When a node in a B-tree is split, are keys from the original node duplicated in the new nodes? What's the purpose of doing this? Isn't this inefficient?
No. It's all done with pointers. Half of the pointers are moved to the new node.
Of course, there's no such thing as 'a B-tree'. There are a myriad of different implementations. I could imagine one in which the keys are actually stored in the nodes, such as a tree where the keys are ints. But they still wouldn't be 'duplicated', just the data copied.
If your beef is the storage left behind in the node that gets split, well, that's another optimization choice whether to free and reallocate smaller or not. Probably not, since more insertions could arrive that go into that node's 1/2 of the key space.
I think that you mean a B+ tree.
In a B+ tree that I wrote, I did duplicate the key values in the parent node during a split. key[pos] in the parent was set to the left node's lowest value and pointed to the left node. The right node's lowest value became key[pos+1] in the parent.
Related
Given an array of binary trees find whether any two trees share a node, not value wise, but "pointer" wise. At the bottom I provided an example.
My approach was to iterate through all the trees and store all the leaves (pointers) from each tree into a list, then check if list has any duplicates, but that's a rather slow approach. Is there perhaps a quicker way to solve this?
In the worst case you will have to traverse all nodes (all pointers) to find a shared node (pointer), as it might happen to be the last one visited. So the best time complexity we can expect to have is O(𝑚+𝑛) where 𝑚 and 𝑛 represent the number of nodes in either tree.
We can achieve this time complexity if we store the pointers from the first tree in a hash set and then traverse the pointers of the second tree to see if any of those is in the set. Assuming that get/set operations on a hash set have an amortized constant time complexity, the overal time complexity will be O(𝑚+𝑛).
If the same program is responsible for constructing the trees, then a reuse of the same node can be detected upon insertion. For instance, reuse of the same node in multiple trees can be completely avoided by having the insert method of your tree only take a value as argument, never a node instance. The method will then encapsulate the actual creation of the node, guaranteeing its uniqueness.
An idea for O(#nodes) time and O(1) space. It does more traversal work than simple traversals using a hash table, but it doesn't have the cost of using a hash table. I don't know what's better. Might depend on the language.
For two trees
Create one extra node. Do a Morris traversal of the first tree. It only modifies right child pointers, so we can use left child pointers for marking nodes as seen. For every tree node without left child, set our extra node as left child. Whenever checking a left child pointer, treat our extra node like a null pointer, i.e., don't visit it. After the traversal, the tree structure is restored, and all originally left-child-less tree nodes now point to our extra node as left child. That includes all leaf nodes.
Do a Morris traversal of the second tree. Again treat pointers to our extra node like null pointers. If we ever do encounter our extra node, we know the trees share a node. If not, then we know the trees don't share a node, since if they did share any, they'd also share a leaf node (just go down from any shared node to a leaf node, that's also shared), and all leafs nodes of the first tree are marked. After the traversal, the second tree is restored.
Do a Morris traversal of the first tree again, this time removing our extra node, restoring the original null pointers.
For an array of more than two trees
Mark the first tree as above. Check the second tree as above. Mark the second tree. Check the third. Mark the third. Check the fourth. Mark the fourth. Etc. When you found a shared node or there are no more trees, unmark the marked trees.
Every shared node must have two parents, or an ancestor with two parents.
LOOP over nodes
IF node has two parents
MARK node as shared
Mark all descendants as shared.
I'm aware of ways to keep binary search trees balanced/self-balancing using rotations.
I am not sure if my case needs to be that complicated. I don't need to maintain any sorted order property like with self-balancing BSTs. I just have an ordinary binary tree that I may need to delete nodes or insert nodes. I need try to maintain balance in the tree. For simplicity, my binary tree is similar to a segment tree, and every time a node is deleted, all the nodes along the path from the root to this node will be affected (in my case, it's just some subtraction of the nodal values). Similarly, every time a node is inserted, all the nodes from the root to the inserted node's final location will be affected (an addition to nodal values this time).
What would be the most straightforward way to keep a tree such as this balanced? It doesn't need to be strictly as height balanced as AVL trees, but something like RB trees or maybe slightly less balanced is acceptable as well.
If a new node does not have to be inserted at a particular spot -- possibly determined by its own value and the values in the tree -- but you are completely free to choose its location, then you could maintain the shape of the tree as a complete tree:
In a complete binary tree every level, except possibly the last, is completely filled, and all nodes in the last level are as far left as possible.
An array is a very efficient data structure for a complete tree, as you can store the nodes in their order in a breadth-first traversal. Because the tree is given to be complete, the array has no gaps. This structure is commonly used for heaps:
Heaps are usually implemented with an array, as follows:
Each element in the array represents a node of the heap, and
The parent / child relationship is defined implicitly by the elements' indices in the array.
Example of a complete binary max-heap with node keys being integers from 1 to 100 and how it would be stored in an array.
In the array, the first index contains the root element. The next two indices of the array contain the root's children. The next four indices contain the four children of the root's two child nodes, and so on. Therefore, given a node at index i, its children are at indices 2i + 1 and 2i + 2, and its parent is at index floor((i-1)/2). This simple indexing scheme makes it efficient to move "up" or "down" the tree.
Operations
In your case, you would define the insert/delete operations as follows:
Insert: append the node to the end of the array. Then perform the mutation needed to its ancestors (as you described in your question)
Delete: replace the node to be deleted with the node that currently sits at the very end of the array, and shorten the array by 1. Make the updates needed that follow from the change at these two locations -- so two paths from root-to-node are impacted.
