Develop Reference

Tree traversing doesn't print desired solution

(java)
I have class called Node, which has following fields:
value (integer)
connectedNodes (array of Node objects, always has same size = 2)
combination (object of Combination class)
Combination class has one field called messageContext, let's just say that it's a message which needs to be shown on the screen when something happens (described later).
Also, we have one Tree object, which has only one field: root (Node object)
Now, let's say that we have one String called combinationStr = "1121". Now, we use Tree's method called addCombination:
public void addCombination(Combination finalCombination, Node current, String combination, int counter) {
if(counter==combination.length()) {
return;
}
int value = combination.charAt(counter)-48;
if(current.connectedNodes[value-1]==null) {
current.connectedNodes[value-1] = new Node(value);
}
if(counter==combination.length()-1) {
current.combination = finalCombination;
return;
}
addCombination(finalCombination,current.connectedNodes[value-1],combination,counter+1);
}
finalCombination object is an object that is going to be assigned to the last Node's combination field, added to the Tree for one combinationStr. So, we use this function to create the Tree-like structure that has path: -1 (root) -> 1 -> 1 -> 2 -> 1
When we come to the last one, traversing the Tree, we should see message appear. This is the messageContext of finalCombination.
Okay so, now let's use while(true) loop that will let us input a number, which will be like a path-chooser. If we input 1, we will go to node 1 and have other options to choose.
While loop looks like this:
Scanner scanner = new Scanner(System.in);
Node currentNode = tree.root;
while(true) {
for(Node node: currentNode.connectedNodes) {
if(node!=null) {
System.out.print(node.value + " ");
continue;
}
System.out.print("nullnode ");
}
System.out.println("");
if(currentNode.combination!=null) {
System.out.println(currentNode.combination.messageContext);
}
if(currentNode.connectedNodes[0]==null && currentNode.connectedNodes[1]==null) {
currentNode = tree.root;
System.out.println("root");
}
int x = scanner.nextInt();
currentNode = tree.takeStep(currentNode,x);
}
So, what are we doing here is actually printing the value of current Node, then printing values of Node's we can go to. If Node doesn't exist, we print nullnode.
The takeStep() method looks like this:
public Node takeStep(Node current, int value) {
if(current.connectedNodes[value-1]!=null) {
return current.connectedNodes[value-1];
}
return this.root;
}
It just checks if there is a node we want to go to and returns that node, if it does. If it doesn't exist, it will return us to root.
But, what's the problem with this code ?
Well, look at the whole main class:
Tree tree = new Tree(new Node(-1));
String[] combination = {"1121","11","2212"};
for(String s: combination) {
Combination tempCombination = new Combination();
tempCombination.messageContext = s + " ova poruka";
tree.addCombination(tempCombination,tree.root,s,0);
tree.traverse(tree.root);
System.out.println("END");
}
Scanner scanner = new Scanner(System.in);
Node currentNode = tree.root;
while(true) {
System.out.println(currentNode.value);
for(Node node: currentNode.connectedNodes) {
if(node!=null) {
System.out.print(node.value + " ");
}
else {
System.out.print("nullnode ");
}
}
int x = scanner.nextInt();
if(currentNode.combination!=null) {
System.out.println(currentNode.combination.messageContext);
if(currentNode.connectedNodes[0]==null && currentNode.connectedNodes[1]==null) {
currentNode = tree.root;
break;
}
}
currentNode = tree.takeStep(currentNode,x);
}
When we enter number x, we will call takeStep and check if that node exists connected to current one. But the problem is: When we input 1, it prints everything normally, when we input 1 again, it prints everything normally, when we input 2, it prints everything normally... but when we input 1 again, it says there are 2 nullnodes, and for some reason it doesn't change to root. Can anyone help me please? Here are the full classes:
NODE:
public class Node {
int value;
Node[] connectedNodes = {null,null};
Combination combination;
public Node(int value) {
this.value = value;
this.combination = null;
}
}
TREE:
public class Tree {
Node root;
public Tree(Node root) {
this.root = root;
}
public void addCombination(Combination finalCombination, Node current, String combination, int counter) {
if(counter==combination.length()) {
return;
}
int value = combination.charAt(counter)-48;
if(current.connectedNodes[value-1]==null) {
current.connectedNodes[value-1] = new Node(value);
}
if(counter==combination.length()-1) {
current.combination = finalCombination;
return;
}
addCombination(finalCombination,current.connectedNodes[value-1],combination,counter+1);
}
public void traverse(Node current) {
System.out.print(current.value+ " ");
for(Node node: current.connectedNodes) {
if(node!=null) {
traverse(node);
}
}
}
public Node takeStep(Node current, int value) {
if(current.connectedNodes[value-1]!=null) {
return current.connectedNodes[value-1];
}
return this.root;
}}
COMBINATION:
public class Combination {
String messageContext;
}
Can you please help me ? I just want to reset to root when it hasn't anywhere to go else ? Thank you in advance!

