257. 二叉樹的所有路徑 | Binary Tree Paths
阿新 • • 發佈:2018-10-07
lse ive pen 所有路徑 dfs sdfs pat var let
Given a binary tree, return all root-to-leaf paths.
Note: A leaf is a node with no children.
Example:
Input: 1 / 2 3 5 Output: ["1->2->5", "1->3"] Explanation: All root-to-leaf paths are: 1->2->5, 1->3
給定一個二叉樹,返回所有從根節點到葉子節點的路徑。
說明: 葉子節點是指沒有子節點的節點。
示例:
輸入: 1 / 2 3 5 輸出: ["1->2->5", "1->3"] 解釋: 所有根節點到葉子節點的路徑為: 1->2->5, 1->3
16ms
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 binaryTreePaths(_ root: TreeNode?) -> [String] { 16 var list:[String] = [String]() 17 recuesive(root,&list,String())18 return list 19 } 20 func recuesive(_ root:TreeNode?,_ list:inout [String],_ str:String) 21 { 22 if root == nil {return} 23 var strNew:String = str 24 var strRoot:String = String(root!.val) 25 if root?.left == nil && root?.right == nil 26 { 27 strNew = strNew + strRoot 28 list.append(strNew) 29 return 30 } 31 strRoot = strNew + strRoot + "->" 32 recuesive(root?.left,&list,strRoot) 33 recuesive(root?.right,&list,strRoot) 34 } 35 }
16ms
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 binaryTreePaths(_ root: TreeNode?) -> [String] { 16 guard let root = root else { 17 return [] 18 } 19 var result = [String]() 20 binaryTreePathsDFS(root, "", &result) 21 return result 22 } 23 24 func binaryTreePathsDFS(_ root: TreeNode, _ out: String, _ result: inout [String]) { 25 if root.left == nil && root.right == nil { 26 result.append(out + String(root.val)) 27 } 28 29 if root.left != nil { 30 binaryTreePathsDFS(root.left!, out + String(root.val) + "->", &result) 31 } 32 if root.right != nil { 33 binaryTreePathsDFS(root.right!, out + String(root.val) + "->", &result) 34 } 35 36 } 37 }
16ms
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 binaryTreePaths(_ root: TreeNode?) -> [String] { 16 var ans = [String]() 17 binaryTreePaths(root, "", &ans) 18 return ans 19 } 20 21 func binaryTreePaths(_ node: TreeNode?, _ path: String, _ ans: inout [String]) { 22 guard let node = node else { return } 23 24 let path = path + String(node.val) 25 26 if node.left == nil && node.right == nil { 27 ans.append(path) 28 return 29 } 30 31 binaryTreePaths(node.left, path + "->", &ans) 32 binaryTreePaths(node.right, path + "->", &ans) 33 } 34 }
257. 二叉樹的所有路徑 | Binary Tree Paths