Is O(logn) always a tree?

Question

We always see operations on a (binary search) tree has O(logn) worst case running time because of the tree height is logn. I wonder if we are told that an algorithm has running time as a function of logn, e.g m + nlogn, can we conclude it must involve an (augmented) tree?

EDIT: Thanks to your comments, I now realize divide-conquer and binary tree are so similar visually/conceptually. I had never made a connection between the two. But I think of a case where O(logn) is not a divide-conquer algo which involves a tree which has no property of a BST/AVL/red-black tree.

That's the disjoint set data structure with Find/Union operations, whose running time is O(N + MlogN), with N being the # of elements and M the number of Find operations.

Please let me know if I'm missing sth, but I cannot see how divide-conquer comes into play here. I just see in this (disjoint set) case that it has a tree with no BST property and a running time being a function of logN. So my question is about why/why not I can make a generalization from this case.

Answer 1

What you have is exactly backwards. O(lg N) generally means some sort of divide and conquer algorithm, and one common way of implementing divide and conquer is a binary tree. While binary trees are a substantial subset of all divide-and-conquer algorithms, the are a subset anyway.

In some cases, you can transform other divide and conquer algorithms fairly directly into binary trees (e.g. comments on another answer have already made an attempt at claiming a binary search is similar). Just for another obvious example, however, a multiway tree (e.g. a B-tree, B+ tree or B* tree), while clearly a tree is just as clearly not a binary tree.

Again, if you want to badly enough, you can stretch the point that a multiway tree can be represented as sort of a warped version of a binary tree. If you want to, you can probably stretch all the exceptions to the point of saying that all of them are (at least something like) binary trees. At least to me, however, all that does is make "binary tree" synonymous with "divide and conquer". In other words, all you accomplish is warping the vocabulary and essentially obliterating a term that's both distinct and useful.

Answer 2

No, you can also binary search a sorted array (for instance). But don't take my word for it http://en.wikipedia.org/wiki/Binary_search_algorithm

Answer 3

As a counter example:

given array 'a' with length 'n'
y = 0
for x = 0 to log(length(a))
    y = y + 1
return y

The run time is O(log(n)), but no tree here!

Answer 4

Answer is no. Binary search of a sorted array is O(log(n)).

Answer 5

Algorithms taking logarithmic time are commonly found in operations on binary trees.

Examples of O(logn):

• Finding an item in a sorted array with a binary search or a balanced search tree.

• Look up a value in a sorted input array by bisection.

Answer 6

As O(log(n)) is only an upper bound also all O(1) algorithms like function (a, b) return a+b; satisfy the condition.

But I have to agree all Theta(log(n)) algorithms kinda look like tree algorithms or at least can be abstracted to a tree.

Answer 7

Short Answer:

Just because an algorithm has log(n) as part of its analysis does not mean that a tree is involved. For example, the following is a very simple algorithm that is O(log(n)

for(int i = 1; i < n; i = i * 2)
  print "hello";

As you can see, no tree was involved. John, also provides a good example on how binary search can be done on a sorted array. These both take O(log(n)) time, and there are of other code examples that could be created or referenced. So don't make assumptions based on the asymptotic time complexity, look at the code to know for sure.

More On Trees:

Just because an algorithm involves "trees" doesn't imply O(logn) either. You need to know the tree type and how the operation affects the tree.

Some Examples:

• Example 1)

Inserting or searching the following unbalanced tree would be O(n).

• Example 2)

Inserting or search the following balanced trees would both by O(log(n)).

Balanced Binary Tree:

Balanced Tree of Degree 3:

Additional Comments

If the trees you are using don't have a way to "balance" than there is a good chance that your operations will be O(n) time not O(logn). If you use trees that are self balancing, then inserts normally take more time, as the balancing of the trees normally occur during the insert phase.