[Swift]LeetCode105. 从前序与中序遍历序列构造二叉树 | Construct Binary Tree from Preorder and Inorder Traversal
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➤微信公众号：山青咏芝（shanqingyongzhi）
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Given preorder and inorder traversal of a tree, construct the binary tree.
Note:
You may assume that duplicates do not exist in the tree.
For example, given
preorder = [3,9,20,15,7] inorder = [9,3,15,20,7]
Return the following binary tree:
3
/ \
9 20
/ \
15 7
根据一棵树的前序遍历与中序遍历构造二叉树。
注意:
你可以假设树中没有重复的元素。
例如，给出
前序遍历 preorder = [3,9,20,15,7] 中序遍历 inorder = [9,3,15,20,7]
返回如下的二叉树：
3
/ \
9 20
/ \
15 7
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 15 extension Array { 16 subscript(safe index: Int) -> Element? { 17 guard index >= 0 && index < self.count else { 18 return nil 19 } 20 21 return self[index] 22 } 23 } 24 25 class Solution { 26 func buildTree(_ preorder: [Int], _ inorder: [Int]) -> TreeNode? { 27 guard let rootValue = preorder[safe: 0] else { 28 return nil 29 } 30 31 var inorderMap: [Int: Int] = [:] 32 for index in 0..<inorder.count { 33 inorderMap[inorder[index]] = index 34 } 35 let root = TreeNode(rootValue) 36 37 buildNextNode(inorder: inorder, preorder: preorder, inorderMap: inorderMap, currentPreOrderIndex: 0, root: root, tooFarValue: nil) 38 39 return root 40 } 41 42 func buildNextNode(inorder: [Int], preorder: [Int], inorderMap: [Int: Int], currentPreOrderIndex: Int, root: TreeNode, tooFarValue: Int?) -> Int { 43 var currentPreOrderIndex = currentPreOrderIndex 44 guard let nextPreOrderValue = preorder[safe: currentPreOrderIndex + 1] else { 45 return currentPreOrderIndex 46 } 47 48 let childDirection = whichChildDirection(inorder: inorder, inorderMap: inorderMap, knownValue: root.val, potentialChildValue: nextPreOrderValue, tooFarValue: tooFarValue) 49 50 guard let direction = childDirection else { 51 // not a child 52 return currentPreOrderIndex 53 } 54 55 let childNode = TreeNode(nextPreOrderValue) 56 57 if direction { 58 // right child 59 root.right = childNode 60 return buildNextNode(inorder: inorder, preorder: preorder, inorderMap: inorderMap, currentPreOrderIndex: currentPreOrderIndex + 1, root: childNode, tooFarValue: tooFarValue) 61 62 } else { 63 // left child 64 root.left = childNode 65 currentPreOrderIndex = buildNextNode(inorder: inorder, preorder: preorder, inorderMap: inorderMap, currentPreOrderIndex: currentPreOrderIndex + 1, root: childNode, tooFarValue: root.val) 66 return buildNextNode(inorder: inorder, preorder: preorder, inorderMap: inorderMap, currentPreOrderIndex: currentPreOrderIndex, root: root, tooFarValue: tooFarValue) 67 } 68 69 return currentPreOrderIndex 70 } 71 72 /** 73 false: left child 74 true: right child 75 nil: neither child 76 */ 77 func whichChildDirection(inorder: [Int], inorderMap: [Int: Int], knownValue: Int, potentialChildValue: Int, tooFarValue: Int?) -> Bool? { 78 let potentialChildIndex = inorderMap[potentialChildValue]! // guarenteed to be in the map 79 let knownIndex = inorderMap[knownValue]! // guarenteed to be in the map 80 81 if potentialChildIndex < knownIndex { 82 return false 83 } 84 85 if let tooFarValue = tooFarValue, let tooFarIndex = inorderMap[tooFarValue] { 86 if tooFarIndex < potentialChildIndex { 87 return nil 88 } 89 } 90 91 return true 92 } 93 }
36ms
