leetcode--从中序与后序遍历序列构造二叉树

leeocode地址：从中序与后序遍历序列构造二叉树
给定两个整数数组 inorder 和 postorder ，其中 inorder 是二叉树的中序遍历， postorder 是同一棵树的后序遍历，请你构造并返回这颗 二叉树 。

示例 1:

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

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

提示:
1 <= inorder.length <= 3000
postorder.length == inorder.length
-3000 <= inorder[i], postorder[i] <= 3000
inorder 和 postorder 都由 不同 的值组成
postorder 中每一个值都在 inorder 中
inorder 保证是树的中序遍历
postorder 保证是树的后序遍历

实现思路

中序遍历（Inorder）：左子树 -> 根节点 -> 右子树
后序遍历（Postorder）：左子树 -> 右子树 -> 根节点
通过给定的中序遍历和后序遍历数组，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：后序遍历的最后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，可以在后序遍历中确定左子树和右子树的后序遍历。
递归地应用以上步骤，即可构造整棵二叉树。

代码实现

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

go实现

package mainimport "fmt"type TreeNode struct {Val   intLeft  *TreeNodeRight *TreeNode
}func buildTree(inorder []int, postorder []int) *TreeNode {if len(inorder) == 0 || len(postorder) == 0 {return nil}rootVal := postorder[len(postorder)-1]root := &TreeNode{Val: rootVal}idx := indexOf(inorder, rootVal)root.Left = buildTree(inorder[:idx], postorder[:idx])root.Right = buildTree(inorder[idx+1:], postorder[idx:len(postorder)-1])return root
}func indexOf(arr []int, val int) int {for i := range arr {if arr[i] == val {return i}}return -1
}func inorderTraversal(root *TreeNode) []int {var result []intvar inorder func(node *TreeNode)inorder = func(node *TreeNode) {if node == nil {return}inorder(node.Left)result = append(result, node.Val)inorder(node.Right)}inorder(root)return result
}func main() {// Exampleinorder := []int{9, 3, 15, 20, 7}postorder := []int{9, 15, 7, 20, 3}root := buildTree(inorder, postorder)// Verify the constructed tree by printing its inorder traversalfmt.Println("Inorder traversal of constructed tree:", inorderTraversal(root))
}

kotlin实现

class TreeNode(var `val`: Int) {var left: TreeNode? = nullvar right: TreeNode? = null
}fun buildTree(inorder: IntArray, postorder: IntArray): TreeNode? {if (inorder.isEmpty() || postorder.isEmpty()) {return null}val rootVal = postorder.last()val root = TreeNode(rootVal)val idx = inorder.indexOf(rootVal)root.left = buildTree(inorder.sliceArray(0 until idx), postorder.sliceArray(0 until idx))root.right = buildTree(inorder.sliceArray(idx + 1 until inorder.size), postorder.sliceArray(idx until postorder.size - 1))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 inorder = intArrayOf(9, 3, 15, 20, 7)val postorder = intArrayOf(9, 15, 7, 20, 3)val root = buildTree(inorder, postorder)// Verify the constructed tree by printing its inorder traversalprintln("Inorder traversal of constructed tree: ${inorderTraversal(root)}")
}