Extended version of Binary tree

A node can be binary. Two children. Left and Right.
Example:
A
B C
class Node
   public int data;              // data item (key)
   public Node leftChild; 
   public Node rightChild;  But how would I create a tree with the node can contain as much children?
A
B C R T
and so on? I want to insert a node in A and have it expand each time. Node "A" will contain node {B,C,R,T} for example.
How would I code this?

Much depends on how you see the relation between a Node and it's children. In the simplest form you could just change
class Node
public int data;              // data item (key)
public Node leftChild; 
public Node rightChild;  to
class Node
    public int data;
    public Node[] children;
}or
class Node
    public int data;
    public ArrayList children;
}BUT
what you probably want is for the ordering to be equivalent to
class Node
    public Node leftChild; 
    public int data;              // data item (key)
    public Node rightChild;
} i.e. the data of the current node is sandwiched between that of the left and right nodes.
A BTree is one of the best examples of this (Google is your friend). An approach (and I stress AN) is to have an array of say 'n' values and an array of 'n+1' child nodes with the implied sandwich ordering
[child 0][data0][child 1][data 1] ... [child n][data n][child n+1].
The result is then
class Node
    int[] data = ...;
    Node[] childrean = ...
}where you would dynamically size the 'data' and 'children' arrays and the length of 'children' is always 1 greater than the order of 'data'.

Similar Messages

  • How to extend  breadth first Search for Binary Tree to any kind of Tree??

    Dear Friends,
    I am thinking a problem, How to extend breadth first Search for Binary Tree to any kind of Tree?? ie each node has more than 2 leaves such as 1, 2,3,4 or any,
    I have following code to successfully apply for breadth first Search in Binary Tree as follows,
    package a.border;
    import java.util.ArrayList;
    import java.util.LinkedList;
    public class Tree
        int root;
        Tree left;
        Tree right;
        static ArrayList<Integer> list = new ArrayList<Integer>();
        static ArrayList<Tree> treeList = new ArrayList<Tree>();
        private static LinkedList<Tree> queue = new LinkedList<Tree>();
         * @param root root value
         * @param left left node
         * @param right right node
        public Tree(int root, Tree left, Tree right)
            this.root = root;
            this.left = left;
            this.right = right;
        /** Creates a new instance of Tree
         * You really should know what this does...
         * @param root
        public Tree(int root)
            this.root = root;
            this.left = null;
            this.right = null;
         * Simply runs a basic left then right traversal.
        public void basicTraversal()
            //Check if we can go left
            if (left != null)
                left.basicTraversal();
            //Add the root
            list.add(root);
            //Check if we can go right
            if (right != null)
                right.basicTraversal();
        public ArrayList<Integer> getBreadthTraversal(ArrayList<Integer> list)
            //Add the root to the arraylist, we know it is always the first entry.
            list.add(root);
            //Basically we add the first set of nodes into the queue for
            //traversing.
            //Query if left exists
            if (left != null)
                //Then add the node into the tree for traversing later
                queue.add(left);
            //Same for right
            if (right != null)
                queue.add(right);
            //Then we call the traverse method to do the rest of the work
            return traverse(list);
        private ArrayList<Integer> traverse(ArrayList<Integer> list)
            //Keep traversing until we run out of people
            while (!queue.isEmpty())
                Tree p = queue.remove();
                //Check if it has any subnodes
                if (p.left != null)
                    //Add the subnode to the back of the queue
                    queue.add(p.left);
                //Same for left
                if (p.right != null)
                    //Same here, no queue jumping!
                    queue.add(p.right);
                //Append to the ArrayList
                list.add(p.root);
            //And return
            return list;
         * Makes a tree and runs some operations
         * @param args
        public static void main(String[] args)
             *                             4
             *          t =           2       6
             *                      1   3    5   7
            Tree leaf6 = new Tree(1);
            Tree leaf7 = new Tree(3);
            Tree leaf8 = new Tree(5);
            Tree leaf9 = new Tree(7);
            Tree t4 = new Tree(2, leaf6, leaf7);
            Tree t5 = new Tree(6, leaf8, leaf9);
            Tree t = new Tree(4, t4, t5);
            t.basicTraversal();
            System.out.println("Here is basicTraversal ="+list.toString());
            list.clear();
            t.getBreadthTraversal(list);
            System.out.println("getBreadthTraversal= " +list.toString());
            list.clear();
        }Can Guru help how to update to any kind of tree??
    here this code is for the tree like:
             *                             4
             *          t =           2       6
             *                      1   3    5   7
             */But i hope the new code can handle tree like:
             *                             4
             *                           /   | \
             *                          /     |   \
             *          t =            2     8   6
             *                        / |  \    |    /| \
             *                      1 11  3 9   5 10  7
             */Thanks

