I have a lab where I need to edit the given code for a binary heap-ordered tree into a triary heap ordered tree. (This is all in Java)
My thought for this is...
With a binary tree, you simple compare the left and right child.
For a triary tree, it seems that you should compare left child to middle child, then middle child to right child... does this sound correct?
Now to the important question.. I can NOT remember what the algorithm for a triary heap ordered tree was... I believe for a binary tree, its
parent = "child - 1 / 2"
so wouldnt triary just be...?
parent = "child - 1 / 3"
In binary tree it's not very clear when it's either left or right however in ternary search tree you can use greater than, less then and equal.
So I have seen a few examples such as
How to validate a Binary Search Tree?
http://www.geeksforgeeks.org/check-if-a-binary-tree-is-subtree-of-another-binary-tree/
They return 1, or true is a tree is null.
Expanding questions a bit - assuming I had to find if TreeSmall is subtree of TreeBig, and my TreeSmall is null, should the return value of checkSubtree(smallTree) true or false ? A true indicates TreeSmall was a tree with value of null. This does not make sense to me.
In pure computer science, null is a valid binary tree. It is called an empty binary tree. Just like an empty set is still a valid set. Furthermore, a binary tree with only a single root node and no children is also valid (but not empty). See this Stack Overflow answer for more information.
In practical implementation, there are two ways to go about it though.
Assume that a valid binary tree must have at least one node and do not allow empty trees. Each node does not have to have children. All recursive methods on this tree do not descend to the level of null. Rather, they stop when they see that the left child or right child of a node is null. This implementation works as long as you don't pass null to any place where a tree is expected.
Assume that null is a valid binary tree (formally, just the empty tree). In this implementation, you first check if the pointer is null before doing any operations on it (like checking for left/right children, etc.) This implementation works for any pointer to a tree. You can freely pass null pointers to methods that are expecting a tree.
Both ways work. The second implementation has the advantage of flexibility. You can pass null to anything that expects a tree and it will not raise an exception. The first implementation has the advantage of not wasting time descending to "child nodes" that are null and you don't have to use null checks at the beginning of every function/method which operates on a node. You simply have to do null checks for the children instead.
That depends on the application and is a question of definition.
Edit:
For example Wikipedia "defines" a BST as follows:
In computer science, a binary search tree (BST), sometimes also called an ordered or sorted binary tree, is a node-based binary tree data structure which has the following properties:
The left subtree of a node contains only nodes with keys less than the node's key.
The right subtree of a node contains only nodes with keys greater than the node's key.
The left and right subtree must each also be a binary search tree.
There must be no duplicate nodes
Let's test those for null:
left subtree doesn't exist, so there are no nodes violating this rule -> check
similiar to first -> check
if all this tests are passed this passes too, since both subtrees are null -> check
of course there aren't duplicates -> check
So by this definition null is a valid BST. You could inverse this by also requiring "there must be one root node". Which doesn't affect any of the practical properties of a BST but might in an explicit application.
Ex falso sequitur quodlibet - since "null" is nothing at all, it can be interpreted to be anything. It is really a matter of design. Some people may claim that checkSubTree() should thrown something like an IllegalArgumentException in this case. Another approach would be to introduce a special kind of typed object or instance which represents an empty tree (cf. NullObjectPattern). Such a null-object would be a tree by all accounts, e.g. EmptyTree instanceof Tree, while null instanceof Tree would always be false.
I have a set of items that are supposed to for a balanced binary tree. Each item is of the form (data,parent), data being the useful information and parent being the index of the parent node in the binary tree.
Nodes in the tree are numbered left-to-right, row-by-row, like this:
1
___/ \___
/ \
2 3
_/\_ _/\_
4 5 6 7
These elements come stored in a linked list. How should I order this list such that it's easier for me to build the tree? Each parent node will be referenced (by index) by exactly two child nodes; if I sort these by parent index, the sorting must be stable.
You can sort the list in any stable sort, according to the parent field, in increasing order.
The result will be a list like that:
[(d_1,nil), (d_2,1), (d_3,1) , (d_4,2), (d_5,2), ...(d_i,x), (d_i+1,x) ]
^
the root has no parent...
Note that in this list, since we used a stable sort - for each two pairs (d_i,x), (d_i+1,x) in the sorted list, d_i is the left leaf!
Now, you can populate the tree in breadth-first traversal,
Since it is homework - I still want you to make sure you understand everything by your own. So I do not want to "feed answer". If you have any specific question, please comment - and I will try to edit and explain the relevant parts with more details.
Bonus: The result of this organization is very common way to implement a binary heap structure, which is a complete binary tree, but for performance, we usually store it as an array, which is very similar to the output generated by this approach.
I don't think I understand what exactly are you trying to achieve. You have to write the function that inserts items in the tree. The red-black tree, for example, has the same complexity for insertions, O(log n), no matter how the input data is sorted. Is there a specific implementation that you have to use or a specific speed target that you must reach for inserts?
