Question

4. Instead of using a linked list to resolve collisions, as in separate chaining, use a binary search tree. That is, create a hash table that is an array of trees. To display a small tree-based hash table, you could use an inorder traversal of each tree. The advantage of a tree over a linked list is that it can be searched in O(logN) instead of O(N) time. This time savings can be a significant advantage if very high load factors are encountered. Checking 15 items takes a maximum of 15 comparisons in a list but only 4 in a tree. Duplicates can present problems in both trees and hash tables, so add some code that prevents a duplicate key from being inserted in the hash table. (Beware: The find()method in Tree assumes a non-empty tree.) To shorten the listing for this program, you can forget about deletion, which for trees requires a lot of MAC286, Assignment 2 Due: May 27, 2019 50 points code. Write algorithm for the scenario. You can write a program for this scenario as well for 5more bonus points. (15 points)

for java

Answer 1

1. Create a object for BST (binary tree item) with key and value as data and left and right as a variable which points to node of the BST type (left pointing to the left child of current node and right pointing to the right child of current node).
2. Create another object BST (for binary tree) with head having the address of root element of binary tree.
3. Now create an array of binary tree of some certain size(10, in this case).
4. A menu is displayed on the screen.
5. User must choose one option from four choices given in the menu.
6. 1st choice: Inserting item into hash table
(a) Take a key and a value as input.
(b) Calculate index as per the given key.
(c) Locate the binary tree at the calculated array index.
(d) If binary tree is present, check whether the key already present in it by finding the node containing key using find() method which returns NULL if key is not present. In that case, we will create a data item and add it into binary tree. If key is already present in the tree, we will update the value of node containing key.
(e) If there is no tree at that particular array index, which means the key is not present in hash table, then create a data item and a binary tree and add the data item to it using insert_element() function.
7. 2nd choice: Removing a key from hash table
(a) Take a key as input which needs to be removed.
(b) Calculate index as per the given key.
(c) Locate the binary tree at the calculated array index.
(d) If tree is present, find the node containing the key in binary tree using find() method which returns NULL if key not present. The key is removed using delete() method if it is present otherwise ‘Key not present’ message will get displayed.
(e) If there is no binary tree at that particular array index, which means the key is not present in hash table, therefore no deletion is required.
8. 3rd choice: Size of hash table
(a) Each time we add a new data item into the hash table, we increment its size by 1.
(b) The size of the hash table can be determined either by size variable or size_of_hashtable() method.

9. 4th choice: Display hash table
(a) Function display() runs for displaying hash table contents.
(b) Here a for loop runs from 0 to array_size-1 with i as iterator.
(c) The code inside loop consists of taking i as index and locating each binary tree at that index of array.
(d) As the content of each node of tree is displayed the variables left and right (pointers) is used to proceed further until all elements are displayed.