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ishtms committed Sep 16, 2023
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7 changes: 1 addition & 6 deletions chapters/ch07.0-ex-implementing-a-trie.md
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[![Read Prev](/assets/imgs/prev.png)](/chapters/ch06.4-the-need-for-a-trie.md)

# Exercise - Implementing a `Trie`
# Exercises - Implementing a `Trie`

> This exercise will motivate you to work on implementing your solution independently. Once you have completed the exercise, you can move on to the next challenge or read the solution to find a different approach.
>
Expand All @@ -25,9 +25,7 @@ The Trie starts with a root node that doesn't hold any character. It serves as t
```

- **Level 1**: You have the characters "O", "T", and "C" branching from the root node.

- **Level 2 and Beyond**: These nodes further branch out to form the words.

- "O" branches to "K", completing the word "OK".
- "T" branches to "O", completing the word "TO".
- "C" branches to "A" and "U":
Expand All @@ -53,15 +51,12 @@ In this first challenge, your task is to implement a Trie data structure with on
### Requirements

1. Create a class called `Trie`.

2. Implement an `insert(word)` method that takes a string `word` and inserts it into the Trie.

### More details

1. **Initialization**: You'll begin with a root node. This node will be the starting point for all word insertions, and it won't store any character itself.

2. **Traversal**: For each character in the word you want to insert, you'll traverse the Trie from the root node, going as far down as the current character sequence allows.

3. **Node Creation**: If a character in the word doesn't match any child node of the current node:

- Create a new node for that character.
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