Trie Node Design and Initialization in DSA Javascript - Build from Scratch

Trie Node Design and Initialization

📖 Scenario: Imagine you are building a simple search feature for a phone contacts app. To do this efficiently, you want to create a special tree called a Trie that helps find names quickly by their letters.

🎯 Goal: You will create the basic building block of this Trie called a TrieNode. This node will hold letters and links to other nodes. You will set it up so it can be used to build the full Trie later.

📋 What You'll Learn

Create a class called TrieNode

Inside the class, create a constructor method

In the constructor, create a property called children that is an empty object

In the constructor, create a property called isEndOfWord and set it to false

💡 Why This Matters

🌍 Real World

Tries are used in search engines, autocomplete features, and spell checkers to quickly find words or prefixes.

💼 Career

Understanding trie node design helps in roles involving data structures, algorithms, and building efficient search or text processing systems.

1

Create the TrieNode class

Create a class called TrieNode with an empty constructor method constructor().

DSA Javascript

class TrieNode {
  constructor() {
    // Your code here
  }
}

Hint

Use the class keyword to create a class named TrieNode. Inside it, write an empty constructor() method.

2

Add the children property

Inside the constructor() of TrieNode, create a property called children and set it to an empty object {}.

DSA Javascript

class TrieNode {
  constructor() {
    // Create children property as empty object
  }
}

Hint

Use this.children = {} inside the constructor to create an empty object for child nodes.

3

Add the isEndOfWord property

Inside the constructor() of TrieNode, create a property called isEndOfWord and set it to false.

DSA Javascript

class TrieNode {
  constructor() {
    this.children = {};
    // Create isEndOfWord property and set to false
  }
}

Hint

Use this.isEndOfWord = false inside the constructor to mark the node as not ending a word yet.

4

Create an instance and print its properties

Create a variable called node and set it to a new instance of TrieNode. Then print node.children and node.isEndOfWord on separate lines.

DSA Javascript

class TrieNode {
  constructor() {
    this.children = {};
    this.isEndOfWord = false;
  }
}

// Create node instance and print properties
// Your code here

Hint

Create const node = new TrieNode(); then use console.log(node.children); and console.log(node.isEndOfWord); to print.

Practice

1. What is the main purpose of a Trie node in a Trie data structure?

easy

A. To store the entire word in one node

B. To store integer values for sorting

C. To hold only the last character of a word

D. To store characters and links to child nodes representing prefixes

Solution

1. Step 1: Understand Trie node role

   A Trie node stores a character and links to child nodes that represent possible next characters in words.

2. Step 2: Compare options

   Only To store characters and links to child nodes representing prefixes correctly describes storing characters and child links for prefixes.

3. Final Answer:

   To store characters and links to child nodes representing prefixes -> Option D

4. Quick Check:

   Trie node = character + child links [OK]

Hint: Trie nodes hold characters and children, not full words [OK]

Common Mistakes:

• Thinking nodes store whole words
• Confusing Trie nodes with array elements
• Assuming nodes only store last characters

2. Which of the following is the correct way to initialize a Trie node in JavaScript with an empty children object and a boolean flag isEndOfWord set to false?

easy

A. const node = { children: null, isEndOfWord: true };

B. const node = { children: [], isEndOfWord: false };

C. const node = { children: {}, isEndOfWord: false };

D. const node = { children: {}, isEndOfWord: true };

Solution

1. Step 1: Identify correct children type

   Children should be an object to map characters to nodes, so {} is correct.

2. Step 2: Check isEndOfWord initial value

   Initially, isEndOfWord should be false because the node does not end a word yet.

