Merge Two Sorted Linked Lists in DSA C - Execution Trace

Concept Flow - Merge Two Sorted Linked Lists

Create dummy node

↓

Set current pointer to dummy

↓

Compare heads of both lists

↓

Smaller node

↓

Move current and chosen list pointer forward

↓

Repeat comparison until one list ends

↓

Attach remaining nodes of non-empty list

↓

Return dummy.next as merged list head

We create a dummy node to start the merged list, then repeatedly pick the smaller node from the heads of the two lists, attach it, and move pointers until one list is empty. Finally, attach the leftover nodes.

Execution Sample

DSA C

ListNode* merge(ListNode* l1, ListNode* l2) {
  ListNode dummy = {0, NULL};
  ListNode* current = &dummy;
  while (l1 && l2) {
    if (l1->val < l2->val) {
      current->next = l1; l1 = l1->next;
    } else {
      current->next = l2; l2 = l2->next;
    }
    current = current->next;
  }
  current->next = l1 ? l1 : l2;
  return dummy.next;
}

This code merges two sorted linked lists by choosing the smaller head node each time and linking nodes to form a new sorted list.

Execution Table

Step	Operation	Nodes in List	Pointer Changes	Visual State
1	Create dummy node and set current to dummy	dummy(0) -> NULL	current = dummy	dummy(0) -> NULL
2	Compare l1(1) and l2(2), attach l1 to current.next	dummy(0) -> 1 -> NULL	current = 1, l1 moves to 3	dummy(0) -> 1 -> NULL
3	Compare l1(3) and l2(2), attach l2 to current.next	dummy(0) -> 1 -> 2 -> NULL	current = 2, l2 moves to 4	dummy(0) -> 1 -> 2 -> NULL
4	Compare l1(3) and l2(4), attach l1 to current.next	dummy(0) -> 1 -> 2 -> 3 -> NULL	current = 3, l1 moves to 5	dummy(0) -> 1 -> 2 -> 3 -> NULL
5	Compare l1(5) and l2(4), attach l2 to current.next	dummy(0) -> 1 -> 2 -> 3 -> 4 -> NULL	current = 4, l2 moves to NULL	dummy(0) -> 1 -> 2 -> 3 -> 4 -> NULL
6	l2 is NULL, attach remaining l1(5)	dummy(0) -> 1 -> 2 -> 3 -> 4 -> 5 -> NULL	current = 5, l1 moves to NULL	dummy(0) -> 1 -> 2 -> 3 -> 4 -> 5 -> NULL
7	Return dummy.next as merged list head	1 -> 2 -> 3 -> 4 -> 5 -> NULL	No pointer changes	1 -> 2 -> 3 -> 4 -> 5 -> NULL

💡 Loop ends when one list is empty; remaining nodes attached to merged list.

Variable Tracker

Variable	Start	After Step 2	After Step 3	After Step 4	After Step 5	After Step 6	Final
l1	1 -> 3 -> 5 -> NULL	3 -> 5 -> NULL	3 -> 5 -> NULL	5 -> NULL	5 -> NULL	NULL	NULL
l2	2 -> 4 -> NULL	2 -> 4 -> NULL	4 -> NULL	4 -> NULL	NULL	NULL	NULL
current	dummy(0)	1	2	3	4	5	5
merged list	empty	dummy(0) -> 1 -> NULL	dummy(0) -> 1 -> 2 -> NULL	dummy(0) -> 1 -> 2 -> 3 -> NULL	dummy(0) -> 1 -> 2 -> 3 -> 4 -> NULL	dummy(0) -> 1 -> 2 -> 3 -> 4 -> 5 -> NULL	1 -> 2 -> 3 -> 4 -> 5 -> NULL

Key Moments - 3 Insights

Why do we use a dummy node instead of starting directly with l1 or l2?

Using a dummy node simplifies linking nodes because it avoids special cases for the head. As seen in step 1 and 7, dummy.next points to the merged list head, so we always have a stable starting point.

What happens when one list becomes empty before the other?

When either l1 or l2 becomes NULL (step 5), we attach the remaining nodes of the other list directly to current->next. This ensures no nodes are lost and the merged list remains sorted.

Why do we move the current pointer after attaching a node?

Moving current forward (steps 2-5) keeps track of the last node in the merged list, so the next chosen node can be linked correctly. Without moving current, nodes would overwrite the same next pointer.

Visual Quiz - 3 Questions

Look at the execution table at step 4. Which node does current point to after attaching?

A Node with value 2

B Node with value 3

C Node with value 4

D Dummy node

💡 Hint

Check the 'Pointer Changes' column at step 4 in the execution_table.

At which step does the loop end because one list becomes empty?

A Step 7

B Step 6

C Step 5

D Step 3

💡 Hint

Look for when l2 moves to NULL in the variable_tracker and execution_table.

If l1 started with values all greater than l2, how would the merged list start?

A With nodes from l2 first

B Alternating nodes from l1 and l2

C With nodes from l1 first

D Empty list

💡 Hint

Recall the comparison step picks the smaller node each time, as shown in the concept_flow.

Concept Snapshot

Merge Two Sorted Linked Lists:
- Use a dummy node to start merged list
- Compare heads of both lists
- Attach smaller node to current->next
- Move current and chosen list pointer forward
- Repeat until one list ends
- Attach remaining nodes
- Return dummy.next as merged list head

Full Transcript

To merge two sorted linked lists, we start by creating a dummy node to simplify linking. We use a pointer 'current' starting at dummy. Then, we compare the values at the heads of both lists. The smaller node is attached to current->next, and we move current and that list's pointer forward. We repeat this until one list is empty. After that, we attach the remaining nodes of the other list to current->next. Finally, we return dummy.next, which points to the head of the merged sorted list. This approach avoids special cases and ensures all nodes are included in order.