Advantages of linked lists over binary trees?

4
votes

The title is mostly self-explanatory: what are the advantages of linked lists over binary trees? The only case I can think of in which a linked list is more efficient is for iterating over every element, in which case it's still pretty close. It looks like binary trees are faster at both accessing data and inserting new elements. So why use a linked list at all?

7
data-structureslinked-listbinary-tree
What are the advantages of a binary tree over a 17-ary tree? If 2 is better than 1, then 17 is a whole lot better than 2, right? :-) – Ken
@Ken: Only if you can make 17 comparisons in one operation. – P Daddy
On top of everything mentioned already, linked lists are useful for implementing other data-structures, like stacks and queues. – BlueRaja - Danny Pflughoeft
What are the advantages of apples over oranges? – starblue
starblue: the lack of individual slices makes infix consumption O(1) rather than O(n), and the edible peel tends to lower the constant factor, too. – Ken

7 Answers

6
votes

If the tail of a linked list is stored, then insertion into a linked list is decidedly faster than insertion into a binary tree. Insertion into a binary tree is O(N) at worst case (O(log N) at best) if it's unbalanced. If it's balanced, then insertions are O(log N), but there is house keeping involved in keeping it balanced. Insertion into a linked list is O(1) if the tail is kept.

Also, as BillyONeal mentioned, a binary tree is usually an associative structure, whereas linked lists are not.

4
votes

It's easier/faster to delete items from a linked list compared to a binary tree which could require few operations to fix the tree.

2
votes

A linked list is not generally used as an associative container (READ: Not to be used as a dictionary) -- only as a literal list of items, like an array. Binary trees' performance when such a simple data structure is desired is poor.

2
votes

In linked list the objects are ordered by the container itself, so you don't need to have a comparison function for the objects.

0
votes

A fair question. I like to use the container that most "tightly" fits my needs. For example, you can use a list when all you need is a queue with no real consequence either... but... queues are highly optimized for this one specific task of popping off the front, and inserting at the back, with no extra pointers, or anything. By using the most appropriate class, you can be sure you're not getting any extra fluff, even if it has the same Big-O. Sometimes those hidden constants do matter.

0
votes

It mostly depends on scenario. If tail of linked list is maintained then insertion is fast in linked list. Deletion is quite fast in linked list but in case of searching it is better in trees( o(log(n) for height balance tree ) while o(n) in linked list.

0
votes

I assume you're talking about actual binary search trees where the nodes are added using algorithms to maximize retrieval performance. As opposed to a simple tree where each node has a maximum of 2 child nodes.

A linked list is often unsorted and so addition of new nodes is simply an O(1) operation normally by appending to the tail of the list.

On the other hand a binary tree must store nodes in a specific ordering mechanism (and possibly enforce balancing) to provide for a more efficient searching/retrieval operation.

If your algorithm does NOT need to retrieve items very efficiently while also provide efficient sorting of the items, a linked list is probably all you need.

Queues and Stacks are examples of data structures that can be happily implemented using a linked list.

Note: Insertion to a linked list is a different (slower) operation than basic addition/append. Insertion often requires traversal along the list until the correct position is found, O(n) where n is the list length. An append is simply adding to the tail of the list (hence O(1))