do i have to have a root node? what am i going to do with this class?

that's exactly why i'm confused too. In a node, there's a root for the info and a left and right link. and what i know is that a binary tree uses nodes.

In this exercise, this is what was given...

yes, i know. but i don't know how

I don't think i've fix it.

here's what i did



public void BTreeInsert(BinaryTree myTree, int newItem) {
if ( myTree == null ) {
myTree = new BinaryTree(newItem);
} else {

public void BTreeInsert(BinaryTree myTree, int newItem) {
if ( myTree == null ) {
myTree = new BinaryTree(newItem);
} else {
if ( newItem < myTree.root) {
...

Hello every one.
Basically i am trying to construct a binary tree. but there's a problem. When i run the program, there is an error. somewhere in the treeInsert method(). I am having difficulty...

pardon me, you said treeInsert(node). but mine is treeInsert(newItem). Am i wrong in doing this? because further you said treeInsert(node.left) which i can't do.. are your saying that it should be...

Recursions the part where i break up.. the above method i posted should be recursive but it seems that isn't. I'm having a hard time figuring out because there are a lot of considerations in...

No, the tree isn't for sorting. I want to put the numbers example 1,2,3,4,5 in the binary tree. then it should be like this:

1, is in the root.
2, is in the left of the root
3, is in the right...

the structure that i want to use is node or link..
do i use recursion on this?

I have a problem:
1.) How do you insert a sequence of integers into a binary tree? does anybody have an algorithm for it?

like input : 1, 5, 3, 4 ,2

binary tree:
1
...

I guess it's just the matter of what you like, and on which is a lot easier. I have learn buffered reader. My prof. didn't teach us scanner. I heard about it from others that it is much easier to...

Here's my idea, i have 2 queues, in which i kepp track of them by:
1st queue:
front and rear;

2nd queue:
front2 and rear.

before i had read your reply stephen84s, i decided to have two...

ok i made revisions on the code i just posted:


public void compareTo(int item){
QueueNode temp = front;
while(temp != null){
if (temp.info > item)
temp.info = item;
...

here is what i had so far:


public int compareTo(int x){
QueueNode temp = front;
while(temp != null){
if (temp.info > x)
temp.info = new queueNode(x);
...

In a queue, where elements in integers are added, how do i do this:

Inside the queue:

1 2 3 5 6 8

when I want to add a number, say 4, the output should be like:

1 2 3 4 5 6 8

Thanks!..:)

Yes!.. :)




public String toString(){
String s = "[ ";
if (front.info != null){
s += "( " + peek().toString() + " )" + " ]";
}else{

When you put an item on a queue, how do you show the items inside the queue? do you use peek method? the situation is a linked queue...

I can't do that, cause my job has to be linked list. :)
I'm pretty sure i am clear of what i want, but can't understand how to do it.. :|
thanks for your replies.

here's the edited code;


...

That's what im trying to do in the first place. :)



i did this:

public Node[] stack; // this is supposedly my array
stack = new Node[size2]; // this is what i had to...

Actually , i also didn't undertsand much about the T[] and oldT[] discussed int our lectures. As i know, T[] refers to the new tops that are computed using the garwick also in allocating memery....

Because i was thinking that if the index is 1, then it would refer to the stack 1 and it would push the item in that stack.

Those T[], B[], and oldT[], keeps track of which stack you want, either stack1, stack2, and stack3.
If i chose stack1 to put an item, it pushes and is ok,but if i try to pop the item in which i put...

when you put an item in the shelf, is done by push method, and then try to retreive the item, done by pop, i doesn't have error, only just print's stack empty.. because i did specify in the code that...

T[] refers to he top and oldT[] is the old top.. becuase in the garwick's algo, oldT will get new T when new base pointers, represented as B[], are computed.

hello.. i really need help.. I don't know why i keep on getting an empty stack if i pop a token from stack even if user has already push a token into it..


public class MStack implements Stack{
...

i mean how can a link stack be in a sequential vector/ isn't a vector here an array? as in sequential representation of stacks are in array and when you say linked representation, its node. for me,...

can anyone clarify to me what this means? it's kind of comfusing..:)
linked stack embedded in a sequencial vector

here is what i did to compute the formula:


public void computeBase(){
int n = size;
int m = 3;
for (int i = 0; i <= 3; i++){
Base[i]= Math.floor(n/m) * i -1;
Object base =...

i think the code i posted isn't any good. (frown)

HAPPY NEW YEAR! every one.. hehe

I want to greet everyone here a happy new year!:)

that's supposedly be like this:
B[i] = [n/m] * i - 1,
Math.floor(n/m) => it should be th floor of the result..
i understand how to compute for the base pointers, but how do i get into which stack?...

Thanks fro that

hello.
can you change my user name to "ezkonekgal"

a lot of people know me by my current username, especially my classmate.. I don't want them to recognize me so much when i post to forums...

This is Garwick's algorithm:
1. Strip all the stacks of unused cells and consider all of the unused cells as comprising the available or free space.

2. Reallocate one to ten percent of the...

To have the bookshelf, i first created and array and this array has a stack. My problem is that i have to subdivide this array to have a multiple stack using base pointer.. I can't quite get the base...

This is what i have so far:



import java.io.*;
import java.util.*;

public class MStack {

/* Default length of the array */

A book shop has bookshelves with adjustable dividers. When one divider becomes full, the divider could be adjusted to make space. Create a Java program that will reallocate bookshelf space using...

How do you change your user i.d in this site?

pls help