Building Binary Searh Tree with multiple types (hint hint Generics?)
I have built binary search trees before and even wanted to build a B+ tree for primary indexes but my Professor suggested limiting to BST.
My question is can I build a BST with a generic type parameterized code using Comparable and Generics to help me insert numeric and non-numeric (String) type nodes using the requirement that anything less than the root go left, greater than or equal to the root go right. There will be no duplicates.
This is definitely new territory for me so any help is appreciated.
Always Learning wrote:
I have a Comparable<Object> x I would like to compareTo(Some other Comparable<Object>) but I know I am going about this all wrong.
Essentially I want to be able to compare String lexicographically or integers/floats in order to properly insert them into a binary search tree and I wanted to use the Java language constructs of the Comparable interface and possibly generics to do it.
This would result in a simple String1.compareTo(String2) or int1.compareTo(int2) etc.So it sounds like you actually have a Comparable<T>; or possibly a Comparable<String>, where all your types are first converted to Strings (however, in the case of ints, that probably means adding leading '0's).
The generics tutorial suggests that you use Comparable<? super T> in cases of arbitrarily comparable items; however, I think you first need to decide how your comparisons are going to be done. For example, how does a String compare to an Integer in your scenario?
Winston
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Hi,
I have a requirement where i have to show a af:tree with child records from two different view Iterators, I created one tree but facing a issue. It shows the name of the Child view iterators and then the record inside it.
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Vectors with multiple types of objects
Ok, this is what i need. I have a main class, EquationVector, and i have four smaller classes, MathPrimitive, which is abstract and all of the following classes extend, Number which just holds numbers, Operator which takes care of what type it is and what to do, and Variable, which knows its value, its coef and its power. I use the MathPrimitive class because all of the other classes need to have a common getType() method, plus it holds static values for all the possible types. What i need is some sort of storage method that is dynamically resizeable, and can handle multiple types of objects. I tried vector, but it doesnt work, even when i cast. I want to be able to do something to the effect of
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.... -
Referencing another class with multiple types.
Here's my code:
import java.io.*;
import java.util.Scanner;
class CH3_13
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// Invoice MyInvoice = new Invoice();
/* myInvoice.displayInvoice();*/ }
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{ private String partNumber;
private String partDescription;
int quantity;
double pricePerItem;
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... because it seems I cannot reference it because the Invoice class has multiple types defined in it. As this is the case, is there anyway, I'd be able to get that line to compile?I am not sure of this is what you are looking for but I would definitely suggest some real refactoring here. First of all the part no. name description donot belong to the invoice class so they need to go into a different class named part no. Secondly you need to identify the relationship which I would say would be has a relationship between invoice and part. A many to many relationship. SInce its many to many I would guess it would need to be biderectional associaition.
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Hi,
Is there any way, so that I can specify multiple "Type" to Get All Descendents Method invoke node.
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@ tst
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That option I had figured out, but I need the references of All Descendents (Folders and VIs) in the order of they appear in the project.
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Hi, I'm using Jdev 9i.
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<FORM ACTION="target.jsp" >
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<TR>
<TD>
DeptNo
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<TD>
DName
</TD>
<TR>
<%
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%>
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<%
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String sDname = "Dname" + nRow;
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%>
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<TR>
<TD>
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<TD>
<input type="text" name="<%=sDname%>" value='<jbo:ShowValue dataitem="Dname" />' />
<input type="text" name="<%=sRowKey%>" value='<jbo:ShowValue dataitem="RowKey" />' />
</TD>
<TR>
</jbo:Row>
</jbo:RowsetIterate>
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</TABLE>
</FORM>
NOTE: This should get you started!!! -
Tree with multiple root nodes does not display
I am trying to create a tree to display some hierarchical data which contains several root nodes.
I have specifed that the TREE_ROOT value is always Null as the root nodes have a null value for the PARENT_ID. The page displays yet none of the data is displayed.
I have manage to get it to work by creating a dummy single root that I reference from the existing root values?Phattam,
The TREE_ROOT identifies the root node's ID, not its PARENT_ID.
For example, if we build a tree on the oehr_employees table, the ID is EMPLOYEE_ID and the PARENT_ID is MANAGER_ID. therefore, to identify the start of the tree, we could populate the TREE_ROOT with a select statement e.g.
SELECT employee_id FROM oehr_employees WHERE employee_id = 100This identifies a single value and the tree starts from employee_id 100.
