[Swift]LeetCode114. 二叉树展开为链表 | Flatten Binary Tree to Linked List
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Given a binary tree, flatten it to a linked list in-place.
For example, given the following tree:
1 / \ 2 5 / \ \ 3 4 6
The flattened tree should look like:
1
\
2
\
3
\
4
\
5
\
6
给定一个二叉树，原地将它展开为链表。
例如，给定二叉树
1 / \ 2 5 / \ \ 3 4 6
将其展开为：
1
\
2
\
3
\
4
\
5
\
6
20ms
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 flatten(_ root: TreeNode?) { 16 guard let root = root else { return } 17 var l = root.left != nil; 18 var r = root.right != nil; 19 flatten(root.left) 20 flatten(root.right) 21 if (l && r) { 22 var left = root.left!; 23 let right = root.right; 24 root.right = left; 25 root.left = nil; 26 while (left.right != nil) { 27 left = left.right!; 28 } 29 left.right = right; 30 } else if (l) { 31 root.right = root.left; 32 root.left = nil; 33 } 34 } 35 }
20ms
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 flatten(_ root: TreeNode?) { 16 if root == nil { 17 return 18 } else { 19 flatten(root!.left) 20 flatten(root!.right) 21 if var node = root!.left { 22 while node.right != nil { 23 node.left = nil 24 node = node.right! 25 } 26 node.right = root!.right 27 root!.right = root!.left 28 root!.left = nil 29 } 30 } 31 } 32 }
24ms
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 pre : TreeNode? 16 func flatten(_ root: TreeNode?) { 17 if root == nil{ 18 return 19 } 20 flatten(root?.right) 21 flatten(root?.left) 22 root?.right = pre 23 root?.left = nil 24 pre = root 25 } 26 }
28ms
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 flatten(_ root: TreeNode?) { 16 helper(root) 17 } 18 19 private func helper(_ node: TreeNode?) -> TreeNode? { 20 var node = node 21 if node == nil { 22 return node 23 } 24 if node!.left == nil && node!.right == nil { 25 return node 26 } 27 28 let left = node!.left, right = node!.right 29 node!.left = nil 30 31 if let left = left { 32 node!.right = left 33 node = helper(left) 34 } 35 if let right = right { 36 node!.right = right 37 node = helper(right) 38 } 39 40 return node 41 } 42 }
40ms
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 array = [Int]() 16 func flatten(_ root: TreeNode?) { 17 if(root == nil){ return } 18 traverse(root) 19 root!.left = nil 20 root!.right = nil 21 var currentNode = root 22 for index in 1 ..< array.count{ 23 currentNode!.right = TreeNode(array[index]) 24 currentNode = currentNode!.right 25 } 26 } 27 func traverse(_ root: TreeNode?){ 28 if(root == nil){ return } 29 self.array.append(root!.val) 30 traverse(root!.left) 31 traverse(root!.right) 32 } 33 }