I ran your code and found out that you are storing the message context in the parent node instead of the actual node which marks the end of the combination. So I changed this piece of code in addCombination.
public void addCombination(Combination finalCombination, Node current, String combination, int counter) {
if (counter == combination.length()) {
//Storing at the original node.
current.combination = finalCombination;
return;
}
int value = combination.charAt(counter) - 48;
if (current.connectedNodes[value - 1] == null) {
current.connectedNodes[value - 1] = new Node(value);
}
addCombination(finalCombination, current.connectedNodes[value - 1], combination, counter + 1);
}
And changed following in the main code.
while (true) {
System.out.println(currentNode.value);
//Moved it up now as the node it self has the message context.
if (currentNode.combination != null) {
System.out.println(currentNode.combination.messageContext);
if (currentNode.connectedNodes[0] == null && currentNode.connectedNodes[1] == null) {
currentNode = tree.root;
continue;
}
}
for (Node node : currentNode.connectedNodes) {
if (node != null) {
System.out.print(node.value + " ");
} else {
System.out.print("nullnode ");
}
}
int x = scanner.nextInt();
currentNode = tree.takeStep(currentNode, x);
}
Now try the code it is resetting to root as expected.

Tree iterator in Java

I have implemented a code which adds elements in a tree and prints them in increasing order. However my aim is to learn iterators and want to replace the inOrder() function with an iterator function. How can I do this?
import java.util.InputMismatchException;
import java.util.Scanner;
import javax.xml.soap.Node;
class Tree
{
public final int mVal;
public Tree mLeft;
public Tree mRight;
public Node next;
public Tree(int val)
{
mVal = val;
}
public void add(int val)
{
if (val < mVal)
{
if (mLeft == null)
mLeft = new Tree(val);
else
mLeft.add(val);
}
else
{
if (val > mVal)
{
if (mRight == null)
mRight = new Tree(val);
else
mRight.add(val);
}
}
}
public String inOrder()
{
return ((mLeft == null) ? "" : mLeft.inOrder())
+ mVal + " "
+ ((mRight == null) ? "" : mRight.inOrder());
}
public static void main(String[] args)
{
Tree t = new Tree(8);
Scanner scanner = new Scanner(System.in);
boolean continueLoop = true; // determines if more input is needed
for (int i = 1; i < 9; ++i)
{
try // read two numbers and calculate quotient
{
System.out.print("Please enter a random integer : ");
int stackInt = scanner.nextInt();
t.add(Integer.valueOf(stackInt));
} // end try
catch (InputMismatchException inputMismatchException){
System.err.printf("\nException: %s\n", inputMismatchException);
scanner.nextLine(); //discard input so user can try again
System.out.println("You must enter integers. Please try again.\n");
} // end catch
}
System.out.println("Values in order = "+ t.inOrder());
}
}

look at this picture
First Step: if node has a left child, visit left child and do the first step with the child
Second Step: node has no left child (or we visited the left child already), add it to the inorder list
Third Step: first step with right child
i didnt test it
#Override
public String toString() {
return String.valueOf(mVal);
}
public String inOrder(Tree root) {
List<Tree> inOrder = new ArrayList<>();
inOrderRecursively(root, inOrder);
return inOrder.toString();
}
private void inOrderRecursively(Tree Node, List<Tree> inOrder) {
if (Node.mLeft != null) {
inOrderIt(Node.mLeft, inOrder);
}
inOrder.add(Node);
if (Node.mRight != null) {
inOrderIt(Node.mRight, inOrder);
}
}
greetings

