聊聊golang怎么用递归实现反转链表

type Listnode struct {
    Val int
    Next *ListNode
}
func reverseList(head *ListNode) *ListNode {
    if head == nil || head.Next == nil {
        return head
    }
    newHead := reverseList(head.Next)
    head.Next.Next = head
    head.Next = nil
    return newHead
}

// 测试代码
func main() {
    node1 := ListNode{Val: 1}
    node2 := ListNode{Val: 2}
    node3 := ListNode{Val: 3}
    node4 := ListNode{Val: 4}
    node5 := ListNode{Val: 5}
    
    node1.Next = &node2
    node2.Next = &node3
    node3.Next = &node4
    node4.Next = &node5
    
    fmt.Println("原链表：")
    printList(&node1)
    newHead := reverseList(&node1)
    fmt.Println("反转后的链表：")
    printList(newHead)
}
func printList(head *ListNode) {
    for p := head; p != nil; p = p.Next {
        fmt.Printf("%d ",p.Val)
    }
    fmt.Println()
}

原链表：
1 2 3 4 5 
反转后的链表：
5 4 3 2 1