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We are given a binary tree (with root node root), a targetnode, and an integer value K.
Return a list of the values of all nodes that have a distance Kfrom the target node. The answer can be returned in any order.
Example 1:
Input: root = 2
Output: [7,4,1]
Explanation:
The nodes that are a distance 2 from the target node (with value 5)
have values 7, 4, and 1.
![[Swift]LeetCode863. 二叉树中所有距离为 K 的结点 | All Nodes Distance K in Binary Tree](/default/index/img?u=L2RlZmF1bHQvaW5kZXgvaW1nP3U9YUhSMGNITTZMeTl6TXkxc1l5MTFjR3h2WVdRdWN6TXVZVzFoZW05dVlYZHpMbU52YlM5MWNHeHZZV1J6THpJd01UZ3ZNRFl2TWpndmMydGxkR05vTUM1d2JtYz0= "[Swift]LeetCode863. 二叉树中所有距离为 K 的结点 | All Nodes Distance K in Binary Tree")
Note that the inputs "root" and "target" are actually TreeNodes.
The descriptions of the inputs above are just serializations of these objects.
Note:
- The given tree is non-empty.
- Each node in the tree has unique values
0 <= node.val <= 500. - The
targetnode is a node in the tree. -
0 <= K <= 1000.
给定一个二叉树（具有根结点 root）， 一个目标结点 target ，和一个整数值 K 。
返回到目标结点 target 距离为 K 的所有结点的值的列表。 答案可以以任何顺序返回。
示例 1：
输入：root = [3,5,1,6,2,0,8,null,null,7,4], target = 5, K = 2 输出：[7,4,1] 解释： 所求结点为与目标结点（值为 5）距离为 2 的结点， 值分别为 7，4，以及 1
注意，输入的 "root" 和 "target" 实际上是树上的结点。 上面的输入仅仅是对这些对象进行了序列化描述。
提示：
- 给定的树是非空的，且最多有
K个结点。 - 树上的每个结点都具有唯一的值
0 <= node.val <= 500。 - 目标结点
target是树上的结点。 -
0 <= K <= 1000.
16ms
1 class Solution { 2 func distanceK(_ root: TreeNode?, _ target: TreeNode?, _ K: Int) -> [Int] { 3 guard let root = root else { return [] } 4 guard let target = target else { return [] } 5 6 let result = dfs(root, 0, target: target) 7 let path = result.paths.reversed() 8 let level = result.level 9 10 var results = [Int]() 11 var node = root 12 13 for (currentLevel, step) in path.enumerated() { 14 let t = K - (level - currentLevel) - 1 15 if t == -1 { 16 results.append(node.val) 17 } 18 if step { 19 if let right = node.right, t >= 0 { 20 results += bfs(right, t) 21 } 22 node = node.left! 23 } else { 24 if let left = node.left, t >= 0 { 25 results += bfs(left, t) 26 } 27 node = node.right! 28 } 29 } 30 31 results += bfs(node, K) 32 33 return results 34 } 35 36 func bfs(_ root: TreeNode, _ targetLevel: Int) -> [Int] { 37 var queue = [root] 38 var level = 0 39 40 while !queue.isEmpty && level < targetLevel { 41 var nextQueue = [TreeNode]() 42 43 for node in queue { 44 if let n = node.left { 45 nextQueue.append(n) 46 } 47 48 if let n = node.right { 49 nextQueue.append(n) 50 } 51 } 52 53 queue = nextQueue 54 level += 1 55 } 56 57 if level == targetLevel { 58 return queue.map { $0.val } 59 } else { 60 return [] 61 } 62 } 63 64 func dfs(_ root: TreeNode, _ level: Int, target: TreeNode) -> (paths: [Bool], level: Int) { 65 if root.left === target { 66 return ([true], level + 1) 67 } else if root.right === target { 68 return ([false], level + 1) 69 } 70 71 if let left = root.left { 72 let result = dfs(left, level + 1, target: target) 73 if result.level > 0 { 74 return (result.paths + [true], result.level) 75 } 76 } 77 78 if let right = root.right { 79 let result = dfs(right, level + 1, target: target) 80 if result.level > 0 { 81 return (result.paths + [false], result.level) 82 } 83 } 84 85 return ([], -1) 86 } 87 }
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 func distanceK(_ root: TreeNode?, _ target: TreeNode?, _ K: Int) -> [Int] { 16 guard let root = root, let target = target else { return [] } 17 guard K != 0 else { return [target.val] } 18 19 var graph: [Int : [Int]] = [:] 20 toGraph(&graph, parent: nil, child: root) 21 var distances: [Int : Int] = [:] 22 distances[target.val] = 0 23 var queue: [Int] = [target.val] 24 var status: [Int : Bool] = [:] 25 var result: [Int] = [] 26 27 while !queue.isEmpty { 28 let current = queue.removeFirst() 29 status[current] = true 30 guard let connectedV = graph[current] else { break } 31 for v in connectedV { 32 guard (status[v] ?? false) == false else { continue } 33 queue.append(v) 34 let distanceV = distances[current]! + 1 35 distances[v] = distanceV 36 if distanceV == K { 37 result.append(v) 38 } 39 } 40 } 41 42 return result 43 } 44 45 func toGraph(_ result: inout [Int : [Int]], parent: TreeNode?, child: TreeNode) { 46 if let parent = parent { 47 if result[child.val] == nil { 48 result[child.val] = [parent.val] 49 }else{ 50 result[child.val]?.append(parent.val) 51 } 52 if result[parent.val] == nil { 53 result[parent.val] = [child.val] 54 }else{ 55 result[parent.val]?.append(child.val) 56 } 57 } 58 if let left = child.left { 59 toGraph(&result, parent: child, child: left) 60 } 61 if let right = child.right { 62 toGraph(&result, parent: child, child: right) 63 } 64 } 65 }