    sunnymanman wrote:
    Dear Friends,
    I am thinking a problem, How to extend breadth first Search for Binary Tree to any kind of Tree?? ...The answer is interfaces.
    What do all trees have in common? And what do all nodes in trees have in common?
    At least these things:
    interface Tree<T> {
        Node<T> getRoot();
    interface Node<T> {
        T getData();
        List<Node<T>> getChildren();
    }Now write concrete classes implementing these interfaces. Let's start with a binary tree (nodes should have comparable items) and an n-tree:
    class BinaryTree<T extends Comparable<T>> implements Tree<T> {
        protected BTNode<T> root;
        public Node<T> getRoot() {
            return root;
    class BTNode<T> implements Node<T> {
        private T data;
        private Node<T> left, right;
        public List<Node<T>> getChildren() {
            List<Node<T>> children = new ArrayList<Node<T>>();
            children.add(left);
            children.add(right);
            return children;
        public T getData() {
            return data;
    class NTree<T> implements Tree<T> {
        private NTNode<T> root;
        public Node<T> getRoot() {
            return root;
    class NTNode<T> implements Node<T> {
        private T data;
        private List<Node<T>> children;
        public List<Node<T>> getChildren() {
            return children;
        public T getData() {
            return data;
    }Now with these classes, you can wite a more generic traversal class. Of course, every traversal class (breath first, depth first) will also have something in common: they return a "path" of nodes (if the 'goal' node/data is found). So, you can write an interface like this:
    interface Traverser<T> {
        List<Node<T>> traverse(T goal, Tree<T> tree);
    }And finally write an implementation for it:
    class BreathFirst<T> implements Traverser<T> {
        public List<Node<T>> traverse(T goal, Tree<T> tree) {
            Node<T> start = tree.getRoot();
            List<Node<T>> children = start.getChildren();
            // your algorithm here
            return null; // return your traversal
    }... which can be used to traverse any tree! Here's a small demo of how to use it:
    public class Test {
        public static void main(String[] args) {
            Tree<Integer> binTree = new BinaryTree<Integer>();
            // populate your binTree
            Tree<Integer> nTree = new NTree<Integer>();
            // populate your nTree
            Traverser<Integer> bfTraverser = new BreathFirst<Integer>();
            // Look for integer 6 in binTree
            System.out.println("bTree bfTraversal -> "+bfTraverser.traverse(6, binTree));
            // Look for integer 6 in nTree
            System.out.println("bTree bfTraversal -> "+bfTraverser.traverse(6, nTree));
    }Good luck!