PS: Sounds like a homework to me :)
It sounds like you want a binary tree that allows you to go from a leaf node to its ancestors, using an array.
Usually sorting a list before putting it into a binary tree causes an unbalanced binary tree, unless you use a treap or other O(logn) datastructure.
The usual way of stashing a (complete) binary tree in an array, is to make node i have two children 2i and 2i+1.
Given this organization (not sorting but organization), you can go to a parent node from a leaf node by dividing the array index by 2 using integer arithmetic which will truncate fractions.
if your binary trees are not always complete, you'll probably be better served by forgetting about using an array, and instead using a more traditional tree structure with pointers/references.
hi i have implemented a bst in mips and i need to print this tree.
Each node has following four information
its value
parent's address
left child's address
right child's address
i should print the tree in the following format.
(x means no child)
12
8-16
x-9 13-17
x-x x-11 x-x x-x
Could you please suggest a way to implement this print method?
The ordering in which you are printing the tree is a breadth-first (level-by-level) traversal. One option would be as follows: maintain a work list, initially seeded with the root of the tree. Then, repeatedly dequeue from the work list, print the current element (or x if none is present), then add the two children to the work list. You would need some way to track when you're done traversing the tree, perhaps by counting the number of nodes first and stopping once you've printed that many nodes.
That said, since you're doing this in MIPS, one simpler option is to linearize the tree into an array, then print the array. If you number the nodes in a fashion similar to how you number the nodes in an implicit binary heap, you can recursively/iteratively walk the tree, fill in the array with the tree nodes, then walk over the array printing everything out.
Hope this helps!
As you need to print level by level of your binary tree, the most obivous way to print the information is to traverse the tree using breadth-first search method.
The rest is straightforward and shouldn't be a problem. :)
What are the main differences between a Linked List and a BinarySearchTree? Is BST just a way of maintaining a LinkedList? My instructor talked about LinkedList and then BST but did't compare them or didn't say when to prefer one over another. This is probably a dumb question but I'm really confused. I would appreciate if someone can clarify this in a simple manner.
Linked List:
Item(1) -> Item(2) -> Item(3) -> Item(4) -> Item(5) -> Item(6) -> Item(7)
Binary tree:
Node(1)
/
Node(2)
/ \
/ Node(3)
RootNode(4)
\ Node(5)
\ /
Node(6)
\
Node(7)
In a linked list, the items are linked together through a single next pointer.
In a binary tree, each node can have 0, 1 or 2 subnodes, where (in case of a binary search tree) the key of the left node is lesser than the key of the node and the key of the right node is more than the node. As long as the tree is balanced, the searchpath to each item is a lot shorter than that in a linked list.
Searchpaths:
------ ------ ------
key List Tree
------ ------ ------
1 1 3
2 2 2
3 3 3
4 4 1
5 5 3
6 6 2
7 7 3
------ ------ ------
avg 4 2.43
------ ------ ------
By larger structures the average search path becomes significant smaller:
------ ------ ------
items List Tree
------ ------ ------
1 1 1
3 2 1.67
7 4 2.43
15 8 3.29
31 16 4.16
63 32 5.09
------ ------ ------
A Binary Search Tree is a binary tree in which each internal node x stores an element such that the element stored in the left subtree of x are less than or equal to x and elements stored in the right subtree of x are greater than or equal to x.
Now a Linked List consists of a sequence of nodes, each containing arbitrary values and one or two references pointing to the next and/or previous nodes.
In computer science, a binary search tree (BST) is a binary tree data structure which has the following properties:
each node (item in the tree) has a distinct value;
both the left and right subtrees must also be binary search trees;
the left subtree of a node contains only values less than the node's value;
the right subtree of a node contains only values greater than or equal to the node's value.
In computer science, a linked list is one of the fundamental data structures, and can be used to implement other data structures.
So a Binary Search tree is an abstract concept that may be implemented with a linked list or an array. While the linked list is a fundamental data structure.
I would say the MAIN difference is that a binary search tree is sorted. When you insert into a binary search tree, where those elements end up being stored in memory is a function of their value. With a linked list, elements are blindly added to the list regardless of their value.
Right away you can some trade offs:
Linked lists preserve insertion order and inserting is less expensive
Binary search trees are generally quicker to search
A linked list is a sequential number of "nodes" linked to each other, ie:
public class LinkedListNode
{
Object Data;
LinkedListNode NextNode;
}
A Binary Search Tree uses a similar node structure, but instead of linking to the next node, it links to two child nodes:
public class BSTNode
{
Object Data
BSTNode LeftNode;
BSTNode RightNode;
}
By following specific rules when adding new nodes to a BST, you can create a data structure that is very fast to traverse. Other answers here have detailed these rules, I just wanted to show at the code level the difference between node classes.
It is important to note that if you insert sorted data into a BST, you'll end up with a linked list, and you lose the advantage of using a tree.
Because of this, a linkedList is an O(N) traversal data structure, while a BST is a O(N) traversal data structure in the worst case, and a O(log N) in the best case.