If we used a select statement that retrieves more than one row, e.g.
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Hope this helps,
Bryan. -
Burning CDs with multiple types of files
How do you burn iTunes songs and pdf documents to the same disk?
This would be done on My MacBook, not a Nano. Haha
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Creating Profile type report that hold fields with multiple values
Really hoping someone can please help me out as I am very new to Crystal Reports.
We use Maximizer CRM and we have been in need of some custom reports to rule out risk for regulators. I contacted Max and they suggested the only possible way is to create through Crystal. Its been almost one month already and I still cannot for the likes of me get this report operating properly. I have been inside and out on all sorts of forums, posted topics but no luck! So I will give it one more attempt in hopes that one of you geniuses can show me the way.
In Maximizer CRM there is date, numeric, alphanumric and table. Our table fields items can be set to either single value or multi-value. So in crystal i did a default join of Client.tbl and the user-defined fields from view and joined the client id and contact number from all the view fields to client table. See Image:
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The formula field you see in the image is from when I posted a discussion and Abhilash assisted me by providing the formulas I should add:
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whileprintingrecords;
stringvar s := s + {field_with_multiple_values} + ", ";
2) Next, move all the fields (except the formula field above) from the Details section to the Report Footer
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whileprintingrecords;
stringvar s;
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This method is not working out for this type of report. When I add the formula Crystal is still counting my 2 records as 5 records but I can only view it as a single record and the multi-field has all the values for both records and displaying as a single record. See image:
Can anyone please assist and advise where I am going wrong?
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Thanks for taking down memory lane that is Maximizer. Nice to see their table structure hasn't been simplified in the last 20 years.
If I understand what's happening, you should only see 2 records and not 5. That means your joins are creating duplicate records. For now I'm going to skip over trying to optimize your query because I still have bad dreams of linking Maximizer tables.
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Brian -
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I would be grateful if you someone could explain how to handle such binary file stream with LabView and send an example to i
llustrate it.
Many thanks in advance for your help.
Donat-Pierre
Attachments:
Demo.acq 248 KB
SCR_EKG_test1b.acq 97 KBThe reading of double is also straight forward : just use a dble float wired to the type cast node, after inverting the string (indian conversion).
See the attached example.
The measure of skin thickness is based on OCT (optical coherent tomography = interferometry) : an optical fiber system send and received light emitted to/back from the skin at a few centimeter distance. A profile of skin structure is then computed from the optical signal.
CC
Chilly Charly (aka CC)
E-List Master - Kudos glutton - Press the yellow button on the left...
Attachments:
Read_AK_time_info.vi.zip 9 KB -
How to build a form with multiple tables in oracle application express
Hi everyone,
I have got problem in building a form with multiple tables.I have a main table with (20) columns and this main table is related to the other tables with the primary key-foreign key relation ship.My requirement is i have to build a form which has fields from many tables and all the fields are related to the main table using (ID) column.In that form if i enter ID field i have to get information from differnt tables.
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Thank you
sansSans,
I am no Apex expert, but with a situation as "complex" as yours, have you thought about creating a VIEW that joins these 7/8 tables, placing an INSTEAD OF trigger on that view to do all the business logic in the database, and base your application on the view?
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ORACLE EXPRESS: build a page with multiple forms linked to one table
hi,
im using oravle application express. APEX
i would like to build a page with multiple forms linked to one table (orders) , the page has 4 from each one with different order_id number (depending on filtering), and if the order is prepared click yes for each order and this 'YES' should be UPDATED AND SAVED to each order number in the same table with the press of one button.
i created all the form as (sql query)
and create one update process
(UPDATE ORDERS
SET TRAY_PREPARED =:P10_TRAY_PREPARED_1
WHERE ORDER_ID =:P10_ORDER_ID_1;
UPDATE ORDERS
SET TRAY_PREPARED =:P10_TRAY_PREPARED_2
WHERE ORDER_ID =:P10_ORDER_ID_2;
UPDATE ORDERS
SET TRAY_PREPARED =:P10_TRAY_PREPARED_3
WHERE ORDER_ID =:P10_ORDER_ID_3;
UPDATE ORDERS
SET TRAY_PREPARED =:P10_TRAY_PREPARED_4
WHERE ORDER_ID =:P10_ORDER_ID_4;
i dont know really if i can do that, but it appear hat it actually saving according to order_id number , but not all the time some time it saved the value as "null".
please guide me what is the correct way to do this.
