leetcode地址：从前序与中序遍历序列构造二叉树
给定两个整数数组 preorder 和 inorder ，其中 preorder 是二叉树的先序遍历， inorder 是同一棵树的中序遍历，请构造二叉树并返回其根节点。

示例 1:

输入: preorder = [3,9,20,15,7], inorder = [9,3,15,20,7]
输出: [3,9,20,null,null,15,7]
示例 2:

输入: preorder = [-1], inorder = [-1]
输出: [-1]

提示:

1 <= preorder.length <= 3000
inorder.length == preorder.length
-3000 <= preorder[i], inorder[i] <= 3000
preorder 和 inorder 均 无重复 元素
inorder 均出现在 preorder
preorder 保证 为二叉树的前序遍历序列
inorder 保证 为二叉树的中序遍历序列

实现思路

先序遍历（Preorder）：根节点 -> 左子树 -> 右子树
中序遍历（Inorder）：左子树 -> 根节点 -> 右子树
通过给定的先序遍历和中序遍历数组，我们可以确定二叉树的根节点以及左右子树的范围。具体步骤如下：
步骤1：先序遍历的第一个元素是根节点的值。
步骤2：在中序遍历中找到根节点的位置，其左侧为左子树的中序遍历，右侧为右子树的中序遍历。
步骤3：根据步骤2中左右子树的大小，可以在先序遍历中确定左子树和右子树的先序遍历。
递归地应用以上步骤，即可构造整棵二叉树。

代码实现

class TreeNode:
    def __init__(self, val=0, left=None, right=None):
        self.val = val
        self.left = left
        self.right = right

def buildTree(preorder, inorder):
    if not preorder or not inorder:
        return None
    
    root_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 root

def 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实现

package main

import "fmt"

type TreeNode struct {
	Val   int
	Left  *TreeNode
	Right *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 int
	for i, v := range inorder {
		if v == rootVal {
			idx = i
			break
		}
	}

	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() {
	// Example
	preorder := []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 traversal
	fmt.Println("Inorder traversal of constructed tree:", inorderTraversal(root))
}

kotlin实现

class TreeNode(var `val`: Int) {
    var left: TreeNode? = null
    var 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) return
        inorder(node.left)
        result.add(node.`val`)
        inorder(node.right)
    }
    inorder(root)
    return result
}

fun main() {
    // Example
    val 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 traversal
    println("Inorder traversal of constructed tree: ${inorderTraversal(root)}")
}