[Swift]LeetCode671. 二叉樹中第二小的節點 | Second Minimum Node In a Binary Tree

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Given a non-empty special binary tree consisting of nodes with the non-negative value, where each node in this tree has exactly two or zero sub-node. If the node has two sub-nodes, then this node's value is the smaller value among its two sub-nodes.

Given such a binary tree, you need to output the second minimum value in the set made of all the nodes' value in the whole tree.

If no such second minimum value exists, output -1 instead.

Example 1:

Input: 
    2
   / \
  2   5
     / \
    5   7

Output: 5
Explanation: The smallest value is 2, the second smallest value is 5. 

Example 2:

Input: 
    2
   / \
  2   2

Output: -1
Explanation: The smallest value is 2, but there isn't any second smallest value.

---

給定一個非空特殊的二叉樹，每一個節點都是正數，而且每一個節點的子節點數量只能爲 2 或 0。若是一個節點有兩個子節點的話，那麼這個節點的值不大於它的子節點的值。 

給出這樣的一個二叉樹，你須要輸出全部節點中的第二小的值。若是第二小的值不存在的話，輸出 -1 。

示例 1:

輸入: 
    2
   / \
  2   5
     / \
    5   7

輸出: 5
說明: 最小的值是 2 ，第二小的值是 5 。

示例 2:

輸入: 
    2
   / \
  2   2

輸出: -1
說明: 最小的值是 2, 可是不存在第二小的值。

---

4ms

 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 findSecondMinimumValue(_ root: TreeNode?) -> Int {
16         guard let root = root else{
17             return -1
18         }
19         var q = [TreeNode]()
20         var set = Set<Int>()
21         q.append(root)
22         while(q.count>0){
23             for _ in q{
24                 var d = q.removeFirst()
25                 if(d.val != root.val){
26                      set.insert(d.val)
27                 }
28            
29                 if let r = d.right{
30                     q.append(r)
31                 }
32                 if let l = d.left{
33                     q.append(l)
34                 }
35             }
36             
37         }
38         var min = Int.max
39      for item in set{
40          if(item < min ){
41              min = item
42             }
43         }
44         return min == Int.max ? -1 : min
45     }
46 }

---

8ms

 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 findSecondMinimumValue(_ root: TreeNode?) -> Int {
16         guard let root = root else {
17             return -1 
18         }
19         
20         var firstMin: Int = Int.max 
21         var secondMin: Int = firstMin 
22         inorderTraversal(root) {
23             value in 
24             firstMin = min(firstMin, value)
25         }    
26         
27         inorderTraversal(root) {
28             value in 
29             if value != firstMin {
30                 secondMin = min(secondMin, value)
31             }
32         }
33         return secondMin == Int.max ? -1 : secondMin
34     }
35 }
36 
37 func inorderTraversal(_ root: TreeNode?, _ visit: (Int) -> Void) {
38     guard let root = root else {
39         return 
40     }
41     
42     inorderTraversal(root.left, visit)
43     visit(root.val)
44     inorderTraversal(root.right, visit)
45 }

---

Runtime: 12 ms

Memory Usage: 18.9 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 findSecondMinimumValue(_ root: TreeNode?) -> Int {
16         return helper(root, root!.val)
17     }
18     
19     func helper(_ node: TreeNode?,_ first:Int) -> Int
20     {
21         if node == nil {return -1}
22         if node!.val != first {return node!.val}
23         var left:Int =  helper(node?.left, first)
24         var right:Int = helper(node?.right, first)
25         return (left == -1 || right == -1) ? max(left, right) : min(left, right)
26     }
27 }

---

24ms

 1 class Solution {
 2     func findSecondMinimumValue(_ root: TreeNode?) -> Int 
 3     {
 4         guard let root = root else { return -1 }
 5         var first = root.val
 6         var second = Int.max
 7         recursion(root, &first, &second)
 8         recursion(root, &second, &first)
 9         return second == Int.max ? -1 : second
10     }
11     
12     func recursion(_ root: TreeNode?, _ first: inout Int, _ second: inout Int)
13     {
14         guard let root = root else { return }
15         if first > root.val && root.val != second
16         {
17             first = root.val
18         }
19         recursion(root.left, &first, &second)
20         recursion(root.right, &first, &second)
21     }
22 }