What are other ways than "Natural" to define trees ? - data-structures

I am reading Data Structure and Algorithm Analysis in Java Chapter 4 - Trees.
I am quoting:
A tree can be defined in several ways. One natural way to define a
tree is recursively
What are other ways to define trees, other than natural way ?

Here several ways contrasts against recursion, not natural - meaning that there are other ways than recursion to generate a tree instance. You could use loops, for example. Recursion is natural because it's an obvious and elegant strategy for this task, as every subtree is itself a tree. Here, recursive code will be much cleaner and easier to understand (assuming you understand recursion) than a loop implementation.

You could define a tree using the terminology of graph theory:
A tree is a connected, acyclic, undirected graph.
Or you could describe it by the elements of which it consists:
A tree is a (possibly non-linear) data structure made up of nodes or vertices and edges without having any cycle. The tree with no nodes is called the null or empty tree. (Wikipedia)
Your quote refers to the fact that for a node of a tree all of its children are individual trees, hence the simple (recursive) definition.

Related

Is there a tree structure with multiple root nodes, if so, what is it called?

A classical tree has one root node. Example:
Is there a tree with multiple initial roots like in the picture?:
As #Joe Sewell noted in the comments, a collection of independent trees is called a forest. This term applies both to collections of directed rooted trees like the one you showed above, plus collections of undirected, unrooted trees as well.
Many data structures and algorithms make use of forests. The binomial and Fibonacci heap data structures store their items in a collection of smaller independent trees. Link/cut trees, which are used in some maximum flow algorithms, work with independent collections of trees as well.

Joining of binary trees

Suppose we have a set of binary trees with their inorder and preorder traversals given,and where no tree is a subtree of another tree in the given set. Now another binary tree Q is given.find whether it can be formed by joining the binary trees from the given set.(while joining each tree in the set should be considered atmost once) joining operation means:
Pick the root of any tree in the set and hook it to any vertex of another tree such that the resulting tree is also a binary tree.
Can we do this using LCA (least common ancestor)?or does it needs any special datastructure to solve?
I think a Binary tree structure should be enough. I don't think you NEED any other special data structure.
And I don't understand how you would use LCA for this. As far as my knowledge goes, LCA is used for knowing the lowest common Ancestor for two NODES in the same tree. It would not help in comparing two trees. (which is what I would do to check if Q can be formed)
My solution in words.
The tree Q that has to be checked if it can be made from the set of trees, So I would take a top-down approach. Basically comparing Q with the possible trees formed from the set.
Logic:
if Q.root does not match with any of the roots of the trees in the set (A,B,C....Z...), No solution possible.
if Q.root matches a Tree root (say A) check corresponding children and mark A as used/visited. (Which is what I understand from the question: a tree can be used only once)
We should continue with A in our solution only if all of Q's children match the corresponding children of A. (I would do Depth First traversal, Breadth First would work as well).
We can add append a new tree from the set (i.e. append a new root (tree B) only at leaf nodes of A as we have to maintain binary tree). Keep track of where the B was appended.
Repeat same check with corresponding children comparison as done for A. If no match, remove B and try to add C tree at the place where B was Added.
We continue to do this till we run out of nodes in Q. (unless we want IDENTICAL MATCH, in which case we would try other tree combinations other than the ones that we have, which match Q but have more nodes).
Apologies for the lengthy verbose answer. (I feel my pseudo code would be difficult to write and be riddled with comments to explain).
Hope this helps.
An alternate solution: Will be much less efficient (try only if there are relatively less number of trees) : forming all possible set of trees ( first in 2s then 3s ....N) and and Checking the formed trees if they are identical to Q.
the comparing part can be referred here:
http://www.geeksforgeeks.org/write-c-code-to-determine-if-two-trees-are-identical/

Data structure vocabulary about trees

What adjectives are used to characterize the following 3 data structures :
a k-ary tree where all nodes have exaclty 0 or k children ?
a tree where all leaves are at the same level ?
a tree where only leaves contain data ("empty" internal nodes) ?
I am searching for widespread and well-established adjectives whose every one in graph theory will understand.
i think this can not be done let say that we have 3-ary tree, when you put the first element the first rule is broken, so you do not have 3 children.
also can be proven that thing what you want is not possible

Can a non binary tree be tranversed in order?

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.

How to modify preorder tree traversal algorithm to handle nodes with multiple parents?

I've been searching for a while now and can't seem to find an alternative solution. I need the tree traversal algorithm in such a way that a node can have more than 1 parent, if it's possible (found a great article here: Storing Hierarchical Data in a Database). Are there any algorithms so that, starting from a root node, we can determine the sequence and dependencies of nodes (currently reading topological sorting)?
The structure you described isn't a tree, it's a directed graph. As it would be suitable for hierarchical drawing you might be tempted to think of it as a tree (which itself is an acyclic connected graph).
Typical traversal algorithms for graphs are depth-first and breadth-first. The graph implementation is only different as it records the nodes it has already visited in order to avoid visiting certain nodes multiple times. However, if your data structure guarantees that it's acyclic, you can use tree algorithms on your graph by simply treating "parents" as "children".
I made a simple sketch to illustrate what I mean (the perfect chance to try Google Docs' new drawing feature):
As you see, it's possible to treat any graph that has an acyclic directed form as a tree and apply tree algorithms on it. As soon as you can't guarantee this property you'll have to go for dedicated graph algorithms.
A tree is basically a directed unweighted graph, where each vertice has N or less edges, and no cycles can happen.
If your'e certain there are no cycles in your tree, you could just treat a parent as another child of the specified node, and preform a preorder traversal normally.
However, if cycles might happen, you need graph algorithms.
Specifically: Breadth first search.
Just checking for maybe a simple case: can the two parents have different parents?
If no you could turn them into single node (conceptually) and have a tree again.
Otherwise you will have to split the child node and duplicate a branch for the other parent.
(This can of course lead to inconsistency and/or inneficient algorithms later, depending if you will need to maintain the data structure).
The above options hold if you insist on having the tree structure, which by definition can have only one parent.
So maybe you need to step back and explain what are you trying to accomplish and why it must be a tree structure if nodes can have two parents.
You aren't describing a tree here. You can NOT call your graph a tree.
A tree is an undirected graph without cycles. Parent/child relationship is NOT an interpretation of directions drawn on the edges. They are the result of naming one vertex the root.
We name a vertex "parent" to current, because it's the next one to the path to root. All other vertexes adjacent to current one are "children".
You can't just lay out an arbitrary graph in such a way that "parents" are "above" or "point to vertex", and children are "below" or "vertex points to them". A tree is a tree because a root is picked. What you depict in your question is not a tree. And tree traversal algorithms are NOT applicable to traversing arbitrary graphs.
There are several graph traversal algorithms, such as breadth-first search or depth-first search (check side notes in those pages for more). Use them instead of trying to tie your full-featured graph into your knowledge about trees.

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