[Swift]LeetCode106. 從中序與後序遍歷序列構造二叉樹 | Construct Binary Tree from Inorder and Postorder Traversal
阿新 • • 發佈:2018-11-13
Given inorder and postorder traversal of a tree, construct the binary tree.
Note:
You may assume that duplicates do not exist in the tree.
For example, given
inorder = [9,3,15,20,7] postorder = [9,15,7,20,3]
Return the following binary tree:
3 / \ 9 20 / \ 15 7
根據一棵樹的中序遍歷與後序遍歷構造二叉樹。
注意:
你可以假設樹中沒有重複的元素。
例如,給出
中序遍歷 inorder = [9,3,15,20,7] 後序遍歷 postorder = [9,15,7,20,3]
返回如下的二叉樹:
3 / \ 9 20 / \ 15 7
108ms
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(_ inorder: [Int], _ postorder: [Int]) -> TreeNode? { 16 let instart = 017 let inend = inorder.count - 1 18 let poststart = 0 19 let postend = postorder.count - 1 20 return buildTree(inorder, instart, inend, postorder, poststart, postend) 21 } 22 private func buildTree(_ inorder: [Int], _ instart: Int, _ inend: Int, _ postorder: [Int], _ poststart: Int, _ postend: Int )-> TreeNode? { 23 if instart > inend || poststart > postend { 24 return nil 25 } 26 let rootValue = postorder[postend] 27 let root = TreeNode(rootValue) 28 var k = 0 29 for i in 0..<inorder.count { 30 if inorder[i] == rootValue { 31 k = i 32 break 33 } 34 } 35 root.left = buildTree(inorder, instart, k - 1, postorder, poststart, poststart + k - 1 - instart) 36 root.right = buildTree(inorder, k + 1, inend, postorder, poststart + k - instart, postend - 1) 37 return root 38 } 39 }
116ms
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(_ inorder: [Int], _ postorder: [Int]) -> TreeNode? { 16 guard inorder.count > 0 && postorder.count > 0 && inorder.count == postorder.count else { 17 return nil 18 } 19 20 return _buildHelper(inorder, 0, inorder.count - 1, postorder, 0, postorder.count - 1) 21 } 22 23 func _buildHelper(_ inorder: [Int], _ inStart: Int, _ inEnd: Int, _ postorder: [Int], _ postStart: Int, _ postEnd: Int) -> TreeNode? { 24 guard inStart <= inEnd && postStart <= postEnd else { 25 return nil 26 } 27 28 let root = TreeNode(postorder[postEnd]) 29 30 var mid = 0 31 for i in inStart...inEnd where inorder[i] == root.val { 32 mid = i 33 break 34 } 35 36 root.left = _buildHelper(inorder, inStart, mid - 1, postorder, postStart, mid - 1 - inStart + postStart) 37 root.right = _buildHelper(inorder, mid + 1, inEnd, postorder, mid - inStart + postStart, postEnd - 1) 38 39 return root 40 } 41 }
156ms
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 // 根據中序和後序遍歷構建二叉樹 16 func buildTree(_ inorder: [Int], _ postorder: [Int]) -> TreeNode? { 17 if inorder.count <= 0 || postorder.count <= 0{ 18 return nil 19 } 20 /* 21 * 後序遍歷的最後一個節點肯定是整個樹的根節點 22 */ 23 return buildTrees(inorder, postorder) 24 } 25 26 func buildTrees(_ inorder: [Int], _ postorder: [Int]) -> TreeNode{ 27 28 let postorderRight = postorder.count - 1 29 // 根據後續遍歷的最後一個節點作為根節點 30 let root = TreeNode.init(postorder[postorderRight]) 31 32 // 根據中序遍歷計算根節點左右子節點的個數 33 var indexRoot: Int = -1 34 35 for index in 0..<inorder.count{ 36 if inorder[index] == postorder[postorderRight]{ 37 indexRoot = index 38 break 39 } 40 } 41 42 // 左子節點的個數 43 let leftNum = indexRoot 44 // 右子節點的個數 45 let rightNum = postorderRight - indexRoot 46 47 // 生成新的左右子樹 48 if indexRoot > 0{ 49 let leftNode = buildTrees([Int](inorder[0...indexRoot - 1]), [Int](postorder[0...leftNum - 1])) 50 root.left = leftNode 51 } 52 if indexRoot < inorder.count - 1{ 53 let rightNode = buildTrees([Int](inorder[indexRoot + 1...inorder.count - 1]), [Int](postorder[leftNum...leftNum + rightNum - 1])) 54 root.right = rightNode 55 } 56 57 return root 58 } 59 }
