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class Node {
    constructor(value){
        this.value = value
        this.left = null
        this.right = null
    
    }

}

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

class SinglyLinkedList{
    constructor(){
        this.head = null;

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// an undirected graph
class Graph{
  constructor(){
      // Graph has an adjacencyList attribute set as an object
      this.adjacencyList = {}
  
  }

 addVertex(vertex){
   // to avoid overwriting the existing vertex, we need if statement

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function ListNode(val, next) {
     this.val = (val===undefined ? 0 : val)
     this.next = (next===undefined ? null : next)
 }

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class Node {
  constructor(value){
      this.value = value
      this.left = null
      this.right = null
  }
}

class BinarySeachTree {
      constructor(){

GAierken

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

}

class SinglyLinkedList{
    constructor(){

GAierken

// Problem link: https://leetcode.com/problems/valid-parentheses/

const isValid = (string) => {
    // create a hashMap to store the valid match   
    const bracketsMap = {
            '(': ')',
            '{': '}',
            '[': ']'
          }
    // create a stack data structure with array

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inOrderDFS(){
    // a variable to store the visited nodes
    let result = []
    // helper function -- accepts a node
    function traverse(node){
        // if node has left, recursion to find the leftest leaf node
     if(node.left) traverse(node.left)
        // push the node to the result
        result.push(node)
        // if node has right, recurstion to find the rightest leaf node

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const reverseList = (head, previous = null) => {

    //base case
    if(!head) return previous
    //variable to store next node
    let temp = head.next
    //reverse the node
        head.next = previous
    //recursion starts 
    return reverseList(temp, head)

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const reverseList = (head) => {
        //initiate previous as null
        let previous = null
        //initiate next without assigning
        let next 
       //keep iteration while head is not null
    while(head){
        //store the current node's next node    
        next = head.next
        //reassign head's next with previous node