When balancing non-BSTs, the big question to ask is
Can your tree efficiently support rotations?
Some types of binary trees, like k-d trees, have a specific layer-by-layer structure that makes rotations infeasible. Others, like range trees, have auxiliary metadata in each node that's expensive to update after a rotation. But if you can handle rotations, then you can use just about any of the balancing strategies out there. The simplest option might be to model your tree on a treap: put a randomly-chosen weight field into each node, and then, during insertions, rotate your newly-added leaf up until its weight is less than its parent. To delete, repeatedly rotate the node with its lighter child until it's a leaf, then delete it.
If you cannot support rotations, you'll need a rebalancing strategy that does not require them. Perhaps the easiest option there is to model your tree after a scapegoat tree, which works by lazily detecting a node that's too deep for the tree to be balanced, then rebuilding the smallest imbalanced subtree possible into a perfectly-balanced tree to get everything back into order. Deletions are handled by rebuilding the whole tree once the number of nodes drops by some constant factor.
Investigate what must happen to delete a key from a Binary Search
Tree. Is deletion always as fast as insertion?
I looked up insertions and deletions in a BST. It appears that a deletion is more complex because the nodes need to be re-routed, which also means that the keys need to be reassigned and reorganized.
As far as speed is concerned, based on the complexity of the deletion, I assume that this means that a deletion is not as fast as an insertion.
Is this a correct assumption? Thanks
Although it may initially appear that insertion should be faster, I'm not at all convinced that this is really true, at least to any significant degree.
When we do an insertion, we always insert the new node as a child of a leaf node. We have to traverse the tree to the leaf node to do the insertion.
When we do a deletion, we have three cases to consider. The simplest is that we're deleting a leaf node. In such a case, we set the parent's pointer to that leaf node to a null pointer, and we release the memory occupied by the leaf. Not really any different from the insertion.
If the node to be deleted is a non-leaf node with one child, the task is only marginally more difficult: we set the parent of the current node to point to the child of the node to be deleted, and (again) release the memory occupied by the node we're deleting.
The only time we encounter anything that could be considered extra work at all is when we have to delete a non-leaf node that has two children. In this case, we need to find a leaf node that's a child of that node--either the right-most descendant of its left child, or the left-most descendant of its right child. We swap that node into the place of the one we're deleting and release its memory.
The thing to keep in mind here is that for insertion, we started by traversing the tree to a leaf, then we insert. In the case of deletion, it's possible that we reach the node to delete before traversing all the way to a leaf--but even in the worst case, we still just continue traversing until we reach the leaf (something we do for insertion anyway), the assign pointers to move that node into the place of the one being deleted.
There might be one or two extra assignments here (depending primarily on how you implement things), but at most the difference is extremely small.
From a practical viewpoint, any real performance difference is likely to come down to one question: whether the memory management you're using attempts to balance the costs of allocation and deletion, or favors one over the other (and of so, which).
In short, depending on how your heap is managed, the slowest part of this may well be allocation or deletion of the memory for the node, and the tree manipulation is basically lost in the noise.
I was writing a function to delete a node in binary tree built on key value of each node and satisfies search property.
it turns out that when the node has both childeren non-empty i have to swap the node with his inorder successer.
for small data in nodes i can swap two nodes by swapping the data and key values. but my node contain huge data.is there any logic to do better becoz i dont want to swap all the data b/w nodes.thnx in advance.
If you are worried about the size of data to be swapped, then it is not an issue. When we swap nodes, we only swap the addresses and not the actual data. So this is a non-issue.
This picture from Wikipedia article has three nodes of a Fibonacci heap marked in blue . What is the purpose of some of the nodes being marked in this data structure ?
Intuitively, the Fibonacci heap maintains a collection of trees of different orders, coalescing them when a delete-min occurs. The hope in constructing a Fibonacci heap is that each tree holds a large number of nodes. The more nodes in each tree, the fewer the number of trees that need to be stored in the tree, and therefore the less time will be spent coalescing trees on each delete-min.
At the same time, the Fibonacci heap tries to make the decrease-key operation as fast as possible. To do this, Fibonacci heaps allow subtrees to be "cut out" of other trees and moved back up to the root. This makes decrease-key faster, but makes each tree hold fewer nodes (and also increases the number of trees). There is therefore a fundamental tension in the structure of the design.
To get this to work, the shape of the trees in the Fibonacci heap have to be somewhat constrained. Intuitively, the trees in a Fibonacci heap are binomial trees that are allowed to lose a small number of children. Specifically, each tree in a Fibonacci heap is allowed to lose at most two children before that tree needs to be "reprocessed" at a later step. The marking step in the Fibonacci heap allows the data structure to count how many children have been lost so far. An unmarked node has lost no children, and a marked node has lost one child. Once a marked node loses another child, it has lost two children and thus needs to be moved back to the root list for reprocessing.
The specifics of why this works are documented in many introductory algorithms textbooks. It's not obvious that this should work at all, and the math is a bit tricky.
Hopefully this provides a useful intuition!
A node is marked when one of its child nodes is cut because of a decrease-key. When a second child is cut, the node also cuts itself from its parent. Marking is done so that you know when the second cut occurs.
Good Explanation from Wiki: Operation decrease key will take the node, decrease the key and if the heap property becomes violated (the new key is smaller than the key of the parent), the node is cut from its parent. If the parent is not a root, it is marked. If it has been marked already, it is cut as well and its parent is marked. We continue upwards until we reach either the root or an unmarked node. Now we set the minimum pointer to the decreased value if it is the new minimum.