JAVA PLEASE : Assume the Tree consists of your ID digits, inserted in the fashion to minimize the height of the tree. Remove duplicates if needed. Trace the delete() method above in a similar fashion as the insert was traced, delete 3 last digits of your ID from the tree in the same order as they are present in you ID.
JAVA PLEASE : Assume the Tree consists of your ID digits, inserted in the fashion to minimize the height of the tree. Remove duplicates if needed. Trace the delete() method above in a similar fashion as the insert was traced, delete 3 last digits of your ID from the tree in the same order as they are present in you ID.
Database System Concepts
7th Edition
ISBN:9780078022159
Author:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan
Publisher:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan
Chapter1: Introduction
Section: Chapter Questions
Problem 1PE
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Question
JAVA PLEASE :
Assume the Tree consists of your ID digits, inserted in the fashion to minimize the height of the tree. Remove duplicates if needed. Trace the delete() method above in a similar fashion as the insert was traced, delete 3 last digits of your ID from the tree in the same order as they are present in you ID.
![The two methods of our most interest are insert and delete. The pseudocode of these is below
boolean insert(E e)
if (root == null)
root = createNewNode(e) // Create root
else
// Locate the parent node
TreeNode<E> parent ← null
TreeNode<E> current root
while (current!= null)
if (c.compare(e, current.element) < 0)
parent ← current
current
else if (c.compare(e, current.element) > 0)
current
parent
current current.right
else
end if
end while
current.left
return false // no Duplicates! Skip it!
else
// Create the new node and attach it
if (c.compare(e, parent.element) < 0)
parent.left ← createNewNode(e)
createNewNode(e)
parent.right
end if
end if
size++
return true // Element inserted
end insert
public boolean delete(E e) {
// find the node to delete and its parent
TreeNode<E> parent - null
TreeNode<E> current root
while (current != null)
if (c.compare(e, current.element) < 0)
parent
current ←
else if (c.compare(e, current.element) > 0)
parent
current
current ← current.right
else
break // It is pointed at by current
end if
end while
if (current == null)
return false // Element is not in the tree
end if
// Case 1: current has no left child
if (current.left == null)
// Connect the parent with the right child
of current
if (parent
root
current
current.left
else
else
if (c.compare(e, parent.element) < 0)
parent.left
current.right
end if
end if
null)
current.right
parent.right
current.right
else
// Case 2: The current node has a left child
// Locate the rightmost node in the left](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Ff88c6847-e01c-43ba-a093-2849d98113a4%2Fe3db578c-7808-4143-99c0-bd334886123a%2Flxxyo6_processed.png&w=3840&q=75)
Transcribed Image Text:The two methods of our most interest are insert and delete. The pseudocode of these is below
boolean insert(E e)
if (root == null)
root = createNewNode(e) // Create root
else
// Locate the parent node
TreeNode<E> parent ← null
TreeNode<E> current root
while (current!= null)
if (c.compare(e, current.element) < 0)
parent ← current
current
else if (c.compare(e, current.element) > 0)
current
parent
current current.right
else
end if
end while
current.left
return false // no Duplicates! Skip it!
else
// Create the new node and attach it
if (c.compare(e, parent.element) < 0)
parent.left ← createNewNode(e)
createNewNode(e)
parent.right
end if
end if
size++
return true // Element inserted
end insert
public boolean delete(E e) {
// find the node to delete and its parent
TreeNode<E> parent - null
TreeNode<E> current root
while (current != null)
if (c.compare(e, current.element) < 0)
parent
current ←
else if (c.compare(e, current.element) > 0)
parent
current
current ← current.right
else
break // It is pointed at by current
end if
end while
if (current == null)
return false // Element is not in the tree
end if
// Case 1: current has no left child
if (current.left == null)
// Connect the parent with the right child
of current
if (parent
root
current
current.left
else
else
if (c.compare(e, parent.element) < 0)
parent.left
current.right
end if
end if
null)
current.right
parent.right
current.right
else
// Case 2: The current node has a left child
// Locate the rightmost node in the left
![Let's trace inserting 1,2,3,4 in this order
Step 2: //we have only root with 1, insert 2
if (root == null) false, root = 1
// Locate the parent node
TreeNode<E> parent
TreeNode<E> current
//subfree of the current node and its parent
TreeNode<E> parentOfRightMost ← current
TreeNode<E> rightMost current.left
while (current!= null), true,
Iteration 1: current is 1 is not null
while (rightMost.right != null)
parentOfRight Most - right Most
rightMost rightMost.right // go right
end while
Step 1: //the tree is empty, we need to insert 1
if (root == null) is true, so we createNewNode with value 1 and it is our root // Created root with1
size of Tree is 0+1=1
if (2-1=1) < 0, false
else if (2-1=1) > 0, true
our parent is root with 1
our current is a right child of root
// Replace current by rightMost
current.element ← rightMost.element
// Eliminate rightmost node
if (parent OfRight Most.right == rightMost)
parent Of Right Most.right
else // parentOfRight Most == current
rightMost.left
parent OfRight Most.left ← - rightMost.left
end if
size-- // Reduce the size of the tree
return true // Element deleted
end delete
null, we created empty TreeNode for root
root with value 1, current is 1
Iteration 2: current was just created, it is null, end while
if (2-1=1) < 0, false
else
parent.right createNewNode(2), we created a node with 2 and assigned as a right child of root](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Ff88c6847-e01c-43ba-a093-2849d98113a4%2Fe3db578c-7808-4143-99c0-bd334886123a%2Fxmwsgsa_processed.png&w=3840&q=75)
Transcribed Image Text:Let's trace inserting 1,2,3,4 in this order
Step 2: //we have only root with 1, insert 2
if (root == null) false, root = 1
// Locate the parent node
TreeNode<E> parent
TreeNode<E> current
//subfree of the current node and its parent
TreeNode<E> parentOfRightMost ← current
TreeNode<E> rightMost current.left
while (current!= null), true,
Iteration 1: current is 1 is not null
while (rightMost.right != null)
parentOfRight Most - right Most
rightMost rightMost.right // go right
end while
Step 1: //the tree is empty, we need to insert 1
if (root == null) is true, so we createNewNode with value 1 and it is our root // Created root with1
size of Tree is 0+1=1
if (2-1=1) < 0, false
else if (2-1=1) > 0, true
our parent is root with 1
our current is a right child of root
// Replace current by rightMost
current.element ← rightMost.element
// Eliminate rightmost node
if (parent OfRight Most.right == rightMost)
parent Of Right Most.right
else // parentOfRight Most == current
rightMost.left
parent OfRight Most.left ← - rightMost.left
end if
size-- // Reduce the size of the tree
return true // Element deleted
end delete
null, we created empty TreeNode for root
root with value 1, current is 1
Iteration 2: current was just created, it is null, end while
if (2-1=1) < 0, false
else
parent.right createNewNode(2), we created a node with 2 and assigned as a right child of root
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