store a binary search tree graph into an array using recursion and print it out

I already build a binary search tree. The primitive data type I store in the tree is integer. I try to store it on a 2-D char array and then print it out as the graph shown below(the numbers represent row numbers and column numbers and I do not need to print it, ignoring "-" symbol please, I only use it to indicate the exact position)
-----0---1---2---3---4---5---6---7---8---9---10---11---12---13---14---15---16
0---------------------------------------12
1--------------------------------/-------------------\
2----------------------8--------------------------------------14
3-----------------/----------\ -----------------------------------------\
4-------------5----------------9-------------------------------------------34
5--------/-------------------------------------------------------------/-------------\
6---2---------------------------------------------------------24------------------------35
number 12 need to store on location [0][8], the middle of the first row.
number 4 store on[2][4], number 14=[2][12], 5=[4][2], 9=[4][9] and so on.
row number 1 which is second row, "/" is on position[1][6] and "\" is on position[1][10] etc.they are also on the middle between two numbers
following is my code
public class MainClass {
public static void main(String[] args) {
//level represents row number;
// start indicates the column I am going to
//store number in, and end is a fixed column number
// BinarySearchTree is a BinaryTree type instance,
// I already story integers on it and follow with the format
// of binary search trees, and I did tested it.
int level=0; int start=0; int end=80;
BinaryTree.plot(BinarySearchTree, level, start, end);
}
private static class BinaryTree {
private BinaryNode root;
static char[][] offset = new char [10][20];
public BinaryTree(){
root = null;
}
public BinaryTree(Object x){
root = new BinaryNode(x);
}
public boolean isEmpty(){
return root == null;
}
public Object getRootobj() throws BinaryTreeException{
if(root == null)
throw new BinaryTreeException("Empty Tree");
else
return root.element;
}
public BinaryTree getLeft() throws BinaryTreeException{
if(root == null)
throw new BinaryTreeException("Empty Tree");
else {
BinaryTree t = new BinaryTree();
t.root = root.left;
return t;
}
}
public BinaryTree getRight() throws BinaryTreeException{
if(root == null)
throw new BinaryTreeException("Empty Tree");
else {
BinaryTree t = new BinaryTree();
t.root = root.right;
return t;
}
}
public static void plot(BinaryTree t, int level, int start, int end){
if(!t.isEmpty()){
plot(t.getLeft(), level+2, start/2, end/2);
String string = Integer.toString((Integer)t.getRootobj());
for(char c: string.toCharArray())
offset[level][start++]=c;
if(!(t.getLeft().isEmpty()))
offset[++level][start/4*3] = '/';
if(!(t.getRight().isEmpty()))
offset[++level][((start+end)/2+start)/2] = '\\';
plot(t.getRight(), level+2, end/2, end);
}
for(int i = 0; i<10; i++){
for(int j= 0; j<20; j++)
System.out.print(offset[i][j]);
}
}
}
private static class BinaryNode {
Object element;
BinaryNode left,right;
BinaryNode() {
this(0);
}
BinaryNode(Object e) {
this(e, null, null);
}
BinaryNode(Object e, BinaryNode ln, BinaryNode m){
element=e;
left=ln;
right=m;
}
}
}
Question: the method plot I used to store and print out binarysearchtree did not work, which causes a java.lang.ArrayIndexOutOfBoundsException:
can anyone take a look at it. appreciated for the help.

