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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 }