Print elements of linkedlist - Golang

 This is 3rd post on linked list. For previous posts, pls check the links below:

https://sreesindhusruthi.blogspot.com/2021/07/create-single-linked-list-in-golang.html

https://sreesindhusruthi.blogspot.com/2021/07/length-of-linkedlist-implementation-in.html

Writing print function for linked list to print elements.

The function name is Print and elements data is available in Node.data .

Step1: Check if linked list is empty. If empty print "NULL"

Step2: Print Node.data and update head to next Node in the linkedlist.

The function code is :

func (ll *LinkedList) Print() {
   if ll.head == nil{
      fmt.Print("NULL")
      return
   }
   //No Intermediate variable is used so we are traversing on
   //the actual object created and it's references gets updated on updating ll.head.
   for ll.head != nil {
      fmt.Printf("%+v -->",ll.head.data)
      ll.head = ll.head.next
      ll.Print()
   }
}

Complete code:

	package main
	import "fmt"
	type Node struct{
	data int
	next *Node
	}
	func (n *Node) NewNode(data int, next *Node){
	n.data = data
	n.next = next
	}
	type LinkedList struct{
	head *Node
	}
	func (ll *LinkedList) Len() int{
	count := 0
	var temp *Node
	if ll.head == nil{
	return 0
	}
	temp = ll.head
	//fmt.Println(temp)
	for temp != nil {
	count ++
	temp = temp.next
	//fmt.Println(temp)
	}
	return count
	}
	func (ll *LinkedList) Print() {
	if ll.head == nil{
	fmt.Print("NULL")
	return
	}
	for ll.head != nil {
	fmt.Printf("%+v -->",ll.head.data)
	ll.head = ll.head.next
	ll.Print()
	}
	}
	func main(){
	llist := new(LinkedList)
	node1 := new(Node)
	node1.NewNode(3, nil)
	llist.head = node1
	//fmt.Println(llist.Len())
	node2 := new(Node)
	node2.NewNode(7, nil)
	llist.head.next = node2
	//fmt.Println(llist.Len())
	node3 := new(Node)
	node3.NewNode(10, nil)
	llist.head.next.next = node3
	llist.Print()
	}
	/*
	output:
	3 -->7 -->10 -->NULL
	*/

view raw printllrecursive.go hosted with ❤ by GitHub