Your fixed-size char-Array cannot cope with your dynamic sized BinaryTree. For your given example alone you need way more then 20 characters per line! That's where your Exception is coming from.
But to give you an idea of an alternative approach - even though it took a while, made the following additions to your code:
First, I added a method to the BinaryNode class:
int getDepth() {
int subTreeDepth;
if (left == null && right == null) {
subTreeDepth = 0;
} else if (left == null) {
subTreeDepth = right.getDepth();
} else if (right == null) {
subTreeDepth = left.getDepth();
} else {
subTreeDepth = Math.max(left.getDepth(), right.getDepth());
}
return 1 + subTreeDepth;
}
Second, I removed your fixed char-Array and replaced the whole plotting algorithm in your BinaryTree (I just couldn't wrap my head around all those relative array-index manipulations):
public void plot() {
if (root == null) {
throw new BinaryTreeException("Empty Tree");
}
int lineCount = 2 * root.getDepth() - 1;
StringBuilder[] lines = new StringBuilder[lineCount];
for (int lineIndex = 0; lineIndex < lineCount; lineIndex++) {
lines[lineIndex] = new StringBuilder();
}
// get the right most node (which contains the largest element value)
BinaryNode rightMostNode = root;
while (rightMostNode.right != null) {
rightMostNode = rightMostNode.right;
}
// check how many characters we have to reserve for a single node element
int maxElementLength = String.valueOf(rightMostNode.element).length();
plot(root, 0, 0, maxElementLength, lines);
for (StringBuilder singleLine : lines) {
System.out.println(singleLine.toString());
}
}
private void plot(BinaryNode subTreeRoot, int offset, int lineIndex, int elementLength, StringBuilder[] lines) {
int actualOffset;
if (subTreeRoot.left == null) {
actualOffset = offset;
} else {
actualOffset = offset + (int) Math.pow(2, subTreeRoot.left.getDepth() - 1) * elementLength;
}
StringBuilder currentLine = lines[lineIndex];
String elementValue = String.valueOf(subTreeRoot.element);
for (int lineFillIndex = currentLine.length() + elementValue.length() / 2; lineFillIndex < actualOffset; lineFillIndex++) {
currentLine.append(' ');
}
currentLine.append(elementValue);
if (subTreeRoot.left != null) {
// draw connection to left sub tree
int connectPosition = (actualOffset - offset) * 3 / 4 + offset;
StringBuilder connectLine = lines[lineIndex + 1];
for (int lineFillIndex = connectLine.length(); lineFillIndex < connectPosition; lineFillIndex++) {
connectLine.append(' ');
}
connectLine.append('/');
// insert the left part of the next value line
plot(subTreeRoot.left, offset, lineIndex + 2, elementLength, lines);
}
if (subTreeRoot.right != null) {
// draw connection to right sub tree
int connectPosition = actualOffset + elementLength - elementValue.length() / 2;
if (subTreeRoot.right.left != null) {
connectPosition += (int) Math.pow(2, subTreeRoot.right.left.getDepth() - 1) * elementLength / 2;
}
StringBuilder connectLine = lines[lineIndex + 1];
for (int lineFillIndex = connectLine.length(); lineFillIndex < connectPosition; lineFillIndex++) {
connectLine.append(' ');
}
connectLine.append('\\');
// insert the right part of the next value line
plot(subTreeRoot.right, actualOffset + elementLength, lineIndex + 2, elementLength, lines);
}
}
For a tree similar to the one, you included in your question:
BinaryTree binarySearchTree = new BinaryTree(
new BinaryNode(12,
new BinaryNode(8,
new BinaryNode(5,
new BinaryNode(3),
null),
new BinaryNode(9)),
new BinaryNode(14,
null,
new BinaryNode(34,
new BinaryNode(24),
new BinaryNode(35)))));
binarySearchTree.plot();
I get the following output:
12
/ \
8 14
/ \ \
5 9 34
/ / \
3 24 35

