leetcode地址：從前序與中序遍歷序列構造二叉樹
給定兩個整數數組 preorder 和 inorder ，其中 preorder 是二叉樹的先序遍歷， inorder 是同一棵樹的中序遍歷，請構造二叉樹并返回其根節點。

示例 1:

輸入: preorder = [3,9,20,15,7], inorder = [9,3,15,20,7]
輸出: [3,9,20,null,null,15,7]
示例 2:

輸入: preorder = [-1], inorder = [-1]
輸出: [-1]

提示:

1 <= preorder.length <= 3000
inorder.length == preorder.length
-3000 <= preorder[i], inorder[i] <= 3000
preorder 和 inorder 均 無重復 元素
inorder 均出現在 preorder
preorder 保證 為二叉樹的前序遍歷序列
inorder 保證 為二叉樹的中序遍歷序列

實現思路

先序遍歷（Preorder）：根節點 -> 左子樹 -> 右子樹
中序遍歷（Inorder）：左子樹 -> 根節點 -> 右子樹
通過給定的先序遍歷和中序遍歷數組，我們可以確定二叉樹的根節點以及左右子樹的范圍。具體步驟如下：
步驟1：先序遍歷的第一個元素是根節點的值。
步驟2：在中序遍歷中找到根節點的位置，其左側為左子樹的中序遍歷，右側為右子樹的中序遍歷。
步驟3：根據步驟2中左右子樹的大小，可以在先序遍歷中確定左子樹和右子樹的先序遍歷。
遞歸地應用以上步驟，即可構造整棵二叉樹。

代碼實現

class TreeNode:def __init__(self, val=0, left=None, right=None):self.val = valself.left = leftself.right = rightdef buildTree(preorder, inorder):if not preorder or not inorder:return Noneroot_val = preorder[0]root = TreeNode(root_val)idx = inorder.index(root_val)root.left = buildTree(preorder[1:idx + 1], inorder[:idx])root.right = buildTree(preorder[idx + 1:], inorder[idx + 1:])return rootdef inorderTraversal(root):if not root:return []return inorderTraversal(root.left) + [root.val] + inorderTraversal(root.right)# Example
preorder = [3,9,20,15,7]
inorder = [9,3,15,20,7]root = buildTree(preorder, inorder)# Verify the constructed tree by printing its inorder traversal
print("Inorder traversal of constructed tree:", inorderTraversal(root))

go實現

package mainimport "fmt"type TreeNode struct {Val   intLeft  *TreeNodeRight *TreeNode
}func buildTree(preorder []int, inorder []int) *TreeNode {if len(preorder) == 0 || len(inorder) == 0 {return nil}rootVal := preorder[0]root := &TreeNode{Val: rootVal}var idx intfor i, v := range inorder {if v == rootVal {idx = ibreak}}root.Left = buildTree(preorder[1:idx+1], inorder[:idx])root.Right = buildTree(preorder[idx+1:], inorder[idx+1:])return root
}func inorderTraversal(root *TreeNode) []int {if root == nil {return []int{}}left := inorderTraversal(root.Left)right := inorderTraversal(root.Right)return append(append(left, root.Val), right...)
}func main() {// Examplepreorder := []int{3, 9, 20, 15, 7}inorder := []int{9, 3, 15, 20, 7}root := buildTree(preorder, inorder)// Verify the constructed tree by printing its inorder traversalfmt.Println("Inorder traversal of constructed tree:", inorderTraversal(root))
}

kotlin實現

class TreeNode(var `val`: Int) {var left: TreeNode? = nullvar right: TreeNode? = null
}fun buildTree(preorder: IntArray, inorder: IntArray): TreeNode? {if (preorder.isEmpty() || inorder.isEmpty()) {return null}val rootVal = preorder[0]val root = TreeNode(rootVal)val idx = inorder.indexOf(rootVal)root.left = buildTree(preorder.sliceArray(1..idx), inorder.sliceArray(0 until idx))root.right = buildTree(preorder.sliceArray(idx + 1 until preorder.size), inorder.sliceArray(idx + 1 until inorder.size))return root
}fun inorderTraversal(root: TreeNode?): List<Int> {val result = mutableListOf<Int>()fun inorder(node: TreeNode?) {if (node == null) returninorder(node.left)result.add(node.`val`)inorder(node.right)}inorder(root)return result
}fun main() {// Exampleval preorder = intArrayOf(3, 9, 20, 15, 7)val inorder = intArrayOf(9, 3, 15, 20, 7)val root = buildTree(preorder, inorder)// Verify the constructed tree by printing its inorder traversalprintln("Inorder traversal of constructed tree: ${inorderTraversal(root)}")
}