  • MorseCode Binary Tree

    So I have an assignment to write a program that deciefers basic morse code using a binary tree. A file was provided with the information for the binary tree and I am using the text books author's version of BinaryTree class with a few methods of my own thrown in.
    For starters .. why would you ever use a tree for this .. just make a look up table. But my real problem is i keep getting null pointers at 38, 24, 55 and I have tried just about everything I know to do at this point. My mind is exhausted and I think I need a fresh pair of eyes to tell me what I have done wrong.
    I am sorry for such sloppy code.. I have been changing and rearranging too much.
    import java.io.*;
    import java.util.*;
    public class MorseCode extends BinaryTree{
         static BufferedReader in;
         static BinaryTree<String> bT = new BinaryTree<String>();
    /*     public static void loadTree() throws FileNotFoundException, IOException{
              in = new BufferedReader(new FileReader("MorseCode.txt"));
              bT = readBinaryTree(in);
         public static String decode(Character c) throws IOException{
              if(c.equals("null")){
                   return "";
              }else if(c.equals(" ")){
                   return " ";
              }else{
                   return (":" + find(c));
         public static String find(Character c) throws IOException, FileNotFoundException{
              in = new BufferedReader(new FileReader("MorseCode.txt"));
              bT = readBinaryTree(in);
              Queue<BinaryTree> data = new LinkedList<BinaryTree>();
              BinaryTree<String> tempTree = bT;
              String temp = null;
              Character tempChar = null;
              data.offer(tempTree);
              while(!data.isEmpty()){
                   tempTree = data.poll();
                   temp = tempTree.getData();
                   tempChar = temp.charAt(0);
                   if(tempChar.equals(c)){
                        break;
                   data.offer(tempTree.getRightSubtree());
                   data.offer(tempTree.getLeftSubtree());
              return temp.substring(2);               
         public static void main(String[] args) throws FileNotFoundException, IOException{
              Scanner scan = new Scanner(new FileReader("encode.in.txt"));
              String s = "";
              String temp = "";
              while(scan.hasNextLine()){
                    temp = scan.nextLine();
                    for(int i = 0; i < temp.length(); i++){
                         s = s + decode(temp.charAt(i));
              System.out.println(s);
    /** Class for a binary tree that stores type E objects.
    *   @author Koffman and Wolfgang
    class BinaryTree <E> implements Serializable
      //===================================================
      /** Class to encapsulate a tree node. */
      protected static class Node <E> implements Serializable
        // Data Fields
        /** The information stored in this node. */
        protected E data;
        /** Reference to the left child. */
        protected Node <E> left;
        /** Reference to the right child. */
        protected Node <E> right;
        // Constructors
        /** Construct a node with given data and no children.
            @param data The data to store in this node
        public Node(E data) {
          this.data = data;
          left = null;
          right = null;
        // Methods
        /** Return a string representation of the node.
            @return A string representation of the data fields
        public String toString() {
          return data.toString();
      }//end inner class Node
      //===================================================
      // Data Field
      /** The root of the binary tree */
      protected Node <E> root;
      public BinaryTree()
        root = null;
      protected BinaryTree(Node <E> root)
        this.root = root;
      /** Constructs a new binary tree with data in its root,leftTree
          as its left subtree and rightTree as its right subtree.
      public BinaryTree(E data, BinaryTree <E> leftTree, BinaryTree <E> rightTree)
        root = new Node <E> (data);
        if (leftTree != null) {
          root.left = leftTree.root;
        else {
          root.left = null;
        if (rightTree != null) {
          root.right = rightTree.root;
        else {
          root.right = null;
      /** Return the left subtree.
          @return The left subtree or null if either the root or
          the left subtree is null
      public BinaryTree <E> getLeftSubtree() {
        if (root != null && root.left != null) {
          return new BinaryTree <E> (root.left);
        else {
          return null;
      /** Return the right sub-tree
            @return the right sub-tree or
            null if either the root or the
            right subtree is null.
        public BinaryTree<E> getRightSubtree() {
            if (root != null && root.right != null) {
                return new BinaryTree<E>(root.right);
            } else {
                return null;
         public String getData(){
              if(root.data == null){
                   return "null";
              }else{
                   return (String) root.data;
      /** Determine whether this tree is a leaf.
          @return true if the root has no children
      public boolean isLeaf() {
        return (root.left == null && root.right == null);
      public String toString() {
        StringBuilder sb = new StringBuilder();
        preOrderTraverse(root, 1, sb);
        return sb.toString();
      /** Perform a preorder traversal.
          @param node The local root
          @param depth The depth
          @param sb The string buffer to save the output
      private void preOrderTraverse(Node <E> node, int depth,
                                    StringBuilder sb) {
        for (int i = 1; i < depth; i++) {
          sb.append("  ");
        if (node == null) {
          sb.append("null\n");
        else {
          sb.append(node.toString());
          sb.append("\n");
          preOrderTraverse(node.left, depth + 1, sb);
          preOrderTraverse(node.right, depth + 1, sb);
      /** Method to read a binary tree.
          pre: The input consists of a preorder traversal
               of the binary tree. The line "null" indicates a null tree.
          @param bR The input file
          @return The binary tree
          @throws IOException If there is an input error
      public static BinaryTree<String> readBinaryTree(BufferedReader bR) throws IOException
        // Read a line and trim leading and trailing spaces.
        String data = bR.readLine().trim();
        if (data.equals("null")) {
          return null;
        else {
          BinaryTree < String > leftTree = readBinaryTree(bR);
          BinaryTree < String > rightTree = readBinaryTree(bR);
          return new BinaryTree < String > (data, leftTree, rightTree);
      }//readBinaryTree()
      /*Method to determine the height of a binary tree
        pre: The line "null" indicates a null tree.
          @param T The binary tree
          @return The height as integer
      public static int height(BinaryTree T){
            if(T == null){
               return 0;
          }else{
               return 1 +(int) (Math.max(height(T.getRightSubtree()), height(T.getLeftSubtree())));
      public static String preOrderTraversal(BinaryTree<String> T){
          String s = T.toString();
          String s2 = "";
          String temp = "";
          Scanner scan = new Scanner(s);
          while(scan.hasNextLine()){
               temp = scan.nextLine().trim();
               if(temp.equals("null")){
                   s2 = s2;
              }else{
                   s2 = s2 + " " + temp;
          return s2;
    }//class BinaryTree