Java Binary Tree entered in a specific order

I am trying to complete an assignment where I need to write a Java program to take a string from the command line, and implement it as a Binary Tree in a specific order, then get the depth of the binary tree.
For example: "((3(4))7((5)9))"
would be entered as a tree with 7 as the root, 3 and 9 as the children, and 4 as a right child of 3, and 5 as a left child of 9.
My code is below.. The problem I am having is that, because I am basing my checks off of finding a right bracket, I am unsure how to get the elements correctly when they are not directly preceding the brackets, such as the 3 in the above string. Any direction would be greatly appreciated..
class Node {
int value;
Node left, right;
}
class BST {
public Node root;
// Add Node to Tree
public void add(int n) {
if (root == null) {
root = new Node( );
root.value = n;
}
else {
Node marker = root;
while (true) {
if (n < marker.value) {
if (marker.left == null) {
marker.left = new Node( );
marker.left.value = n;
break;
} else {
marker = marker.left;
}
} else {
if (marker.right == null) {
marker.right = new Node( );
marker.right.value = n;
break;
} else {
marker = marker.right;
}
}
}
}
} // End ADD
//Find Height of Tree
public int height(Node t) {
if (t.left == null && t.right == null) return 0;
if (t.left == null) return 1 + height(t.right);
if (t.right == null) return 1 + height(t.left);
return 1 + Math.max(height(t.left), height(t.right));
} // End HEIGHT
// Check if string contains an integer
public static boolean isInt(String s) {
try {
Integer.parseInt(s);
}
catch(NumberFormatException e) {
return false;
}
return true;
} // End ISINT
public int elementCount(String[] a) {
int count = 0;
for (int i = 0; i < a.length; i++) {
if (isInt(a[i])) count++;
}
return count;
}
} // End BST Class
public class Depth {
public static void main(String[] args) {
String[] a = args[0].split(" ");
BST tree = new BST();
int[] bcount = new int[10];
int[] elements = new int[10];
int x = 0, bracketcount = 0;
// Display entered string
System.out.print("Entered Format: ");
for (int j=0; j < a.length; j++) {
System.out.print(a[j]);
}
for (int i=0; i < a.length; i++) {
char c = a[i].charAt(0);
switch (c)
{
case '(':
bracketcount++;
break;
case ')':
if (isInt(a[i-1])) {
bcount[x] = bracketcount--;
elements[x++] = Integer.parseInt(a[i-1]);
}
break;
case '1':
case '7':
default : // Illegal character
if ( (a[i-1].charAt(0) == ')') && (a[i+1].charAt(0) == '(') ) {
bcount[x] = bracketcount;
elements[x++] = Integer.parseInt(a[i]);
}
break;
}
}
System.out.println("\nTotal elements: " + tree.elementCount(a));
// Display BracketCounts
for (int w = 0; w < x; w++) {
System.out.print(bcount[w] + " ");
}
System.out.println(" ");
// Display Elements Array
for (int w = 0; w < x; w++) {
System.out.print(elements[w] + " ");
}
System.out.println("\nDepth: " + tree.height(tree.root));
// Build the tree
for (int y = 0; y < x-1; y++) {
for (int z = 1; z < tree.height(tree.root); z++) {
if (bcount[y] == z) {
tree.add(elements[y]);
}
}
}
} // End Main Function
public static boolean isInt(String s) {
try {
Integer.parseInt(s);
}
catch(NumberFormatException e) {
return false;
}
return true;
}
} // End Depth Class

I would do a couple of statements to get access to a tree with that kind of shape:
For input string : input= "((3(4))7((5)9))"
You could do :
public class Trial {
/**
* #param args
*/
public static void main(String[] args) {
// TODO Auto-generated method stub
String input = "((3(4))7((5)9))";
String easier = input.replaceAll("\\(\\(", "");
String evenEasier = easier.replaceAll("\\)\\)", "");
System.out.println(evenEasier);
int firstVal = Integer.parseInt(evenEasier.substring(0, 1));
int firstBracketVal = Integer.parseInt(evenEasier.substring(2, 3));
int middleVal = Integer.parseInt(evenEasier.substring(3, 4));
int secondBracketVal = Integer.parseInt(evenEasier.substring(4,5));
int lastVal = Integer.parseInt(evenEasier.substring(6));
System.out.println("First Val:"+firstVal);
System.out.println("First bracket Val:"+firstBracketVal);
System.out.println("Middle Val:"+middleVal);
System.out.println("Second Bracket Val:"+secondBracketVal);
System.out.println("Last Val:"+lastVal);
}
}
This however would only ever work for entries in that specific format, if that were to change, or the length of the input goes up - this would work a bit or break.....If you need to be able to handle more complicated trees as input in this format a bit more thought would be needed on how to best handle and convert into your internal format for processing.

