Hey I'm a junior student major in computer science, I want to ask is the final tree will be the same when you are using successor or predecessor when deleting in red-black tree? Thanks.
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I'm having trouble conceptualising the process of deletion from a splay tree. Given this intial, tree, I want to delete the node 78.
Based on the information from my course (derived from Goodrich, Tamassia and Goldwasser), the deleted node in a BST should be replaced by the next node reached by performing an in-order traversal from the node which should be 91. This node should then be splayed to the top of the tree. However, this is not the case as shown on this visualiser here. https://www.cs.usfca.edu/~galles/visualization/SplayTree.html
The visualizer replaced 78 by its in order predecessor (70) instead and splayed that node. (The in order successor, i.e., the next key in sorted order is 83, not 91.) In general, splay trees are wonderfully malleable and as long as you approximately halve the length of the path you just descended while making every other path at most a little bit longer, you’re doing it right from an asymptotic performance standpoint (your professor may have different ideas, however).
Your textbook description:
the deleted node in a BST should be replaced by the next node reached by performing an in-order traversal from the node which should be 91
That description applies to unbalanced BST (binary search trees) but does not apply to most of the various kinds of balanced binary trees, and also does not apply to Splay Trees. To delete a node in a splay tree do the following:
Splay the node to be deleted to the root and dispose of it. This leaves two trees, call the left tree A and the right tree B.
The new root of the recombined tree will come from A. Splay the largest (rightmost) node of A tree to its root.
Because A's new root has the greatest key in A, it has no right child. Set the right child of A's new root to B.
A is the new combined tree.
This is what the visualization at https://www.cs.usfca.edu/%7Egalles/visualization/SplayTree.html did.
You said in comments to the other answer:
So in practice, the node that you choose to replace the deleted node doesn't really matter, i.e. affect performance etc.
In the typical splay tree deletion algorithm the node to replace will be the predecessor or successor node, in key order.
The rule of thumb is to always splay whenever a specific node is accessed. Find the node to delete, then splay it to the root. Find its predecessor, then splay it to the root. There are variations where you can splay less aggressively, too.
Im trying to solve this problem:" A binary tree is given, check his pre-order visit and build a binary search tree with the same pre-order visit. Demonstrate if it is always possible, if not give an example when this is not possible."
Any help? I need to write pseudocode and give time complexity but i have a lot of doubts about building a binary-search-tree with the same pre-order visit for every possible binary tree.
If you are using the classic algorithm for inserting in a binary search tree, that is, to perform a search and on the found NULL pointer where the search was stopped to put the new node, then just to insert in an empty tree the preorder sequence will produce exactly a binary tree with exactly the given preorder sequence.
Just try. Traverse any preorder sequence and insert it in an empty tree and you will realize it.
I hope to help you. And welcome to stack overflow!
I am reading cracking the coding interview, and I have a question on the solution of the following problem:
You have two very large binary trees: T1, with millions of nodes, and T2, with hundreds of nodes. Create an algorithm to decide if T2 is a subtree of T1.
The simple solution that it suggests is to create a string representing the in-order and pre-order traversals and check if T2s pre-order/in-order traversal is substring of T1's pre-order/in-order traversal.
What I wonder is why do we need to compare both traversals? And why exactly that two traversals, why not for example in-order and post-order. And mainly won't only one traversal be enough? Say only in-order or pre-order traversal?
One traversal isn't enough. Consider the graphs 1->2->3 and 2<-1->3. If you start with node 1 and do a traversal, you encounter the nodes in the order 1, 2, 3. If you simply create a string showing the pre-order the two give the same result: 1,2,3
On the other hand, if you use a post-order, then the two will give a different result. 3,2,1 and 2,3,1
I bet for any one ordering, you can find two different trees with the same result.
So the question you need to answer for yourself for any other pair you want to look at is: would there be a tree that would give the same order for both traversals? I'm going to leave that as something to think about and come back later to see if you've got it.
I think preorder traversal with sentinel to represent null node is enough.
we can use this approach to serialize/deserialize a binary tree. That means, it is an one-to-one mapping between a binary tree to its preorder+sentinel representation.
After we get strings for both small tree and big tree. then we do a string match using kmp algorithm.
I know people are saying that we have to use both preorder and inorder (or postorder and inorder). but most of them just follow what others are saying, rather than think independently.
Perhaps it is a dumb question but I am confused with the following:
Are n-ary trees and m-way trees the same thing?
I see in various posts talk about m-way trees and other n-ary trees but I am not sure if these are the same data structures or are used for the same problems
Are they the same thing? Could someone please help me clear this out in my head?
From Wikipedia:
In graph theory, a k-ary tree is a rooted tree in which each node has
no more than k children. It is also sometimes known as a k-way tree,
an N-ary tree, or an M-ary tree. A binary tree is the special case
where k=2.
So the answer your question is: Yes, it is the same thing.
If n == m, then there is no difference.
We are dealing with a Most similar neigthbour algorithm here. Part of the algorithm involves searching in order over a tree.
The thing is that until now, we cant make that tree to be binary.
Is there an analog to in order traversal for non binary trees. Particularly, I think there is, just traversing the nodes from left to right (and processing the parent node only once?")
Any thoughts?
update
This tree will have in each node a small graph of n objects. Each node will have n children (1 per each element in the graph), each of which will be another graph. So its "kind of" a b tree, without all the overflow - underflow mechanics. So I guess the most similar in order traversal would be similar to a btree inorder traversal ?
Thanks in advance.
Yes, but you need to define what the order is. Post and Pre order are identical, but inorder takes a definition of how the branches compare with the nodes.
There is no simple analog of the in-order sequence for trees other than binary trees (actually in-order is a way to get sorted elements from a binary search tree).
You can find more detail in "The art of computer programming" by Knuth, vol. 1, page 336.
If breadth-first search can serve your purpose then you can use that.