，敢教日月换新天。为有牺牲多壮志

[Swift]LeetCode617. 合并二叉树 | Merge Two Binary Trees

★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★
➤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
复制代码
 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 }
复制代码

120ms

复制代码
 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 }
复制代码

124ms

复制代码
 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 }
复制代码

136ms

复制代码
 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 }
复制代码

160ms

复制代码
 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 }
复制代码

204ms

复制代码
 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 }
复制代码

 

posted @   为敢技术  阅读(306)  评论(0)  编辑  收藏  举报
编辑推荐：
· 记一次.NET内存居高不下排查解决与启示
· 探究高空视频全景AR技术的实现原理
· 理解Rust引用及其生命周期标识（上）
· 浏览器原生「磁吸」效果！Anchor Positioning 锚点定位神器解析
· 没有源码，如何修改代码逻辑？
阅读排行：
· 全程不用写代码，我用AI程序员写了一个飞机大战
· MongoDB 8.0这个新功能碉堡了，比商业数据库还牛
· DeepSeek 开源周回顾「GitHub 热点速览」
· 记一次.NET内存居高不下排查解决与启示
· 白话解读 Dapr 1.15：你的「微服务管家」又秀新绝活了
点击右上角即可分享
微信分享提示
哥伦布
3°
哥伦布
09:09发布
哥伦布
09:09发布
3°
多云
东南风
3级
空气质量
无
相对湿度
47%
今天
中雨
3°/15°
周三
中雨
3°/13°
周四
小雪
-1°/6°
Powered by 心知天气