pseudocode:
function getNode(Node)
get one char;
if (the char is "(")
getNode(Node.left);
get one char;
end if;
Node.value = Integer(the char);
get one char;
if (the char is "(")
getNode(Node.right);
get one char;
end if;
//Now the char is ")" and useless.
end function
Before calling this function, you should get a "(" first.
In this method, the framwork of a Node in string is "[leftchild or NULL] value [rightchild or NULL])".
"("is not belong to the Node, but ")" is.

BinaryTree implementation in java

I have this code for BinaryTree creation and traversal
class Node
{
Integer data;
Node left;
Node right;
Node()
{
data = null;
left = null;
right = null;
}
}
class BinaryTree
{
Node head;
Scanner input = new Scanner(System.in);
BinaryTree()
{
head = null;
}
public void createNode(Node temp, Integer value)
{
Node newnode= new Node();
value = getData();
newnode.data = value;
temp = newnode;
if(head==null)
{
head = temp;
}
System.out.println("If left child exits for ("+value+") enter y else n");
if(input.next().charAt(0)=='y')
{
createNode(temp.left, value);
}
System.out.println("If right child exits for ("+value+") enter y else n");
if(input.next().charAt(0)=='y')
{
createNode(temp.right, value);
}
}
public Integer getData()
{
out.println("Enter the value to insert:");
return (Integer)input.nextInt();
}
public void print()
{
inorder(head);
}
public void inorder(Node node)
{
if(node!=null)
{
inorder(node.left);
System.out.println(node.data);
inorder(node.right);
}
else
return;
}
}
class BinaryTreeWorker
{
static BinaryTree treeObj = null;
static Scanner input = new Scanner(System.in);
public static void displaymenu()
{
int choice;
do{
out.print("\n Basic operations on a tree:");
out.print("\n 1. Create tree \n 2. Insert \n 3. Search value \n 4. print list\n Else. Exit \n Choice:");
choice = input.nextInt();
switch(choice)
{
case 1:
treeObj = createBTree();
break;
case 2:
treeObj.createNode(null, null);
break;
case 3:
//searchnode();
break;
case 4:
treeObj.print();
break;
default:
return;
}
}while(true);
}
public static BinaryTree createBTree()
{
return new BinaryTree();
}
public static void main(String[] args)
{
displaymenu();
}
}
It compiles and runs. But I think there is something wrong with the inorder traversal.
I created the below tree,
2
1 3
But it prints only 2.