32ms
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 distanceK(_ root: TreeNode?, _ target: TreeNode?, _ K: Int) -> [Int] { 16 guard let root = root else { 17 return [] 18 } 19 20 guard K != 0 else { 21 return [target!.val] 22 } 23 24 let graph = Graph<Int>() 25 inorderTraversal(root) { 26 node in 27 let source = graph.createVertex(node.val) 28 if node.left != nil { 29 let destination = graph.createVertex(node.left!.val) 30 graph.addUndirectedEdge(between: source, to: destination) 31 } 32 33 if node.right != nil { 34 let destination = graph.createVertex(node.right!.val) 35 graph.addUndirectedEdge(between: source, to: destination) 36 } 37 } 38 return breathFirstTraversal(graph, Vertex(target!.val), K) 39 } 40 41 } 42 43 func breathFirstTraversal<T: Hashable>(_ graph: Graph<T>, 44 _ source: Vertex<T>, 45 _ k: Int 46 ) -> [T] { 47 var discovered: Set<Vertex<T>> = [source] 48 var stack: [Vertex<T>] = [source] 49 var k = k 50 while k > 0 { 51 var tmp: [Vertex<T>] = [] 52 while let source = stack.popLast() { 53 for edge in graph.edges(of: source) { 54 let destination = edge.destination 55 if !discovered.contains(destination) { 56 tmp.append(destination) 57 discovered.insert(destination) 58 } 59 } 60 } 61 k -= 1 62 if k == 0 { 63 return tmp.map { $0.val } 64 } 65 if tmp.isEmpty { break } 66 stack = tmp 67 } 68 69 return [] 70 } 71 72 func inorderTraversal(_ root: TreeNode?, _ visit: (TreeNode) -> Void) { 73 guard let root = root else { 74 return 75 } 76 inorderTraversal(root.left, visit) 77 visit(root) 78 inorderTraversal(root.right, visit) 79 } 80 81 struct Vertex<T: Hashable> : Hashable { 82 public var val: T 83 public init(_ val: T) { 84 self.val = val 85 } 86 87 public var hashValue: Int { 88 return val.hashValue 89 } 90 91 public static func ==(lhs: Vertex<T>, rhs: Vertex<T>) -> Bool { 92 return lhs.val == rhs.val 93 } 94 } 95 96 struct Edge<T: Hashable> { 97 public let source: Vertex<T> 98 public let destination: Vertex<T> 99 public let weight: Double? 100 101 public init(source: Vertex<T>, destination: Vertex<T>, weight: Double? = nil) { 102 self.source = source 103 self.destination = destination 104 self.weight = weight 105 } 106 } 107 108 class Graph<T: Hashable> { 109 public var vertice: [Vertex<T>] = [] 110 private var adjacencyList: [Vertex<T>: [Edge<T>]] = [:] 111 112 public func createVertex(_ val: T) -> Vertex<T> { 113 let source = Vertex(val) 114 if adjacencyList[source] == nil { 115 adjacencyList[source] = [] 116 vertice.append(source) 117 } 118 return source 119 } 120 121 public func addDirectedEdge(from source: Vertex<T>, 122 to destination: Vertex<T>, 123 weight: Double? = nil 124 ) { 125 let edge = Edge(source: source, destination: destination, weight: weight) 126 adjacencyList[source]?.append(edge) 127 } 128 129 public func addUndirectedEdge(between source: Vertex<T>, 130 to destination: Vertex<T>, 131 weight: Double? = nil 132 ) { 133 addDirectedEdge(from: source, to: destination, weight: weight) 134 addDirectedEdge(from: destination, to: source, weight: weight) 135 } 136 137 public func edges(of source: Vertex<T>) -> [Edge<T>] { 138 return adjacencyList[source] ?? [] 139 } 140 141 public func weight(from source: Vertex<T>, to destination: Vertex<T>) -> Double? { 142 return adjacencyList[source]?.first{ $0.destination == destination }?.weight 143 } 144 }
Runtime: 32 ms
Memory Usage: 19.2 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 var map:[(TreeNode,String)] = [(TreeNode,String)]() 16 var path:String = String() 17 func distanceK(_ root: TreeNode?, _ target: TreeNode?, _ K: Int) -> [Int] { 18 var list:[Int] = [Int]() 19 getNodeDist(root,target,"") 20 var i:Int = 0 21 for ele in map 22 { 23 var s:String = ele.1 24 var i:Int = 0 25 var arrS:[Character] = Array(s) 26 var arrP:[Character] = Array(path) 27 while(i<s.count && i<path.count && arrS[i] == arrP[i]) 28 { 29 i += 1 30 } 31 if s.count - i + path.count - i == K 32 { 33 list.append(ele.0.val) 34 } 35 } 36 return list 37 } 38 39 func getNodeDist(_ root: TreeNode?,_ target: TreeNode?,_ p:String) 40 { 41 if root != nil 42 { 43 path = root!.val == target!.val ? p : path 44 map.append((root!, p)) 45 getNodeDist(root?.left,target,p+"0") 46 getNodeDist(root?.right,target,p+"1") 47 } 48 } 49 }