    As well as warnerja's point, you say you keep getting these errors. Sometimes it's helpful when illustrating a problem to replace the file based input with input that comes from a given String. That way we all see the same behaviour under the same circumstances.
    public static void main(String[] args) throws FileNotFoundException, IOException{
        //Scanner scan = new Scanner(new FileReader("encode.in.txt"));
        Scanner scan = new Scanner("whatever");The following isn't the cause of an NPE, but it might be allowing one to "slip through" (ie you think you've dealt with the null case when you haven't):
    public static String decode(Character c) throws IOException{
        if(c.equals("null")){
    c is a Character so it will never be the case that it is equal to the string n-u-l-l.
    Perhaps the behaviour of each of these methods needs to be (documented and) tested.

  • How to crossover this binary tree..?

    You can view detail http://www.codeguru.com/forum/showthread.php?s=bb4cf7ad2b18a5115e8bd6ab3a4e9d17&t=470868
    [nha khoa|http://www.sieuthi77.com/main/nhakhoa.html] .com/forum/showthread.php?s=bb4cf7ad2b18a5115e8bd6ab3a4e9d17&t=470868
    I have these classes which model a tree (binary). the thing is i cant figure out how i can set the elements of individual nodes in that tree. i can get individual elements but i cannot set them due to the strucutre of the tree and how it is implemented. The changes that I make to these classes should be as less as possible, because i have an algorithm which generates the tree structure randomly, plus i can evaluate the tree easily by using recursion. The only left thing to do is CROSSOVER but how?!?
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    Code:
    public abstract class Node implements Cloneable{
    abstract double evaluate(VariableInput v);
    abstract String print();
    abstract int getNumberOfNodes();
    abstract ArrayList<Object> getChildren();
    @Override
    public Node clone(){
    Node copy;
    try {
    copy = (Node) super.clone();
    } catch (CloneNotSupportedException unexpected) {
    throw new AssertionError(unexpected);
    //In an actual implementation of this pattern you might now change references to
    //the expensive to produce parts from the copies that are held inside the prototype.
    return copy;
    Code:
    public class UnaryNode extends Node {
    private UnaryFunction operator;
    private Node left;
    public UnaryNode(UnaryFunction op, Node terminal) {
    operator = op;
    this.left = terminal;
    public String print(){
    String r = "(" + operator.toString()+ " " + left.print() + ")";
    return r;
    void setLeft(Node left) {
    this.left = left;
    @Override
    int getNumberOfNodes() {
    return 1 + left.getNumberOfNodes();
    Node getLeft() {
    return left;
    ArrayList<Object> getChildren() {
    ArrayList<Object> arr = new ArrayList<Object>();
    arr.add(this);
    arr.addAll(left.getChildren());
    return arr;
    Code:
    public class BinaryNode extends Node {
    private BinaryFunction operator;
    private Node left;
    private Node right;
    public BinaryNode(BinaryFunction op, Node left, Node right) {
    operator = op;
    this.left = left;
    this.right = right;
    public String print(){
    String r = "(" + operator.toString()+ " " + left.print() + " " + right.print()+")";
    return r;
    public void setLeft(Node left){
    this.left = left;
    public void setRight(Node right){
    this.right = right;
    @Override
    int getNumberOfNodes() {
    return 1 + left.getNumberOfNodes() + right.getNumberOfNodes();
    Node getRight() {
    return right;
    Node getLeft() {
    return left;
    @Override
    ArrayList<Object> getChildren() {
    ArrayList<Object> arr = new ArrayList<Object>();
    arr.add(this);
    arr.addAll(left.getChildren());
    arr.addAll(right.getChildren());
    return arr;
    public class NumericNode extends Node{
    private double value;
    public NumericNode(double v){
    value = v;
    @Override
    double evaluate(VariableInput c) {
    return value;
    public String print(){
    String r = "" + value;
    return r;
    @Override
    int getNumberOfNodes() {
    return 1;
    @Override
    ArrayList<Object> getChildren() {
    ArrayList<Object> arr = new ArrayList<Object>();
    arr.add(new NumericNode(value));
    return arr;
    }p.s. I have this get children method which return a list of REFERENCES to all the nodes in the tree, but if i change any of them it wont have an effect to the tree itself because they are references.
    Any ideas or codes will be much appreciated. Thanks!