I have tried solving the problem your way and I have pasted the solution below.. Though I haven't tested it thoroughly so it might fail in some edge condition.. But I have tested it for one case. Kindly let me know if it fails in some scenario. I would appreciate others help in making this answer better. I agree that this solution is not the most ideal way to code a Binary Tree but it wont hurt this way if some one is just practicing..
import java.util.Scanner;
class Node
{
Integer data;
Node left;
Node right;
Node()
{
data = null;
left = null;
right = null;
}
}
class BinaryTree
{
Node head;
Scanner input = new Scanner(System.in);
BinaryTree()
{
head = null;
}
public void createNode(Node temp,Node newnode)
{
if(head==null)
{
System.out.println("No value exist in tree, the value just entered is set to Root");
head = newnode;
return;
}
if(temp==null)
temp = head;
System.out.println("where you want to insert this value, l for left of ("+temp.data+") ,r for right of ("+temp.data+")");
char inputValue=input.next().charAt(0);
if(inputValue=='l'){
if(temp.left==null)
{
temp.left=newnode;
System.out.println("value got successfully added to left of ("+temp.data+")");
return;
}else {
System.out.println("value left to ("+temp.data+") is occupied 1by ("+temp.left.data+")");
createNode(temp.left,newnode);
}
}
else if(inputValue=='r')
{
if(temp.right==null)
{
temp.right=newnode;
System.out.println("value got successfully added to right of ("+temp.data+")");
return;
}else {
System.out.println("value right to ("+temp.data+") is occupied by ("+temp.right.data+")");
createNode(temp.right,newnode);
}
}else{
System.out.println("incorrect input plz try again , correctly");
return;
}
}
public Node generateTree(){
int [] a = new int[10];
int index = 0;
while(index<a.length){
a[index]=getData();
index++;
}
if(a.length==0 ){
return null;
}
Node newnode= new Node();
/*newnode.left=null;
newnode.right=null;*/
return generateTreeWithArray(newnode,a,0);
}
public Node generateTreeWithArray(Node head,int [] a,int index){
if(index >= a.length)
return null;
System.out.println("at index "+index+" value is "+a[index]);
if(head==null)
head= new Node();
head.data = a[index];
head.left=generateTreeWithArray(head.left,a,index*2+1);
head.right=generateTreeWithArray(head.right,a,index*2+2);
return head;
}
public Integer getData()
{
System.out.println("Enter the value to insert:");
return (Integer)input.nextInt();
}
public void print()
{
inorder(head);
}
public void inorder(Node node)
{
if(node!=null)
{
inorder(node.left);
System.out.println(node.data);
inorder(node.right);
}
else
return;
}
}
public class BinaryTreeWorker
{
static BinaryTree treeObj = null;
static Scanner input = new Scanner(System.in);
public static void displaymenu()
{
int choice;
do{
System.out.print("\n Basic operations on a tree:");
System.out.print("\n 1. Create tree \n 2. Insert \n 3. Search value \n 4. print list\n 5. generate a tree \n Else. Exit \n Choice:");
choice = input.nextInt();
switch(choice)
{
case 1:
treeObj = createBTree();
break;
case 2:
Node newnode= new Node();
newnode.data = getData();
newnode.left=null;
newnode.right=null;
treeObj.createNode(treeObj.head,newnode);
break;
case 3:
//searchnode();
break;
case 4:
System.out.println("inorder traversal of list gives follows");
treeObj.print();
break;
case 5:
Node tempHead = treeObj.generateTree();
System.out.println("inorder traversal of list with head = ("+tempHead.data+")gives follows");
treeObj.inorder(tempHead);
break;
default:
return;
}
}while(true);
}
public static Integer getData()
{
System.out.println("Enter the value to insert:");
return (Integer)input.nextInt();
}
public static BinaryTree createBTree()
{
return new BinaryTree();
}
public static void main(String[] args)
{
displaymenu();
}
}
[Update] : Updated the code to generate a binary tree using an array. This will involve less user interaction.

Best way to implement Binary Tree in Java with all the traverse types and test cases as below
package com.nitin.tree;
public class Tree
{
private Node parent;
private int data;
private int size = 0;
public Tree() {
parent = new Node(data);
}
public void add(int data) {
if (size == 0) {
parent.data = data;
size++;
} else {
add(parent, new Node(data));
}
}
private void add(Node root, Node newNode) {
if (root == null) {
return;
}
if (newNode.data < root.data) {
if (root.left == null) {
root.left = newNode;
size++;
} else {
add(root.left, newNode);
}
} else {
if (root.right == null) {
root.right = newNode;
size++;
} else {
add(root.right, newNode);
}
}
}
public int getLow() {
Node current = parent;
while (current.left != null) {
current = current.left;
}
return current.data;
}
public int getHigh() {
Node current = parent;
while (current.right != null) {
current = current.right;
}
return current.data;
}
private void in(Node node) {
if (node != null) {
in(node.left);
System.out.print(node.data + " ");
in(node.right);
}
}
private void pre(Node node) {
if (node != null) {
System.out.print(node.data + " ");
pre(node.left);
pre(node.right);
}
}
private void post(Node node) {
if (node != null) {
post(node.left);
post(node.right);
System.out.print(node.data + " ");
}
}
public void preorder() {
System.out.print("Preorder Traversal->");
pre(parent);
System.out.println();
}
public void postorder() {
System.out.print("Postorder Traversal->");
post(parent);
System.out.println();
}
public void inorder() {
System.out.print("Inorder Traversal->");
in(parent);
System.out.println();
}
private class Node {
Node left;
Node right;
int data;
public Node(int data) {
this.data = data;
}
}
public String toString() {
Node current = parent;
System.out.print("Traverse From Left ");
while (current.left != null && current.right != null) {
System.out.print(current.data + "->[" + current.left.data + " " + current.right.data + "] ");
current = current.left;
}
System.out.println();
System.out.print("Traverse From Right ");
current = parent;
while (current.left != null && current.right != null) {
System.out.print(current.data + "->[" + current.left.data + " " + current.right.data + "] ");
current = current.right;
}
return "";
}
public static void main(String af[]) {
Tree t = new Tree();
t.add(40);
t.add(25);
t.add(78);
t.add(10);
t.add(32);
t.add(50);
t.add(93);
t.add(3);
t.add(17);
t.add(30);
t.add(38);
System.out.println(t.getLow());
System.out.println(t.getHigh());
System.out.println("Size-" + t.size);
System.out.println(t);
t.inorder();
t.preorder();
t.postorder();
}
}

