Trie Insert Operation in DSA Go - Execution Trace

Concept Flow - Trie Insert Operation

Start at root node

↓

For each character in word

↓

Check if child node exists

↓

Move to child

↓

Repeat for next character

↓

Mark last node as end of word

↓

Done

Start at the root, for each letter check if child exists; if not, create it. Move down nodes until word ends, then mark last node as word end.

Execution Sample

DSA Go

func Insert(word string) {
  current := root
  for _, ch := range word {
    if current.children[ch] == nil {
      current.children[ch] = newNode()
    }
    current = current.children[ch]
  }
  current.isEnd = true
}

Inserts a word into the trie by creating nodes for missing letters and marking the last node as end.

Execution Table

Step	Operation	Current Character	Node Created	Pointer Changes	Trie Visual State
1	Start insertion	-	No	current points to root	root (empty)
2	Check 'c'	c	Yes	root.children['c'] = newNode; current moves to 'c'	root └─ c
3	Check 'a'	a	Yes	'c'.children['a'] = newNode; current moves to 'a'	root └─ c └─ a
4	Check 't'	t	Yes	'a'.children['t'] = newNode; current moves to 't'	root └─ c └─ a └─ t
5	Mark end	-	No	current.isEnd = true	root └─ c └─ a └─ t* (end)
6	Insertion complete	-	No	-	root └─ c └─ a └─ t* (end)

💡 All characters processed; last node marked as end of word.

Variable Tracker

Variable	Start	After Step 2	After Step 3	After Step 4	After Step 5	Final
current	root	node 'c'	node 'a'	node 't'	node 't' (isEnd=true)	node 't' (isEnd=true)
node created	none	node 'c'	node 'a'	node 't'	none	none

Key Moments - 3 Insights

Why do we create a new node only if the child does not exist?

Because if the child node for a character already exists, we just move to it to avoid overwriting existing words. See execution_table rows 2-4 where nodes are created only when missing.

Why do we mark the last node as end of word?

Marking the last node as end tells the trie that a complete word ends here. Without this, the trie can't distinguish between prefixes and full words. See execution_table row 5.

What happens if we insert a word that shares prefix with existing words?

We reuse existing nodes for shared prefix characters and create new nodes only for new suffix characters. This is shown in the flow where existing children are checked before creating new nodes.

Visual Quiz - 3 Questions

Look at the execution_table, what node does 'current' point to after step 3?

A node 'c'

B node 'a'

C node 't'

D root

💡 Hint

Check variable_tracker column 'After Step 3' for 'current' value.

At which step is the last node marked as end of word?

A Step 4

B Step 3

C Step 5

D Step 6

💡 Hint

See execution_table row where 'current.isEnd = true' is set.

If the word inserted was 'car' instead of 'cat', which step would create the node for 'r'?

A Step 4

B Step 3

C Step 5

D Step 2

💡 Hint

Insertion creates nodes for each new character; 'r' is the third character, so step 4.

Concept Snapshot

Trie Insert Operation:
- Start at root node
- For each character:
  - If child node exists, move to it
  - Else create new node and move
- After last char, mark node as end of word
- Efficiently stores prefixes and words

Full Transcript

The Trie Insert Operation starts at the root node. For each character in the word, it checks if a child node exists. If not, it creates a new node. Then it moves the current pointer to that child node. After processing all characters, it marks the last node as the end of a word. This process allows the trie to store words sharing prefixes efficiently. The execution table shows each step, node creation, pointer changes, and the trie state. The variable tracker follows the current pointer and node creation. Key moments clarify why nodes are created only when missing and why marking the end is important. The visual quiz tests understanding of pointer positions and steps.