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


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    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 }
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  • 原文地址:https://www.cnblogs.com/strengthen/p/10470025.html
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