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 buildTree(_ preorder: [Int], _ inorder: [Int]) -> TreeNode? { 16 17 guard preorder.count > 0, inorder.count > 0 else { 18 return nil 19 } 20 21 var inorderDict = [Int: Int]() 22 for (i, item) in inorder.enumerated() { 23 inorderDict[item] = i 24 } 25 26 return create(inorderDict, preorder, 0, preorder.count - 1, inorder, 0 , inorder.count - 1) 27 28 } 29 30 func create(_ inorderDict: [Int: Int], _ preorder: [Int], _ preLow: Int, _ preHi: Int, _ inorder: [Int], _ inLow: Int, _ inHi: Int) -> TreeNode? { 31 32 if preLow > preHi { //debug 33 return nil 34 } 35 36 var root = TreeNode(preorder[preLow]) 37 let rootIndex = inorderDict[preorder[preLow]]! // debug: dict value is optional, must be unwrapped 38 let leftNum = rootIndex - inLow // debug 39 40 root.left = create(inorderDict, preorder, preLow + 1, preLow + leftNum, inorder, inLow, rootIndex - 1) 41 root.right = create(inorderDict, preorder, preLow + leftNum + 1, preHi, inorder, rootIndex + 1, inHi) 42 43 return root 44 } 45 }
44ms
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 buildTree(_ preorder: [Int], _ inorder: [Int]) -> TreeNode? { 16 if preorder.count == 0 { 17 return nil 18 } 19 20 var map = [Int: Int]() 21 for i in 0..<inorder.count { 22 map[inorder[i]] = i 23 } 24 25 var stack = [TreeNode]() 26 var value = preorder[0] 27 let root = TreeNode(value) 28 29 stack.append(root) 30 for i in 1..<preorder.count { 31 value = preorder[i] 32 let node = TreeNode(value) 33 if map[value]! < map[stack.last!.val]! { 34 stack.last!.left = node 35 } else { 36 var parent: TreeNode? = nil 37 while !stack.isEmpty && map[value]! > map[stack.last!.val]! { 38 parent = stack.removeLast() 39 } 40 parent?.right = node 41 } 42 stack.append(node) 43 } 44 45 return root 46 } 47 }
56ms
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 buildTree(_ preorder: [Int], _ inorder: [Int]) -> TreeNode? { 16 17 guard !preorder.isEmpty && !inorder.isEmpty else { 18 return nil 19 } 20 21 return buildTree(preorder, 0, preorder.count-1, inorder, 0, inorder.count-1) 22 } 23 24 func buildTree(_ preorder: [Int], _ preStart: Int, _ preEnd: Int, _ inorder: [Int], _ inStart: Int, _ inEnd: Int) -> TreeNode? { 25 26 guard preStart <= preEnd && inStart <= inEnd else { 27 return nil 28 } 29 30 var rootVal = preorder[preStart] 31 var root = TreeNode(rootVal) 32 var mid: Int = 0 33 34 for i in inStart...inEnd { 35 if inorder[i] == rootVal { 36 mid = i 37 break 38 } 39 } 40 41 root.left = buildTree(preorder, preStart+1, preStart + mid - inStart, inorder, inStart, mid - 1) 42 root.right = buildTree(preorder, preStart + mid - inStart + 1, preEnd, inorder, mid+1, inEnd) 43 return root 44 } 45 }
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 buildTree(_ preorder: [Int], _ inorder: [Int]) -> TreeNode? { 16 let len = preorder.count 17 if len == 0 { return nil } 18 let root = TreeNode(preorder[0]) 19 20 var i = 0 21 while i < len { 22 if inorder[i] == root.val { 23 break 24 } 25 i += 1 26 } 27 28 if i > 0 { 29 root.left = buildTree(Array(preorder[1...i]), Array(inorder[..<i])) 30 } 31 if i < len-1 { 32 root.right = buildTree(Array(preorder[(i+1)...]), Array(inorder[(i+1)...])) 33 } 34 35 return root 36 } 37 }
240ms
