jordanrios94

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jordanrios94

function bubbleSort(arr) {
  for (let i = 0; i < arr.length; i++) {
    for (let j = 0; j < (arr.length - i - 1); j++) {
      if (arr[j] > arr[j+1]) {
        const lesser = arr[j+1];
        arr[j+1] = arr[j];
        arr[j] = lesser;
      }
    }
  }

jordanrios94

/*
Given a node of a binary search tree, validate the binary search tree.
  - Ensure that every node's left hand child is less than the parent node's value
  - Ensure that every node's right hand child is greater than the parent
*/

function validate(node, min = null, max = null) {
  if (max !== null && node.data > max) {
    return false;
  }

jordanrios94

/*
- Implement the Node class to create a binary search tree. The constructor should initialize values 'data', 'left', and 'right'.

- Implement the 'insert' method for the Node class. Insert should accept an argument 'data', then create an insert a new node at the appropriate location in the tree.

- Implement the 'contains' method for the Node class. Contains should accept a 'data' argument and return the Node in the tree with the same value.
*/

class Node {
  constructor(data) {

jordanrios94

/*
1) Create a node class.
  - The constructor should accept an argument that gets assigned to the data property and initialise an empty arroy for storing children.
  - The node class should have methods 'add' and 'remove'.

2) Create a tree class. The tree constructor should initialise a 'root' property to null.

3) Implement 'taverseBFS' and 'traverseDFS' on the tree class.
*/

jordanrios94

/*
Given a linked list and integer n, return the element n spaces from the last node in the list. 
  - Do not call the 'size' method of the linked list
  - Assume that n will always be less than the length of the list.
*/

function fromLast(list, n) {
    let slow = list.getFirst();
    let fast = list.getAt(n);
    

jordanrios94

/*
Given a linked list, return true if the list is circular, false if it is not.
*/

function circular(list) {
    let slow = list.getFirst();
    let fast = list.getFirst();
    
    for (let item of list) {
        if (slow.next && fast.next) {

jordanrios94

/*
Return the 'middle' node of a linked list.
  - If the list has an even number of elements return the node at the end of the first half of the list.
  - Do not use a counter variable
  - Do not retrieve the size of the list
  - Only iterate through the list one time
*/

function midpoint(list) {
  let slow = list.getFirst();

jordanrios94

class Tree {
	constructor(value = null, children = []) {
		this.value = value;
		this.children = children;
	}
	
	*printValues() {
		yield this.value;
		
		for (let child of this.children) {

jordanrios94

class LinkedList {
	constructor() {
		this.head = null;
	}
	
	insertFirst(data) {
		this.head = new Node(data, this.head);
	}
	
	size() {