Challenge 1: Building, Printing, and Freeing a Linked List — Possible Solution
====================================================================

#include <stdio.h>
#include <stdlib.h>

typedef struct Node {
    int value;
    struct Node *next;
} Node;

Node *push_front(Node *head, int value) {
    Node *n = malloc(sizeof(Node));
    n->value = value;
    n->next = head;
    return n;
}

void free_list(Node *head) {
    while (head != NULL) {
        Node *next = head->next;
        free(head);
        head = next;
    }
}

int main() {
    Node *head = NULL;
    head = push_front(head, 3);
    head = push_front(head, 2);
    head = push_front(head, 1);

    for (Node *cur = head; cur != NULL; cur = cur->next) {
        printf("%d\n", cur->value);
    }

    free_list(head);
    return 0;
}

Output:

    1
    2
    3

WHY THIS WORKS AS AN ANSWER
------------------------------
Each push_front call prepends a new node, so the three calls with 3,
then 2, then 1 produce the list in the order 1 -> 2 -> 3 (the most
recently pushed value ends up at the front) -- exactly what the
traversal prints. free_list is called only after every use of the
list is complete, following the chapter's own save-next-before-freeing
pattern to release every node without leaking or reading freed memory.