Linked Structures

Traverse and Print

Walk from the head pointer to null, visiting each node exactly once without random indexing.

Algorithm

The replay labels nodes by value, such as node(20), and never exposes object identity or memory addresses. This Rust DSA implementation uses the same small chain as the rest of the DSA track.

cursor A cursor reference names the node currently being visited.
null stop Traversal ends when the cursor reaches the null marker.

Basic Implementation

basic.rs
Replay: real traced execution (multi-file project)
struct Node {
    value: i32,
    next: Option<Box<Node>>,
}

fn node(value: i32, next: Option<Box<Node>>) -> Option<Box<Node>> {
    Some(Box::new(Node { value, next }))
}

fn render(head: &Option<Box<Node>>) -> String {
    let mut parts: Vec<String> = Vec::new();
    let mut cursor = head.as_ref();
    while let Some(node) = cursor {
        parts.push(node.value.to_string());
        cursor = node.next.as_ref();
    }
    parts.join(" -> ") + " -> null"
}

fn delete_value(head: Option<Box<Node>>, target: i32) -> Option<Box<Node>> {
    match head {
        Some(mut n) => {
            if n.value == target {
                n.next
            } else {
                n.next = delete_value(n.next, target);
                Some(n)
            }
        }
        None => None,
    }
}

fn main() {
    let head = node(10, node(20, node(30, None)));
    println!("{}", render(&head));
}

  1. head ← 10 -> 20 -> 30 -> null

    1struct Node {2    value: i32,
    values this step10 -> 20 -> 30 -> nullhead

  2. output ← 10

    12let mut cursor = head.as_ref();13while let Some(node) = cursor {14    parts.push(node.value.to_string());
    values this step10outputnode(10)cursor

  3. output ← 10 -> 20

    12let mut cursor = head.as_ref();13while let Some(node) = cursor {14    parts.push(node.value.to_string());
    values this step10 -> 20outputnode(20)cursor

  4. output ← 10 -> 20 -> 30 -> null

    12let mut cursor = head.as_ref();13while let Some(node) = cursor {14    parts.push(node.value.to_string());
    values this step10 -> 20 -> 30 -> nulloutputnode(30)cursor

  5. stdout ← 10 -> 20 -> 30 -> null

    35    let head = node(10, node(20, node(30, None)));36    println!("{}", render(&head));37}
    values this step10 -> 20 -> 30 -> nullstdout10 -> 20 -> 30 -> nulloutput

Complexity

  • Time: O(n)
  • Space: O(1)

Implementation notes

  • Keep the explicit node and pointer/reference operations; array shortcuts hide the linked-list state this lesson is meant to replay.
  • The final output prints the chain in a deterministic a -> b -> null form for cross-language comparison.