Question 196830 in Javascript Question posted in
A very good W3school tutorial can be found here.

LinkedList push() method in JavaScript with a this.tail , unable to understand the reference by value

class Node {
  constructor(val) {
    this.val = val;
    this.next = null;
  }
}

class SLL {
  constructor() {
    this.head = null;
    this.tail = null;
    this.length = 0;
  }

  push(val) {
    let newNode = new Node(val)
    if (!this.head) {
      this.head = newNode;
      this.tail = this.head;
    } else {
      console.log(this.tail);
      console.log(this.tail.next);
      debugger
      this.tail.next = newNode;
      this.tail = newNode;
    }
    this.length++;
  }
}

let x = new SLL();
x.push('Hi');
x.push('Hello');
x.push('How');
x.push('Are');
x.push('You');
x.push('Man');

Doubt:
After the 2nd push(‘Hello’), the tail is assigned with the newNode "this.tail = newNode" under else part. But in the 3rd push "push(‘How’)" how the this.tail is still referencing the this.head.

I have done console.log of both "this.tail" and "this.tail.next".The output shows fresh node. Am not able to understand how this still reference to the this.head.

enter image description here

enter image description here

Thanks for the answers!

Tags:

2

Answers


  1. 0 
    class Node {
  constructor() {
    this.value = null;
    this.next = null;
  }
}

class SinglyLinkedList {
  constructor() {
    this.root = null;
  }

  insert(value) {
    // create object of node class
    let newNode = new Node();

    if (!this.root) {
      newNode.value = value;
      this.root = newNode;
      return;
    }

    let current = this.root;

    // traversing to get the node whose next node pointer is null
    // this is to get one previous node to get access of next node whoose next pointer address is null
    while (current.next != null) {
      current = current.next;
    }
    //assign value to newNode
    newNode.value = value;
    //attach newNode to linked list
    current.next = newNode;
  }

  //   To print all nodes
  printAllNodes() {
    let current = this.root;
    while (current != null) {
      console.log(current.value);
      current = current.next;
    }
  }
}

let list = new SinglyLinkedList();

list.insert(5);
list.insert(10);
list.insert(15);
list.insert(20);

list.printAllNodes();

// console.log(list.root);
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  2. 0

    Linked list implementation with Tail address pointer

    class Node {
  constructor() {
    this.value = null;
    this.next = null;
  }
}

class SinglyLinkedList {
  constructor() {
    this.root = null;
    this.tail;
  }

  insert(value) {
    // create object of node class
    let newNode = new Node();

    if (!this.root) {
      newNode.value = value;
      this.root = newNode;
      this.tail = this.root;
      return;
    }

    let current = this.tail;
    //assign value to newNode
    newNode.value = value;

    current.next = newNode;
    //update the tail to newly inserted Node address
    this.tail = current.next;
  }

  //   To print all nodes
  printAllNodes() {
    let current = this.root;
    while (current != null) {
      console.log(current.value);
      current = current.next;
    }
  }
}

let list = new SinglyLinkedList();

list.insert(5);
list.insert(10);
list.insert(15);
list.insert(20);
list.insert(25);

list.printAllNodes();

// console.log(list.root);
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