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 // 根据中序和前序遍历构建二叉树 16 func buildTree(_ inorder: [Int], _ preorder: [Int]) -> TreeNode? { 17 18 if inorder.count <= 0 || preorder.count <= 0{ 19 return nil 20 } 21 /* 22 * 后序遍历的最后一个节点肯定是整个树的根节点 23 */ 24 return buildTrees(preorder, inorder) 25 } 26 27 func buildTrees(_ inorder: [Int], _ preorder: [Int]) -> TreeNode{ 28 29 // 根据前序遍历的第一个节点作为根节点 30 let root = TreeNode.init(preorder[0]) 31 32 // 根据中序遍历计算根节点左右子节点的个数 33 var indexRoot: Int = -1 34 35 for index in 0..<inorder.count{ 36 if inorder[index] == preorder[0]{ 37 indexRoot = index 38 break 39 } 40 } 41 42 // 左子节点的个数 43 let leftNum = indexRoot 44 // 右子节点的个数 45 let rightNum = inorder.count - indexRoot - 1 46 47 // 生成新的左右子树 48 if indexRoot > 0{ 49 let leftNode = buildTrees([Int](inorder[0...indexRoot - 1]), [Int](preorder[1...leftNum])) 50 root.left = leftNode 51 } 52 if indexRoot < inorder.count - 1{ 53 let rightNode = buildTrees([Int](inorder[indexRoot + 1...inorder.count - 1]), [Int](preorder[leftNum + 1...leftNum + rightNum])) 54 root.right = rightNode 55 } 56 return root 57 } 58 }
288ms
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 buildTree(_ preorder: [Int], _ inorder: [Int]) -> TreeNode? { 16 17 if preorder.count == 0 { 18 return nil 19 } 20 21 if preorder.count == 1 { 22 let val = preorder[0] 23 let root = TreeNode(val) 24 return root 25 } 26 27 let val = preorder[0] 28 let root = TreeNode(val) 29 30 var rootIndex = 0 31 for i in 0..<inorder.count { 32 if inorder[i] == val { 33 rootIndex = i 34 } 35 } 36 37 var tmpPreorder = [Int]() 38 if rootIndex > 0 { 39 for i in 1...rootIndex { 40 tmpPreorder.append(preorder[i]) 41 } 42 } 43 44 var tmpInorder = [Int]() 45 for i in 0..<rootIndex { 46 tmpInorder.append(inorder[i]) 47 } 48 49 root.left = buildTree(tmpPreorder, tmpInorder) 50 51 tmpPreorder = [Int]() 52 for i in rootIndex+1..<preorder.count { 53 tmpPreorder.append(preorder[i]) 54 } 55 56 tmpInorder = [Int]() 57 for i in rootIndex+1..<inorder.count { 58 tmpInorder.append(inorder[i]) 59 } 60 61 root.right = buildTree(tmpPreorder, tmpInorder) 62 63 return root 64 } 65 }
328ms
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 buildTree(_ preorder: [Int], _ inorder: [Int]) -> TreeNode? { 16 guard preorder.count != 0, inorder.count != 0, preorder.count == inorder.count else { return nil } 17 let root = TreeNode(preorder[0]) 18 let rootInOrderIndex = inorder.index(of: root.val) ?? -1 19 let leftInOrder = Array(inorder[0..<rootInOrderIndex]) 20 let leftPreOrderStartIndex = 1 21 let leftPreOrderEndIndexExclusive = leftPreOrderStartIndex + leftInOrder.count 22 let leftPreOrder = Array(preorder[leftPreOrderStartIndex..<leftPreOrderEndIndexExclusive]) 23 root.left = buildTree(leftPreOrder, leftInOrder) 24 25 let rightInOrder = Array(inorder[rootInOrderIndex+1..<inorder.count]) 26 let rightPreOrder = Array(preorder[leftPreOrderEndIndexExclusive..<preorder.count]) 27 root.right = buildTree(rightPreOrder, rightInOrder) 28 return root 29 } 30 }
332ms
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 buildTree(_ preorder: [Int], _ inorder: [Int]) -> TreeNode? { 16 var inorder = inorder 17 var preorder = preorder 18 var root: TreeNode 19 if preorder.count < 1{ 20 return nil 21 } 22 let first = preorder.removeFirst() 23 root = TreeNode(first) 24 let index = inorder.index(of: first)! 25 let lin = Array(inorder[0..<index]) 26 let lpro = Array(preorder[0..<index]) 27 root.left = buildTree(lpro, lin) 28 let rin = Array(inorder[(index+1)...]) 29 let rpro = Array(preorder[index...]) 30 root.right = buildTree(rpro, rin) 31 return root 32 } 33 }
