Trie Insert Operation in DSA Javascript

Mental Model

A trie stores words by sharing common prefixes in a tree structure, adding new letters as branches.

Analogy: Imagine a tree of roads where each letter is a street sign; inserting a word means building new streets only where needed.

root
 ↓
 {}  (empty root node with no children)

Dry Run Walkthrough

Input: Insert words: 'cat', then 'car'

Goal: Build a trie that stores both words sharing the prefix 'ca'

Step 1: Insert 'c' from 'cat' at root

root
 ↓
 {c: {}}

Why: Start the branch for the first letter 'c'

Step 2: Insert 'a' after 'c'

root
 ↓
 {c: {a: {}}}

Why: Add next letter 'a' as child of 'c'

Step 3: Insert 't' after 'a' and mark end of word

root
 ↓
 {c: {a: {t: {end: true}}}}

Why: Complete word 'cat' by adding 't' and marking word end

Step 4: Insert 'c' from 'car' (already exists)

root
 ↓
 {c: {a: {t: {end: true}}}}

Why: Reuse existing 'c' node for new word

Step 5: Insert 'a' after 'c' (already exists)

root
 ↓
 {c: {a: {t: {end: true}}}}

Why: Reuse existing 'a' node for new word

Step 6: Insert 'r' after 'a' and mark end of word

root
 ↓
 {c: {a: {t: {end: true}, r: {end: true}}}}

Why: Add new branch 'r' to complete 'car'

Result:

root
 ↓
 {c: {a: {t: {end: true}, r: {end: true}}}}

Annotated Code

DSA Javascript

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

class Trie {
  constructor() {
    this.root = new TrieNode();
  }

  insert(word) {
    let node = this.root;
    for (const char of word) {
      if (!node.children[char]) {
        node.children[char] = new TrieNode();
      }
      node = node.children[char]; // advance to child node
    }
    node.end = true; // mark end of word
  }

  print(node = this.root, prefix = '') {
    if (node.end) {
      console.log(prefix);
    }
    for (const char in node.children) {
      this.print(node.children[char], prefix + char);
    }
  }
}

const trie = new Trie();
trie.insert('cat');
trie.insert('car');
trie.print();

for (const char of word) {

iterate over each letter in the word

if (!node.children[char]) {

check if current letter branch exists

node.children[char] = new TrieNode();

create new node if branch missing

node = node.children[char]; // advance to child node

move down to the child node for next letter

node.end = true; // mark end of word

mark current node as end of a valid word

Output

cat

car

Complexity Analysis

Time: O(m) because we process each of the m letters in the word once

Space: O(m) in worst case when all letters are new and need new nodes

vs Alternative: Compared to storing words in a list, trie insert is faster for prefix queries but uses more space

Edge Cases

Inserting empty string

Marks root as end of word without adding children

DSA Javascript

node.end = true; // mark end of word

Inserting word already in trie

No new nodes created, just marks end again

DSA Javascript

if (!node.children[char]) {

Inserting word with shared prefix

Reuses existing nodes for prefix, adds new nodes only for suffix

DSA Javascript

if (!node.children[char]) {

When to Use This Pattern

When you need to store many words and quickly check prefixes, use a trie insert pattern to build a prefix tree efficiently.

Common Mistakes

Mistake: Not marking the end of word node, so words are not recognized as complete
Fix: Always set node.end = true after inserting all letters

Mistake: Overwriting existing child nodes instead of checking if they exist
Fix: Check if child exists before creating new node

Summary

Inserts words letter by letter into a tree structure sharing prefixes.

Use when you want fast prefix-based word storage and lookup.

The key is to reuse existing nodes for shared prefixes and mark word ends.

Practice

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

easy

A. To count the total words stored in the Trie

B. To delete a word from the Trie

C. To search for a word in the Trie

D. To add a new word character by character into the Trie

Solution

1. Step 1: Understand the insert operation

   The insert operation adds each character of a word into the Trie one by one, creating nodes if they don't exist.

2. Step 2: Compare with other operations

   Deleting removes words, searching finds words, and counting totals words, but insert specifically adds words.

3. Final Answer:

   To add a new word character by character into the Trie -> Option D

4. Quick Check:

   Insert = Add word step-by-step [OK]

Hint: Insert means adding word letters stepwise [OK]

Common Mistakes:

• Confusing insert with search or delete
• Thinking insert adds whole word as one node
• Assuming insert counts words

2. Which of the following is the correct way to start inserting a word "cat" into an empty Trie in JavaScript?

easy

A. for (let char of word) { if (!node.children[char]) node.children[char] = new TrieNode(); node = node.children[char]; }

B. for (let i = 0; i < word.length; i++) { node = node.children[word[i]]; }

C. while (word) { node.children[word[0]] = new TrieNode(); word = word.slice(1); }

D. node.children[word] = new TrieNode();

Solution

1. Step 1: Analyze the loop for insertion

   for (let char of word) { if (!node.children[char]) node.children[char] = new TrieNode(); node = node.children[char]; } loops over each character, checks if child node exists, creates if not, then moves to that child node.