    What? Changes to what a node is referencing will be reflected in the tree, unless your getChildren is returning a copy, like in NumericNode.
    Kaj

  • Bonjour, quand j'utilise mon adobe photoshop cs6 extended (Version Adobe Photoshop : 13.0.6 (13.0.6 20131025.r.54 2013/10/25:21:00:00) x64) avec les calques, le clavier de mon Mac Book Pro ( osx 10,9,5 - 2,64 GHZ intel core i7 -8GO  1333 MHTZ ) se bloque

    Bonjour, quand j'utilise mon adobe photoshop cs6 extended (Version Adobe Photoshop : 13.0.6 (13.0.6 20131025.r.54 2013/10/25:21:00:00) x64) notamment avec les calques, le clavier de mon Mac Book Pro ( osx 10,9,5 - 2,64 GHZ intel core i7 -8GO  1333 MHTZ ) se bloque. Je suis obligé d'enregistrer ce que je fais dans le texte pré inscrit, puis de rouvrir pour enregistrer dans la rubrique et dans les termes voulu.
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    Bonjour! Oui: il faudrait essayer de réinitialiser les préférences, en maintenant CTRL+SHIFT+ALT dès que l'on démarre Photoshop, ce qui supprime le fichier des préférence et le remplace par un nouveau. On doit alors obtenir un message qui demande si on veut recréer les préférences.
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    Les pilotes des cartes graphiques datent de Janvier. Y-a-t-il des mises à jour disponibles?
    De même, Photoshop n'est pas dans sa version 13.0.1.1 il convient de se rendre dans Aide>Mises à Jour