3. Final Answer:

   const node = { children: {}, isEndOfWord: false }; -> Option C

4. Quick Check:

   Empty object + false flag = correct initialization [OK]

Hint: Children is an object, isEndOfWord starts false [OK]

Common Mistakes:

• Using array instead of object for children
• Setting isEndOfWord true at initialization
• Setting children to null

3. Consider this JavaScript code snippet for a Trie node:

class TrieNode {
  constructor() {
    this.children = {};
    this.isEndOfWord = false;
  }
}

const node = new TrieNode();
console.log(node.isEndOfWord);
console.log(Object.keys(node.children).length);

What will be the output?

medium

A. false
0

B. true
0

C. false
undefined

D. undefined
0

Solution

1. Step 1: Check isEndOfWord value after construction

   Constructor sets isEndOfWord to false, so console.log prints false.

2. Step 2: Check children keys count

   Children is an empty object, so Object.keys(node.children).length is 0.

3. Final Answer:

   false 0 -> Option A

4. Quick Check:

   isEndOfWord = false, children keys = 0 [OK]

Hint: Empty children means zero keys; isEndOfWord defaults false [OK]

Common Mistakes:

• Assuming isEndOfWord is true by default
• Expecting children keys length to be undefined
• Confusing object keys count with children length

4. The following code attempts to create a Trie node but has an error:

function TrieNode() {
  this.children = [];
  this.isEndOfWord = false;
}

What is the main problem with this code?

medium

A. children should be an object, not an array

B. isEndOfWord should be true initially

C. Function should return a value

D. Missing semicolon after isEndOfWord assignment

Solution

1. Step 1: Analyze children data type

   Children must be an object to map characters to nodes, not an array.

2. Step 2: Check other parts

   isEndOfWord false is correct; function constructors do not need explicit return; semicolons are optional in JS.

3. Final Answer:

   children should be an object, not an array -> Option A

4. Quick Check:

   Children = object, not array [OK]

Hint: Children must be object for character keys, not array [OK]

Common Mistakes:

• Thinking arrays can map characters
• Expecting function constructors to return explicitly
• Focusing on semicolon errors

5. You want to design a Trie node class in JavaScript that supports Unicode characters as keys in children and initializes isEndOfWord to false. Which of the following implementations correctly achieves this and allows adding child nodes dynamically?

hard

A. class TrieNode {
constructor() {
this.children = {};
this.isEndOfWord = true;
}
addChild(char) {
this.children[char] = new TrieNode();
}
}

B. class TrieNode {
constructor() {
this.children = new Map();
this.isEndOfWord = false;
}
addChild(char) {
if (!this.children.has(char)) {
this.children.set(char, new TrieNode());
}
}
}

C. class TrieNode {
constructor() {
this.children = [];
this.isEndOfWord = false;
}
addChild(char) {
this.children.push(char);
}
}

D. class TrieNode {
constructor() {
this.children = new Set();
this.isEndOfWord = false;
}
addChild(char) {
this.children.add(char);
}
}

Solution

1. Step 1: Check children data structure for Unicode support

   Using Map allows any string keys including Unicode, better than plain objects.

2. Step 2: Verify isEndOfWord initialization and addChild method

   isEndOfWord is false initially, and addChild adds new TrieNode only if missing, correctly updating children Map.

3. Step 3: Compare other options

   class TrieNode { constructor() { this.children = {}; this.isEndOfWord = true; } addChild(char) { this.children[char] = new TrieNode(); } } sets isEndOfWord true initially (wrong). class TrieNode { constructor() { this.children = []; this.isEndOfWord = false; } addChild(char) { this.children.push(char); } } uses array which can't map chars to nodes. class TrieNode { constructor() { this.children = new Set(); this.isEndOfWord = false; } addChild(char) { this.children.add(char); } } uses Set which stores chars but not nodes.

4. Final Answer:

   Implementation using Map with isEndOfWord false and addChild method -> Option B

5. Quick Check:

   Map + false flag + dynamic addChild = correct Trie node [OK]

Hint: Use Map for Unicode keys and false isEndOfWord [OK]

Common Mistakes:

• Using array or Set instead of Map or object
• Setting isEndOfWord true at start
• Not checking if child exists before adding