In an AVL tree, it takes a constant number of single and double rotations every time we rebalance on insertion and deletion since we only have to check the path from point of insertion or deletion to the root.
If we had an unbalanced tree, we would have to check if every possible node is balanced, so it would cost O(n) to rebalance an unbalanced tree. Is this correct?
It does take time O(n) to rebalamce an unbalanced tree, but not for the reason you mentioned. In an AVL tree, insertions and deletions may require Θ(log n) rotations if the tree was maximally imbalanced before the element was inserted. This could potentially require O(n log n) time to rebalance the tree, since you might do O(log n) work per each of the n nodes.
However, using other algorithms, you can rebalance a tree in O(n) time. One simple option is to do an inorder traversal of the tree to get the elements in sorted order, then reconstruct an optimal BST from those elements by recursively building the tree bottom-up. Alternatively, you can use the Day-Stout-Warren algorithm, which balances any tree in O(n) time and O(1) space.
Hope this helps!
Related
I know removing a node in AVL tree takes time complexity of O(logn). That being said, removing an AVL tree with n nodes would take O(nlogn). However, I am wondering if my goal is to have the sorted element of AVL tree that I could remove all elements in O(n) instead of O(nlogn). Possibly by implementing an remove element that would take O(1).
I was not able to find any way to do it in O(n). Is it because we can't or am I missing something?
If you don't need to preserve AVL structure after every deletion, then perform post-order traversal, just deleting every node without balancing instead of "Display the data part"
Is it possible that we could perform m insert and delete operations on a balanced binary search tree such that delete operation deletes a node and the whole subtree below it and after that balance it? The whole process being in done in amortized O(log n) time per step?
Short answer: Yes it is possible.
What you are describing is a self balancing binary tree, like an AVL-tree or a Red-Black tree. Both take O(log n) for deletion, which includes reordering of the nodes. Here is a link to a page describing such trees and how they work in much more detail than I can, including illustrations. You can also check out the Wikipedia page of AVL-trees, they have a decent explanation as well as animations of the insertions. Here is a short version of what you were most interested in:
Deletions in an AVL tree are on average O(log n) and the rebalancing is on average O(log n), worst case O(1). This is done by doing rotations, again, well explained in both the sources.
The Wikipedia page also includes some code there, if you need to implement it.
EDIT:
For removing a subtree, you will still be able to do the same thing. Here is a link to a very good explanation of this. Short version: deleting the subtree is can be done O(log n) (keep in mind that deletion, regardless of the number of nodes deleted is still O(log n) as long as you do not directly rebalance the tree), then the tree would rebalance itself using rotations. This can also change the root of your tree. Removing a whole subtree is of course going to create a bigger height difference than just the deletion of one node at the end of the tree. Still, using rotation the tree can be rebalanced itself by finding the first node imbalance and then doing the AVL rebalancing scheme. Due to the use of the rotations, this should still all be O(log n). Here you will find how the tree rebalances itself after a deletion, which creates a height imbalance.
What I mean by general tree is an unbalanced tree with multiple child nodes (not restricted to 2 child node for each branch like Binary Tree). What is the Big-O complexity of remove node, insert node, find node
The average time complexity of searching in balanced BST in O(log(n)). The worst case complexity of searching in unbalanced binary tree is O(n).
If you're talking about a regular k-ary tree that does nothing special with its data, then to find any one node in the tree would take O(n) time assuming there are n nodes.
Inserting a node would be O(1) since you can store it wherever you want, and removing a node would be O(n) since you'd have to look at every node (worst case) to find the one to delete, and since there's no order to the data you don't have to do anything with the rest of the nodes.
http://www.geeksforgeeks.org/group-multiple-occurrence-of-array-elements-ordered-by-first-occurrence/
Please check this question.
How to do BST method of this problem.
They have mentioned that total time complexity will be O(NLogN).
How is time complexity of tree is LogN for traversal?
Please help
search, delete and insert running time all depend on the height of tree, or O(h) for BST. A degenerate tree almost looks like a linked list can produce a running time of O(N).
On the other hand, consider a self-balancing tree such as AVL tree, the running time for lookup is lower bounded by O(logN) because like Binary Search, we divide the search space by half each time as in left and right subtree have almost identical height.
Fibonacci heaps are efficient in an amortized sense, but how efficient are they in the worst case? Specifically, what is the worst-case time complexity of each of these operations on an n-node Fibonacci heap?
find min
delete-min
insert
decrease-key
merge
The find-min operation on a Fibonacci heap always takes worst-case O(1) time. There's always a pointer maintained that directly points to that object.
The cost of a delete-min, in the worst-case, takes time Θ(n). To see this, imagine starting with an empty heap and doing a series of n insertions into it. Each node will be stored in its own tree, and doing a delete-min the heap will coalesce all these objects into O(log n) trees, requiring Θ(n) work to visit all the nodes at least once.
The cost of an insertion is worst-case O(1); this is just creating a single node and adding it to the list. A merge is similarly O(1) since it just splices two lists together.
The cost of a decrease-key in the worst case is Θ(n). It's possible to build a degenerate Fibonacci heap in which all the elements are stored in a single tree consisting of a linked list of n marked nodes. Doing a decrease-key on the bottommost node then triggers a run of cascading cuts that will convert the tree into n independent nodes.
I almost agree with the great answer from #templatetypedef.
There cannot be a tree of a classical Fibonacci heap with $n$ marked nodes. This would mean that the height of the tree is O(n), but since for each subtree of rank $k$ its children are of ranks $\geq 0, \geq 1, ... , \geq k-1$. It is easy to see that the depth of the tree is at most O(logn). And therefore a single Decrease-key operation can cost O(logn).
I checked this thread, and it takes some modification of the Fibonacci heap, as it has marked node in the root list and does operation which do not belong to the Fibonacci heap.