[Swift]LeetCode671. 二叉树中第二小的节点 | 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.

---

给定一个非空特殊的二叉树，每一个节点都是正数，而且每一个节点的子节点数量只能为 2 或 0。若是一个节点有两个子节点的话，那么这个节点的值不大于它的子节点的值。 

给出这样的一个二叉树，你须要输出全部节点中的第二小的值。若是第二小的值不存在的话，输出 -1 。

示例 1:

输入: 
    2
   / \
  2   5
     / \
    5   7

输出: 5
说明: 最小的值是 2 ，第二小的值是 5 。

示例 2:

输入: 
    2
   / \
  2   2

输出: -1
说明: 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 }