156ms
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(_ inorder: [Int], _ postorder: [Int]) -> TreeNode? { 16 let len = postorder.count 17 if len == 0 { return nil } 18 let root = TreeNode(postorder.last!) 19 var i = 0 20 while i < len { 21 if inorder[i] == root.val { 22 break 23 } 24 i += 1 25 } 26 27 if i > 0 { 28 root.left = buildTree(Array(inorder[..<i]), Array(postorder[..<i])) 29 } 30 31 if i < len-1 { 32 root.right = buildTree(Array(inorder[(i+1)...]), Array(postorder[i...(len-2)])) 33 } 34 35 return root 36 } 37 }
248ms
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(_ inorder: [Int], _ postorder: [Int]) -> TreeNode? { 16 17 if postorder.count == 0 { 18 return nil 19 } 20 21 if postorder.count == 1 { 22 let val = postorder[postorder.count - 1] 23 let root = TreeNode(val) 24 return root 25 } 26 27 let val = postorder[postorder.count - 1] 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 tmpInorder = [Int]() 38 for i in 0..<rootIndex { 39 tmpInorder.append(inorder[i]) 40 } 41 42 43 var tmpPostorder = [Int]() 44 for i in 0..<rootIndex { 45 tmpPostorder.append(postorder[i]) 46 } 47 48 root.left = buildTree(tmpInorder, tmpPostorder) 49 50 tmpInorder = [Int]() 51 for i in rootIndex+1..<inorder.count { 52 tmpInorder.append(inorder[i]) 53 } 54 55 tmpPostorder = [Int]() 56 for i in rootIndex..<inorder.count - 1 { 57 tmpPostorder.append(postorder[i]) 58 } 59 60 root.right = buildTree(tmpInorder, tmpPostorder) 61 62 return root 63 } 64 }
316ms
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(_ inorder: [Int], _ postorder: [Int]) -> TreeNode? { 16 guard inorder.count > 0 && postorder.count > 0 else { return nil } 17 let node = TreeNode(postorder.last!) 18 let index = inorder.firstIndex(of: node.val)! 19 20 let inLeft = Array(inorder[0 ..< index]) 21 let inRight = Array(inorder[index + 1 ..< inorder.count]) 22 23 let postLeft = Array(postorder[0 ..< index]) 24 let postRight = Array(postorder[index ..< postorder.count - 1]) 25 26 node.left = buildTree(inLeft, postLeft) 27 node.right = buildTree(inRight, postRight) 28 29 return node 30 } 31 }
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(_ inorder: [Int], _ postorder: [Int]) -> TreeNode? { 16 if inorder.count == 0 || postorder.count == 0 { 17 return nil 18 } 19 20 let root = TreeNode(postorder[postorder.count - 1]) 21 22 guard let i = inorder.index(of: root.val) else { return nil } 23 if i <= 1 && i > 0 { 24 root.left = TreeNode(inorder[0]) 25 } else if i > 1 { 26 root.left = buildTree(Array(inorder[0..<i]), Array(postorder[0..<i])) 27 } 28 29 if i >= inorder.count - 2 && i < inorder.count - 1 { 30 root.right = TreeNode(inorder[inorder.count - 1]) 31 } else if i <= inorder.count - 1 { 32 root.right = buildTree(Array(inorder[i + 1..<inorder.count]), Array(postorder[i..<postorder.count - 1])) 33 } 34 35 return root 36 } 37 }
392ms
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(_ inorder: [Int], _ postorder: [Int]) -> TreeNode? { 16 guard !inorder.isEmpty && !postorder.isEmpty && inorder.count == postorder.count else { 17 return nil 18 } 19 20 let root = TreeNode(postorder.last!) 21 let index = inorder.index(of: postorder.last!)! 22 root.left = buildTree([Int](inorder[..<index]), [Int](postorder[..<index])) 23 root.right = buildTree([Int](inorder[(index+1)...]), [Int](postorder[index..<(postorder.count-1)])) 24 25 return root 26 } 27 }