I READ THIS ONE
http://stackoverflow.com/questions/7877396/apex-creating-a-page-with-multiple-forms-linked-to-multiple-related-tables
BUT IT WAS FOR MULTIPLE INSERT
thanks.Sans,
I am no Apex expert, but with a situation as "complex" as yours, have you thought about creating a VIEW that joins these 7/8 tables, placing an INSTEAD OF trigger on that view to do all the business logic in the database, and base your application on the view?
This is the "thick-database" approach that has been gaining momentum of late. The idea is to put your business logic in the database wherever possible, and let the application (Form, Apex, J2EE, whatever) concentrate on UI issues, -
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Hi all, just having troubles with a program im writing.
My program is based on a binary search tree full of items which are passed in via an input file and then saved to an output file.
I have written a sellItem method which gets passed in the item and quantity that has been sold which then needs to be changed in the binary tree.
Here if my sell Item method:
public void sellItem(Item item, int quantity){
stockItem = item;
mQuantity = quantity;
if (tree.includes(stockItem)){
int tempQuantity = stockItem.getQuantityInStock();
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ABCD, PENCIL, 1, 0.35, 200, 100, 200
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Therefore if i pass 40 into the method the binary search tree should then store the quantity as 160.
below is a copy of my binary tree :
public class BSTree extends Object {
private class TreeNode extends Object{
public Comparable data;
public TreeNode left;
public TreeNode right;
public TreeNode() {
this(null);
public TreeNode(Comparable barCode){
super();
data = barCode;
left = null;
right = null;
private TreeNode root;
private int nodeCount;
public BSTree(){
super();
root = null;
nodeCount = 0;
public boolean isEmpty() {
return root == null;
public int size() {
return nodeCount;
private TreeNode attach(TreeNode newPointer, TreeNode pointer){
if (pointer == null){
nodeCount++;
return newPointer;
else {
Comparable obj1 = (Comparable) newPointer.data;
Comparable obj2 = (Comparable) pointer.data;
if (obj1.compareTo(obj2) < 0) //Go left
pointer.left = attach(newPointer, pointer.left);
else //Go right
pointer.right = attach(newPointer, pointer.right);
return pointer;
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//Create a new node and initialize
TreeNode newPointer = new TreeNode(item);
//Attach it to the tree
root = attach(newPointer, root);
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TreeNode pointer;
TreeNode parent;
//Find the node to be removed
parent = null;
pointer = root;
while ((pointer != null) && !key.equals(pointer.data)) {
parent = pointer;
if (key.compareTo(pointer.data) <0)
pointer = pointer.left;
else
pointer = pointer.right;
if (pointer == null)
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TreeNode leftSubtree = pointer.left;
TreeNode rightSubtree = pointer.right;
if (parent == null)
root = null;
else if (key.compareTo(parent.data) < 0)
parent.left = null;
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//Reattaching any orphans in the left subtree
if (leftSubtree != null) {
root = attach(leftSubtree, root);
nodeCount--;
//Reattaching any orphans in the right subtree
if (rightSubtree != null) {
root = attach(rightSubtree, root);
nodeCount--;
nodeCount--;
return pointer.data;
private TreeNode search(TreeNode pointer, Comparable key) {
if (pointer == null)
return null;
else if (pointer.data.compareTo(key) == 0)
return pointer;
else if (key.compareTo(pointer.data) < 0)
return search(pointer.left, key);
else
return search(pointer.right, key);
public boolean includes(Comparable key) {
return (search(root, key) != null);
public Comparable retrieve(Comparable key) {
TreeNode pointer;
pointer = search(root, key);
if (pointer == null)
return null;
else
return pointer.data;
public Comparable[] getAllInOrder() {
Comparable[] list = new Comparable[nodeCount];
inOrderVisit(root,list,0);
return list;
private int inOrderVisit(TreeNode pointer, Comparable[] list, int count) {
if (pointer != null) {
count = inOrderVisit(pointer.left, list, count);
list[count++] = pointer.data;
count = inOrderVisit(pointer.right, list, count);
return count;
public String toString() {
StringBuffer result = new StringBuffer(100);
inOrderString(root, result);
return result.toString();
private void inOrderString(TreeNode pointer, StringBuffer result) {
if (pointer != null) {
inOrderString(pointer.left, result);
result.append(pointer.data.toString() + "\n");
inOrderString(pointer.right, result);
}Thanks for everyones help. Keep in mind i'm very new to java.Hi all, just having troubles with a program im writing.
