In C, how can you improve the following linked node declaration and allocation code using typedef so that the type name is shorter and clearer? struct node { int data1; float data2; struct node *left; struct node *right; }; struct node *ptr; ptr = (struct node *) malloc(sizeof(struct node));

Introduction / Context:
C code that uses struct types can become verbose when the keyword struct is repeated many times, especially in pointer declarations and dynamic allocations. The typedef feature allows you to create a shorter alias for a struct type, which improves readability for frequently used structures such as linked nodes. This question asks you to select the version of the code that uses typedef in a clean and idiomatic way when declaring and allocating a node pointer.

Given Data / Assumptions:

• The original definition uses struct node and struct node pointers for left and right children.
• ptr is intended to be a pointer to the node type.
• Dynamic allocation is done using malloc with the desired object size.
• We aim to introduce a typedef without breaking the structure or pointer semantics.

Concept / Approach:
A standard way to simplify usage is to keep the struct tag for debugging or forward declarations and create a typedef name that refers to the complete struct. The typedef can appear immediately after the struct definition: typedef struct node { fields } Node;. From that point on, Node is a synonym for struct node, and pointers can be written as Node *ptr; instead of struct node *ptr;. For allocation, it is safer and clearer to use sizeof(Node) or sizeof *ptr so that changes to the type are automatically reflected in the allocation size.

Step-by-Step Solution:
Step 1: Keep the struct node definition with its tag, because the left and right pointers use struct node * as their type. Step 2: Introduce a typedef that names this type Node, as in } Node; at the end of the definition. Step 3: Declare the pointer ptr using the new alias: Node *ptr; which is equivalent to struct node *ptr;. Step 4: Allocate memory using malloc(sizeof(Node)); so that the allocation matches the full size of the struct. Step 5: Choose the option that implements exactly this pattern without changing the structure layout or pointer types.

Verification / Alternative check:
You can verify equivalence by noting that Node and struct node refer to the same type. Any function that previously took struct node * can now be rewritten to use Node * without changing compiled behaviour. Using sizeof(Node) and sizeof(struct node) yields the same size. This confirms that the chosen refactoring is purely syntactic, improving readability while preserving semantics.

Why Other Options Are Wrong:
Option B introduces the typedef but continues to use struct node *ptr and sizeof(struct node), which misses the opportunity to simplify declarations. Option C defines an anonymous struct and incorrectly declares ptr as a plain Node rather than a pointer, making the malloc call invalid. Option D declares Node as an alias before the struct is defined and misuses Node as a non pointer object, which does not match the original pointer based design.

Common Pitfalls:
A common error is to drop the struct tag entirely and then accidentally refer to struct node elsewhere, causing compilation errors. Another pitfall is to mismatch pointer and non pointer types, for example allocating with malloc for one Node but assigning to a Node rather than Node * pointer. When refactoring with typedef, always ensure that the alias represents the full structure and that pointers are declared explicitly where needed.

Final Answer:
The improved code defines typedef struct node { ... } Node; and then uses Node *ptr = malloc(sizeof(Node)); so that Node becomes a convenient alias for struct node.