2. Step 2: Check other options for correctness

   for (let i = 0; i < word.length; i++) { node = node.children[word[i]]; } does not create nodes if missing. while (word) { node.children[word[0]] = new TrieNode(); word = word.slice(1); } modifies the word string but does not move node properly. node.children[word] = new TrieNode(); tries to add whole word as one child.

3. Final Answer:

   for (let char of word) { if (!node.children[char]) node.children[char] = new TrieNode(); node = node.children[char]; } -> Option A

4. Quick Check:

   Loop chars, create missing nodes, move node [OK]

Hint: Loop chars, create missing nodes, move node [OK]

Common Mistakes:

• Not creating nodes when missing
• Assigning whole word as one child
• Not updating current node pointer

3. What will be the structure of the Trie after inserting the words ["bat", "ball"] in order? Consider each node stores children in an object and a boolean isEnd for word end.

medium

A. {"b":{"a":{"t":{"isEnd":false},"l":{"l":{"isEnd":true}}}}}

B. {"b":{"a":{"t":{"isEnd":true},"l":{"isEnd":true}}}}

C. {"b":{"a":{"t":{"isEnd":true},"l":{"l":{"isEnd":true}}}}}

D. {"b":{"a":{"t":{"isEnd":true}},"l":{"l":{"isEnd":true}}}}

Solution

1. Step 1: Insert "bat" into empty Trie

   Creates nodes b -> a -> t with t's isEnd true.

2. Step 2: Insert "ball" next

   Nodes b -> a exist, then add l -> l with last l's isEnd true.

3. Final Answer:

   {"b":{"a":{"t":{"isEnd":true},"l":{"l":{"isEnd":true}}}}} -> Option C

4. Quick Check:

   Both words share prefix 'ba', separate branches for 't' and 'll' [OK]

Hint: Shared prefix nodes, separate branches for different suffixes [OK]

Common Mistakes:

• Marking wrong nodes as word end
• Not creating second 'l' node for 'ball'
• Overwriting nodes incorrectly

4. Given this insert function snippet, what is the bug?

insert(word) {
  let node = this.root;
  for (let char of word) {
    if (!node.children[char]) {
      node.children[char] = new TrieNode();
    }
    node = node.children;
  }
  node.isEnd = true;
}

medium

A. Missing check for empty word input

B. node is assigned to node.children instead of node.children[char]

C. isEnd should be set before the loop

D. Children should be an array, not an object

Solution

1. Step 1: Check node update inside loop

   Inside loop, node is set to node.children (object), not node.children[char] (child node), so traversal breaks.

2. Step 2: Confirm correct pointer update

   Correct is node = node.children[char] to move down the Trie properly.

3. Final Answer:

   node is assigned to node.children instead of node.children[char] -> Option B

4. Quick Check:

   Update node to child node, not children object [OK]

Hint: Update node to child node, not children object [OK]

Common Mistakes:

• Assigning node to children object instead of child node
• Setting isEnd too early
• Ignoring empty input check

5. You want to insert the words ["apple", "app", "ape"] into a Trie and then check if the word "app" is a complete word in the Trie. Which approach correctly handles this scenario?

hard

A. Insert each word character by character, mark isEnd true at last character; then search for "app" checking isEnd at last node

B. Insert all words, but mark isEnd true only for the longest word; search ignores isEnd

C. Insert words as full strings in one node; search by matching full string keys

D. Insert words without marking isEnd; search by checking if path exists

Solution

1. Step 1: Insert words with isEnd marking

   Each word is inserted letter by letter; isEnd is set true at last letter node to mark complete words.

2. Step 2: Search for "app" with isEnd check

   Search traverses nodes for 'a'->'p'->'p' and checks isEnd true to confirm "app" is a complete word, not just prefix.

3. Final Answer:

   Insert each word character by character, mark isEnd true at last character; then search for "app" checking isEnd at last node -> Option A

4. Quick Check:

   Mark word ends, check isEnd to confirm full word [OK]

Hint: Mark word ends with isEnd, check isEnd when searching [OK]

Common Mistakes:

• Not marking isEnd for shorter words
• Treating prefixes as complete words without isEnd
• Storing whole words as single nodes