Your problem is in public void createNodes(Node temp, T data) function. You pass in a parameter the same name as the class variable temp. First of all I don't think you need the class variable by itself. Second of all assigning to temp in this method has only local effect - you loose the information in the temp parameter, but setting temp, will not infuence its value in the called method. I suggest you rewrite the method so that it returns the pointer to the newly created node and assign this pointer to the left and right of the local temp. That way the changes will propagate out.

another type of outputting the tree:
public void inorder()
{
inorder(root);
}
protected void visit(BSTNode<T> p)
{
System.out.println("Node: " + p.el + "Left Side:" + (p.left!=null?p.left.el:"null") +
"Right Side:" + (p.right!=null?p.right.el:"null"));
}

I've changed the BinaryTree Class as below. See the change on the the createNode method in particular.
The problem, as mentioned in the post before this, is that your reference doesn't persist when it is passed as an argument to the createNode method. That change is only local. You need to return an explicit Node reference in the method itself as you're creating the node.
public Node createNode()
{
Integer value = getData();
Node temp = new Node(value);
if(head==null)
{
head = temp;
}
System.out.println("Do you want to add left branch on node("+value+")? Enter y/n");
if(input.next().charAt(0)=='y')
{
temp.left=createNode();
}
System.out.println("Do you want to add right branch on node("+value+")? Enter y/n");
if(input.next().charAt(0)=='y')
{
temp.right=createNode();
}
return temp;
}
Here is the resulting output:
Basic operations on a tree:
1. Create tree
2. Insert
3. Search value
4. print list
Else. Exit
Choice:1
Basic operations on a tree:
1. Create tree
2. Insert
3. Search value
4. print list
Else. Exit
Choice:2
Enter the value to insert:
10
Do you want to add left branch on node(10)? Enter y/n
y
Enter the value to insert:
20
Do you want to add left branch on node(20)? Enter y/n
n
Do you want to add right branch on node(20)? Enter y/n
n
Do you want to add right branch on node(10)? Enter y/n
y
Enter the value to insert:
30
Do you want to add left branch on node(30)? Enter y/n
n
Do you want to add right branch on node(30)? Enter y/n
n
Basic operations on a tree:
1. Create tree
2. Insert
3. Search value
4. print list
Else. Exit
Choice:4
20
10
30
I hope this will be of some help to someone later (even if this is 3 years late..). I just started learning about Binary Trees today myself. I'm actually planning on using this as a base to doing more involved tasks!

I changed the createNode method so that it works:
public Node createNode(Node temp, Integer value)
{
Node newnode = new Node();
value = getData();
newnode.data = value;
temp = newnode;
if(head == null)
{
head = temp;
}
System.out.println("If left child exits for ("+value+") enter y else n");
if(input.next().charAt(0) == 'y')
{
newnode.left = createNode(newnode.left, value);
}
System.out.println("If right child exits for ("+value+") enter y else n");
if(input.next().charAt(0) == 'y')
{
newnode.right = createNode(newnode.right, value);
}
return newnode;
}