所有的值均只出现一次，以任意顺序返回所有值。

示例 2
输入:

  5
 /  
2   -5

返回 [2]，只有 2 出现两次，-5 只出现 1 次。

提示：假设任意子树元素和均可以用 32 位有符号整数表示。

Runtime: 64 ms

Memory Usage: 21.2 MB

 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 = nil
10  *         self.right = nil
11  *     }
12  * }
13  */
14 class Solution {
15     func findFrequentTreeSum(_ root: TreeNode?) -> [Int] {
16         var res:[Int] = [Int]()
17         var m:[Int:Int] = [Int:Int]()
18         var cnt:Int = 0
19         postorder(root, &m, &cnt)
20         for (key,val) in m
21         {
22             if val == cnt 
23             {
24                 res.append(key)
25             }
26         }
27         return res
28     }
29     
30     func postorder(_ node: TreeNode?,_ m:inout [Int:Int],_ cnt:inout Int) -> Int
31     {
32         if node == nil {return 0}
33         var left:Int = postorder(node!.left, &m, &cnt)
34         var right:Int = postorder(node!.right, &m, &cnt)
35         var sum = left + right + node!.val
36         if m[sum] == nil
37         {
38             m[sum] = 1
39         }
40         else
41         {
42             m[sum]! += 1
43         }        
44         cnt = max(cnt, m[sum]!)
45         return sum
46     }
47 }

64ms

 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 = nil
10  *         self.right = nil
11  *     }
12  * }
13  */
14 class Solution {
15     func findFrequentTreeSum(_ root: TreeNode?) -> [Int] {
16         var result = [Int:Int]()
17         _ = _findFrequentTreeSum(root, &result)
18         
19         let maxRepeat = result.values.max() ?? 0
20         
21         return result.compactMap { key,value in
22             guard value == maxRepeat else { return nil }
23             return key
24         }
25     }
26     
27     func _findFrequentTreeSum(_ root: TreeNode?, 
28                               _ result: inout [Int:Int]) -> Int 
29     {
30         guard let root = root else { return 0 }
31         
32         var value = root.val
33         value += _findFrequentTreeSum(root.left, &result)
34         value += _findFrequentTreeSum(root.right, &result)
35         
36         result[value] = (result[value] ?? 0) + 1
37         
38         return value
39     }
40 }

76ms

 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 = nil
10  *         self.right = nil
11  *     }
12  * }
13  */
14 class Solution {
15     func findFrequentTreeSum(_ root: TreeNode?) -> [Int] {
16         guard let root = root else { return [] }
17         
18         var counter = [Int : Int]()
19         func dfs(_ node: TreeNode?) -> Int {
20             guard let node = node else { return 0 }
21             let s = node.val + dfs(node.left) + dfs(node.right)
22             counter[s, default: 0] += 1
23             return s
24         }
25         
26         dfs(root)
27         let kvs = counter.sorted(by: {$0.1 > $1.1})
28         let freq = kvs[0].value
29         var ret = [Int]()
30         for (s, f) in kvs {
31             if f == freq {
32                 ret.append(s)
33             }
34         }
35         return ret
36     }
37 }

84ms

 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 = nil
10  *         self.right = nil
11  *     }
12  * }
13  */
14 class Solution {
15     var mp = [Int: Int]()
16 
17 func postorder(_ root: TreeNode?) -> Int{
18     if (root == nil) {
19         return 0;
20     }
21     
22     let left = postorder(root?.left);
23     let right = postorder(root?.right);
24     
25     let sum = (root?.val)! + left + right;
26     
27     if mp[sum] != nil {
28         mp[sum]! += 1
29     } else {
30          mp[sum] =  1
31     }
32     return sum;
33 }
34 
35 
36 func findFrequentTreeSum(_ root: TreeNode?) -> [Int] {
37     postorder(root)
38     print(mp)
39     var fq = [Int]()
40     var max = Int.min
41     for number in mp.values {
42         if max < number {
43             max = number
44         }
45     }
46     
47     for (i,number) in mp {
48         if max == number {
49             fq.append(i)
50         }
51     }
52     
53    return fq
54   }
55 }

104ms

 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 = nil
10  *         self.right = nil
11  *     }
12  * }
13  */
14 class Solution {
15     func findFrequentTreeSum(_ root: TreeNode?) -> [Int] {
16         var map = [Int: Int]()
17         traverse(root, &map)
18         
19         var result = [Int]()
20         var max = 0
21         for (sum, count) in map {
22             if count == max {
23                 result.append(sum)
24             } else if count > max {
25                 result.removeAll()
26                 max = count
27                 result.append(sum)
28             }
29         }
30         return result
31     }
32     
33     private func traverse(_ node: TreeNode?, _ map: inout [Int: Int]) -> Int {
34         guard let node = node else { return 0 }
35         
36         let sum = node.val + traverse(node.left, &map) + traverse(node.right, &map)
37         map[sum, default: 0] += 1
38         
39         return sum
40     }
41 }

108ms

 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 = nil
10  *         self.right = nil
11  *     }
12  * }
13  */
14 
15 class Solution {
16     
17     func recordSubTreeSums(_ root: TreeNode?, _ subTreeSumMap: inout Dictionary<Int, Int>) -> Int {
18         
19         guard let node = root else { return 0 }
20         
21         let subTreeSum =    recordSubTreeSums(node.left, &subTreeSumMap) + 
22                             recordSubTreeSums(node.right, &subTreeSumMap) +
23                             node.val
24         
25         if let count = subTreeSumMap[subTreeSum] {
26             subTreeSumMap[subTreeSum] = count + 1
27         } else {
28             subTreeSumMap[subTreeSum] = 1
29         }
30         
31         return subTreeSum
32     }
33     
34     func findFrequentTreeSum(_ root: TreeNode?) -> [Int] {
35         
36         var subTreeSumMap:Dictionary<Int, Int> = [:]
37         var result:[Int] = []
38     
39         recordSubTreeSums(root, &subTreeSumMap)
40         
41         var maxCount = -1
42         
43         for key in subTreeSumMap.keys {
44             if let count = subTreeSumMap[key] {
45                 if (count > maxCount) { maxCount = count }
46             }
47         }
48         
49         for key in subTreeSumMap.keys {
50             if let count = subTreeSumMap[key] {
51                 if (count == maxCount) { result = result + [key] }
52             }
53         }
54         
55         return result
56     }
57 }

相关阅读:
SQL2005四个排名函数（row_number、rank、dense_rank和ntile）的比较
 浅谈数据库索引
 移动网络应用开发中，使用 HTTP 协议比起使用 socket 实现基于 TCP 的自定义协议有哪些优势？
http协议学习系列
 隐藏帧技术
 第2章：标准输入与输出
 第1章：认识Shell脚本
 Cisco配置单臂路由及静态路由
 Cisco交换机端口聚合(EtherChannel)
Cisco配置VLAN+DHCP中继代理+NAT转发上网
原文地址：https://www.cnblogs.com/strengthen/p/10392404.html