They do have similarities, but the main difference is that a Binary Search Tree is designed to support efficient searching for an element, or "key".
A binary search tree, like a doubly-linked list, points to two other elements in the structure. However, when adding elements to the structure, rather than just appending them to the end of the list, the binary tree is reorganized so that elements linked to the "left" node are less than the current node and elements linked to the "right" node are greater than the current node.
In a simple implementation, the new element is compared to the first element of the structure (the root of the tree). If it's less, the "left" branch is taken, otherwise the "right" branch is examined. This continues with each node, until a branch is found to be empty; the new element fills that position.
With this simple approach, if elements are added in order, you end up with a linked list (with the same performance). Different algorithms exist for maintaining some measure of balance in the tree, by rearranging nodes. For example, AVL trees do the most work to keep the tree as balanced as possible, giving the best search times. Red-black trees don't keep the tree as balanced, resulting in slightly slower searches, but do less work on average as keys are inserted or removed.
Linked lists and BSTs don't really have much in common, except that they're both data structures that act as containers. Linked lists basically allow you to insert and remove elements efficiently at any location in the list, while maintaining the ordering of the list. This list is implemented using pointers from one element to the next (and often the previous).
A binary search tree on the other hand is a data structure of a higher abstraction (i.e. it's not specified how this is implemented internally) that allows for efficient searches (i.e. in order to find a specific element you don't have to look at all the elements.
Notice that a linked list can be thought of as a degenerated binary tree, i.e. a tree where all nodes only have one child.
It's actually pretty simple. A linked list is just a bunch of items chained together, in no particular order. You can think of it as a really skinny tree that never branches:
1 -> 2 -> 5 -> 3 -> 9 -> 12 -> |i. (that last is an ascii-art attempt at a terminating null)
A Binary Search Tree is different in 2 ways: the binary part means that each node has 2 children, not one, and the search part means that those children are arranged to speed up searches - only smaller items to the left, and only larger ones to the right:
5
/ \
3 9
/ \ \
1 2 12
9 has no left child, and 1, 2, and 12 are "leaves" - they have no branches.
Make sense?
For most "lookup" kinds of uses, a BST is better. But for just "keeping a list of things to deal with later First-In-First-Out or Last-In-First-Out" kinds of things, a linked list might work well.
The issue with a linked list is searching within it (whether for retrieval or insert).
For a single-linked list, you have to start at the head and search sequentially to find the desired element. To avoid the need to scan the whole list, you need additional references to nodes within the list, in which case, it's no longer a simple linked list.
A binary tree allows for more rapid searching and insertion by being inherently sorted and navigable.
An alternative that I've used successfully in the past is a SkipList. This provides something akin to a linked list but with extra references to allow search performance comparable to a binary tree.
A linked list is just that... a list. It's linear; each node has a reference to the next node (and the previous, if you're talking of a doubly-linked list). A tree branches---each node has a reference to various child nodes. A binary tree is a special case in which each node has only two children. Thus, in a linked list, each node has a previous node and a next node, and in a binary tree, a node has a left child, right child, and parent.
These relationships may be bi-directional or uni-directional, depending on how you need to be able to traverse the structure.
Linked List is straight Linear data with adjacent nodes connected with each other e.g. A->B->C. You can consider it as a straight fence.
BST is a hierarchical structure just like a tree with the main trunk connected to branches and those branches in-turn connected to other branches and so on. The "Binary" word here means each branch is connected to a maximum of two branches.
You use linked list to represent straight data only with each item connected to a maximum of one item; whereas you can use BST to connect an item to two items. You can use BST to represent a data such as family tree, but that'll become n-ary search tree as there can be more than two children to each person.
A binary search tree can be implemented in any fashion, it doesn't need to use a linked list.
A linked list is simply a structure which contains nodes and pointers/references to other nodes inside a node. Given the head node of a list, you may browse to any other node in a linked list. Doubly-linked lists have two pointers/references: the normal reference to the next node, but also a reference to the previous node. If the last node in a doubly-linked list references the first node in the list as the next node, and the first node references the last node as its previous node, it is said to be a circular list.
A binary search tree is a tree that splits up its input into two roughly-equal halves based on a binary search comparison algorithm. Thus, it only needs a very few searches to find an element. For instance, if you had a tree with 1-10 and you needed to search for three, first the element at the top would be checked, probably a 5 or 6. Three would be less than that, so only the first half of the tree would then be checked. If the next value is 3, you have it, otherwise, a comparison is done, etc, until either it is not found or its data is returned. Thus the tree is fast for lookup, but not nessecarily fast for insertion or deletion. These are very rough descriptions.
Linked List from wikipedia, and Binary Search Tree, also from wikipedia.
They are totally different data structures.
A linked list is a sequence of element where each element is linked to the next one, and in the case of a doubly linked list, the previous one.
A binary search tree is something totally different. It has a root node, the root node has up to two child nodes, and each child node can have up to two child notes etc etc. It is a pretty clever data structure, but it would be somewhat tedious to explain it here. Check out the Wikipedia artcle on it.