[Swift]LeetCode617.合并二叉树|MergeTwoBinaryTrees

OStack程序员社区-中国程序员成长平台 › 门户 › 编程› 移动开发›Swift 教程

原作者: [db:作者] 来自: [db:来源] 收藏邀请

★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★
➤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/10470025.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！
★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★

热烈欢迎，请直接点击！！！

进入博主App Store主页，下载使用各个作品！！！

注：博主将坚持每月上线一个新app！！！

Given two binary trees and imagine that when you put one of them to cover the other, some nodes of the two trees are overlapped while the others are not.

You need to merge them into a new binary tree. The merge rule is that if two nodes overlap, then sum node values up as the new value of the merged node. Otherwise, the NOT null node will be used as the node of new tree.

Example 1:

Input: 
	Tree 1                     Tree 2                  
          1                         2                             
         / \                       / \                            
        3   2                     1   3                        
       /                           \   \                      
      5                             4   7                  
Output: 
Merged tree:
	     3
	    / \
	   4   5
	  / \   \ 
	 5   4   7

Note: The merging process must start from the root nodes of both trees.

给定两个二叉树，想象当你将它们中的一个覆盖到另一个上时，两个二叉树的一些节点便会重叠。

你需要将他们合并为一个新的二叉树。合并的规则是如果两个节点重叠，那么将他们的值相加作为节点合并后的新值，否则不为 NULL 的节点将直接作为新二叉树的节点。

示例 1:

输入: 
	Tree 1                     Tree 2                  
          1                         2                             
         / \                       / \                            
        3   2                     1   3                        
       /                           \   \                      
      5                             4   7                  
输出: 
合并后的树:
	     3
	    / \
	   4   5
	  / \   \ 
	 5   4   7

注意: 合并必须从两个树的根节点开始。

Runtime: 100 ms

Memory Usage: 19.8 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 mergeTrees(_ t1: TreeNode?, _ t2: TreeNode?) -> TreeNode? {
16         if t1 == nil {return t2}
17         if t2 == nil {return t1}
18         var t = TreeNode(t1!.val + t2!.val)
19         t.left =  mergeTrees(t1!.left, t2!.left)
20         t.right = mergeTrees(t1!.right, t2!.right)
21         return t
22     }
23 }

120ms

 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 mergeTrees(_ t1: TreeNode?, _ t2: TreeNode?) -> TreeNode? {
16         switch (t1, t2) {
17         case (let t1, nil) where t1 != nil:
18             return t1!
19         case (nil, let t2) where t2 != nil:
20             return t2!
21         case (nil, nil):
22             return nil
23         default:
24             guard let tree1 = t1, let tree2 = t2 else { return nil }
25             let newTree = TreeNode(tree1.val + tree2.val)
26             newTree.left = mergeTrees(tree1.left, tree2.left)
27             newTree.right = mergeTrees(tree1.right, tree2.right)
28             return newTree
29         }
30     }
31 }

124ms

 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 mergeTrees(_ t1: TreeNode?, _ t2: TreeNode?) -> TreeNode? {        
16         guard let t1 = t1 else {
17             return t2
18         }
19         
20         guard let t2 = t2 else {
21             return t1
22         }
23         
24         t1.val += t2.val
25         t1.left = mergeTrees(t1.left, t2.left)
26         t1.right = mergeTrees(t1.right, t2.right)
27         return t1
28     }        
29 }

136ms

 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 mergeTrees(_ t1: TreeNode?, _ t2: TreeNode?) -> TreeNode? {
16             if t1 == nil && t2 == nil { return nil }
17             var newNode = TreeNode((t1?.val ?? 0) + (t2?.val ?? 0))
18 
19             if t1?.left == nil || t2?.left == nil {
20                 newNode.left = t1?.left ?? t2?.left                    
21             } else {
22                 newNode.left = mergeTrees(t1?.left, t2?.left)
23             }
24             
25             if t1?.right == nil || t2?.right == nil {
26                 newNode.right = t1?.right ?? t2?.right                    
27             } else {
28                 newNode.right = mergeTrees(t1?.right, t2?.right)
29             }
30             
31             return newNode
32     }
33 }

160ms

 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 mergeTrees(_ t1: TreeNode?, _ t2: TreeNode?) -> TreeNode? {
16     switch (t1, t2) {
17     case (nil, _): return t2
18     case (_, nil): return t1
19         default:
20             var treeNode: TreeNode
21             treeNode = TreeNode(t1!.val + t2!.val)
22             treeNode.left = mergeTrees(t1!.left, t2!.left)
23             treeNode.right = mergeTrees(t1!.right, t2!.right)
24             return treeNode
25         }
26     }
27 }

204ms

 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 A = TreeNode(0)
16     func mergeTrees(_ t1: TreeNode?, _ t2: TreeNode?) -> TreeNode? {
17         if t1 == nil && t2 == nil{
18             return t1
19         }else{
20             B(t1,t2,A)
21         }
22         
23         return A
24     }
25     func B(_ t1: TreeNode?, _ t2: TreeNode?, _ t3: TreeNode?)
26     {
27         if t1?.left != nil && t2?.left != nil
28         {
29             t3?.left = TreeNode(0)
30             B(t1?.left,t2?.left,t3?.left)
31         }else if t1?.left == nil && t2?.left != nil{
32             t3?.left = TreeNode(0)
33             B(nil,t2?.left,t3?.left)
34         }else if t1?.left != nil && t2?.left == nil{
35             t3?.left = TreeNode(0)
36             B(t1?.left,nil,t3?.left)
37         }
38         
39         if t1 != nil && t2 != nil
40         {
41             let a:Int = t1?.val as! Int
42             let b:Int = t2?.val as! Int
43             t3?.val = a + b
44         }else if t1 == nil && t2 != nil{
45             let b:Int = t2?.val as! Int
46             t3?.val = b
47         }else if t1 != nil && t2 == nil{
48             let a:Int = t1?.val as! Int
49             t3?.val = a
50         }else{
51             return
52         }
53         
54         if t1?.right != nil && t2?.right != nil
55         {
56             t3?.right = TreeNode(0)
57             B(t1?.right,t2?.right,t3?.right)
58         }else if t1?.right == nil && t2?.right != nil{
59             t3?.right = TreeNode(0)
60             B(nil,t2?.right,t3?.right)
61         }else if t1?.right != nil && t2?.right == nil{
62             t3?.right = TreeNode(0)
63             B(t1?.right,nil,t3?.right)
64         }else{
65             return
66         }        
67     }
68 }