My program is based on a binary search tree full of items which are passed in via an input file and then saved to an output file.
I have written a sellItem method which gets passed in the item and quantity that has been sold which then needs to be changed in the binary tree.
Here if my sell Item method:
public void sellItem(Item item, int quantity){
stockItem = item;
mQuantity = quantity;
if (tree.includes(stockItem)){
int tempQuantity = stockItem.getQuantityInStock();
tempQuantity -= mQuantity;
else{
throw new IllegalArgumentException("The Barcode " + mBarCode + " does NOT exist.");
}and here is where i am calling it in a test class :
number1.sellItem(item1, 40);Each item is in this format :
ABCD, PENCIL, 1, 0.35, 200, 100, 200
where 200 is the quantity.
Therefore if i pass 40 into the method the binary search tree should then store the quantity as 160.
below is a copy of my binary tree :
public class BSTree extends Object {
private class TreeNode extends Object{
public Comparable data;
public TreeNode left;
public TreeNode right;
public TreeNode() {
this(null);
public TreeNode(Comparable barCode){
super();
data = barCode;
left = null;
right = null;
private TreeNode root;
private int nodeCount;
public BSTree(){
super();
root = null;
nodeCount = 0;
public boolean isEmpty() {
return root == null;
public int size() {
return nodeCount;
private TreeNode attach(TreeNode newPointer, TreeNode pointer){
if (pointer == null){
nodeCount++;
return newPointer;
else {
Comparable obj1 = (Comparable) newPointer.data;
Comparable obj2 = (Comparable) pointer.data;
if (obj1.compareTo(obj2) < 0) //Go left
pointer.left = attach(newPointer, pointer.left);
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pointer.right = attach(newPointer, pointer.right);
return pointer;
public void insert(Comparable item){
//Create a new node and initialize
TreeNode newPointer = new TreeNode(item);
//Attach it to the tree
root = attach(newPointer, root);
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TreeNode pointer;
TreeNode parent;
//Find the node to be removed
parent = null;
pointer = root;
while ((pointer != null) && !key.equals(pointer.data)) {
parent = pointer;
if (key.compareTo(pointer.data) <0)
pointer = pointer.left;
else
pointer = pointer.right;
if (pointer == null)
return null;
//Orphans
TreeNode leftSubtree = pointer.left;
TreeNode rightSubtree = pointer.right;
if (parent == null)
root = null;
else if (key.compareTo(parent.data) < 0)
parent.left = null;
else
parent.right = null;
//Reattaching any orphans in the left subtree
if (leftSubtree != null) {
root = attach(leftSubtree, root);
nodeCount--;
//Reattaching any orphans in the right subtree
if (rightSubtree != null) {
root = attach(rightSubtree, root);
nodeCount--;
nodeCount--;
return pointer.data;
private TreeNode search(TreeNode pointer, Comparable key) {
if (pointer == null)
return null;
else if (pointer.data.compareTo(key) == 0)
return pointer;
else if (key.compareTo(pointer.data) < 0)
return search(pointer.left, key);
else
return search(pointer.right, key);
public boolean includes(Comparable key) {
return (search(root, key) != null);
public Comparable retrieve(Comparable key) {
TreeNode pointer;
pointer = search(root, key);
if (pointer == null)
return null;
else
return pointer.data;
public Comparable[] getAllInOrder() {
Comparable[] list = new Comparable[nodeCount];
inOrderVisit(root,list,0);
return list;
private int inOrderVisit(TreeNode pointer, Comparable[] list, int count) {
if (pointer != null) {
count = inOrderVisit(pointer.left, list, count);
list[count++] = pointer.data;
count = inOrderVisit(pointer.right, list, count);
return count;
public String toString() {
StringBuffer result = new StringBuffer(100);
inOrderString(root, result);
return result.toString();
private void inOrderString(TreeNode pointer, StringBuffer result) {
if (pointer != null) {
inOrderString(pointer.left, result);
result.append(pointer.data.toString() + "\n");
inOrderString(pointer.right, result);
}Thanks for everyones help. Keep in mind i'm very new to java.
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