  • Binary Tree Question

    Ok, I am making a binary tree. I have a question for my insert method. Firstly, I can't find out why the root node is inserted more than once in the tree. Also, I am having trouble with connecting the nodes that I insert to the tree. I can attach modes to root just fine, but I can't find out how to attach nodes to the existing nodes that are already attached to the tree. When I insert a node that meets the criteria of an already existing node, it replaces the node instead of getting attached to it. The answer is probably trivial, but I can't find it. Here is my insert method.
    public void insert(T obj) {
              int _result1;
              int _result2;
           //TODO: Implement Q1 here
              if(isEmpty() == true){
                   _root = createNode(obj, null, null, null);
                   System.out.println("The root inserted is " + _root.element());
                   insert(obj);
              _node = createNode(obj, _root, null, null);
              _result1 = _node.element().compareTo(_root.element());
              if(_result1 < 0){
                        if(_node.isInternal() == true){
                             _current = (BTreeNode<T>) _node2.element();
                             _result2 = _current.element().compareTo(_current.getParent().element());
                             //System.out.println("The current element is " + _current.element());
                             if(_result2 < 0){
                                  _node1.getLeft();
                                  _current = _node1.getLeft();
                             else{
                                  _node1.getRight();
                                  _current = _node1.getRight();
                             if(_node1.hasLeft() == true){
                                  _node1.getLeft();
                                  _current = _node1.getLeft();
                             else{
                                  _current.setLeft(_node1);
                                  _current = _node2;
                        else{
                             _node1 = createNode(obj, _node, null, null);
                             _node.setLeft(_node1);
                             _node1.setParent(_node);
                             System.out.println("The parent of the left node " + _node.element() + " is " + _node.getParent().element());
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                             _current = _node1;
                             //System.out.println("The current element is " + _node.element());
              else{
                        if(_node.isInternal() == true){
                             _current = (BTreeNode<T>) _current.element();
                             _result2 = _current.element().compareTo(_current.getParent().element());
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                             if(_result2 < 0){
                                  _node1.getLeft();
                                  _current = _node1.getLeft();
                             else{
                                  _node1.getRight();
                                  _current = _node1.getRight();
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                                  _current = _node1.getRight();
                             else{
                                  _current.setLeft(_node1);
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                   else{
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                        _node1.setParent(_node);
                        //_current = _node1;
                        System.out.println("The parent of the right node " + _node.element() + " is " + _node.getParent().element());
                        _root.setRight(_node1);
                        _current = _node1;                         
                        //System.out.println("The current element is " + _current.element());
              }The output I get is:
    The root inserted is 6
    The parent of the right node 6 is 6
    The parent of the right node 6 is 6 ** I can't figure out why the root is inserted two extra times**
    The parent of the left node 3 is 6
    The parent of the right node 11 is 6
    The parent of the right node 12 is 6 ** this node should be attaching to 11 instead of replacing it **
    The parent of the right node 8 is 6
    The parent of the right node 9 is 6
    The parent of the right node 10 is 6
    The parent of the right node 7 is 6
    preorder :(6 (3 )(7 ))
    postorder:(( 3) ( 7) 6)

    IMO, your insert method is way too complicated.
    Have a look at this pseudo code: that's all it takes to insert nodes in a BT:
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        private Node root;
        public void insert(T obj) {
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                let 'root' be the 'newNode'
            ELSE
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            END IF
        public void insert(Node parent, Node newNode) {
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                IF the left child of 'parent' is null
                    let 'newNode' be the left child of 'parent'
                ELSE
                    make a recursive call here: insert('???', 'newNode')
                END IF
            ELSE
                IF the right child of 'parent' is null
                    let 'newNode' be the right child of 'parent'
                ELSE
                    make a recursive call here: insert('???', 'newNode')
                END IF
            END IF
    }

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    Edited by: 815035 on Nov 23, 2010 12:27 PM

    The example of what the OP wants it to look like I thought was quite plain. Its right at the top of the post.
    Unfortunately it is also quite difficult to accomplish using System.out.print statements.
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    So you will need to traverse the tree at least twice.
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    The working out of offsets would have to be a depth search traversal I think
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                                  newNode.dData = dd;
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                                  newNode.dData = dd;
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              } // end else not root
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                        parent.rightChild = null;
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              else if(current.leftChild==null)
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                   else
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                        successor.leftChild = current.leftChild;
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              System.out.println();
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                   preOrder(localRoot.rightChild);
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                   postOrder(localRoot.rightChild);
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                   isRowEmpty = true;
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                             localStack.push(null);
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         } // end displayTree()
    } // end class Tree
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    theTree.insert(25, 1.2);
    theTree.insert(75, 1.7);
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    theTree.insert(43, 1.7);
    theTree.insert(30, 1.5);
    theTree.insert(33, 1.2);
    theTree.insert(87, 1.7);
    theTree.insert(93, 1.5);
    theTree.insert(97, 1.5);
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              theTree.insert(25, 1.2);
              theTree.insert(75, 1.7);
              theTree.insert(12, 1.5);
              theTree.insert(37, 1.2);
              theTree.insert(43, 1.7);
              theTree.insert(30, 1.5);
              theTree.insert(33, 1.2);
              theTree.insert(87, 1.7);
              theTree.insert(93, 1.5);
              theTree.insert(97, 1.5);
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                        break;
                   case 'f':
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    perform print_value using <ltree>. "print right
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