Non recursive preorder traversal of binary tree

hi,
I am trying to implement a non-recursive traversal of binary tree. I already know the recursive one.
I am trying to do it by using a Stack.
I begin by Pushing the root of an element on to a stack, and then run a while loop in which i pop an element of the stack and get its children from right to left. and push it in the same order on to the stack. So during the next iteration of my while loop the top most element gets popped and its children and pushed on to the stack in the above manner.
but when i pop an element from a stack its popped as an object so i dont know how to access its children.
help me i am really stuck.

Hi, I suppose you have something like this :
class Stack {
  public void push( Object object ) throws ... { ... }
  public Object pop() throws ... { ... }
class Element {
  Element elem;
  stack.push(elem);
  /* because pop() method return an object of type Object
  ** if you are sure that your stack only contains Element object
  ** then you need to cast (change the type of) what the pop() method
  ** returns in this way :
  elem = (Element)stack.pop();
  ...further reading on casting will be a good idea anyway.

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  • Non-recursive postorder traversal (bst)

    Hello!
    I'm having difficulty writing a non-recursive postorder traversal for a binary search tree. Not getting an error via compiler but I know that the answer is incorrect. Please provide some insight on what I'm doing wrong!
    Here's an example of what I mean by incorrect..
    correct postorder answer (done recursively) --> 0 4 2 6 10 12 14 8 18 16
    what I'm getting using the function below --> 8 6 2 0 4 14 12 10 18 18
    And here's what I have:
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    LinkedStack<BinarySearchTreeNode<T>> stack = new LinkedStack<BinarySearchTreeNode<T>>();
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    When you post code, wrap it in code tags.
    Anyway, ultimately the solution is to put in debugging tags to see where things are going wrong.
    You don't really need that done variable. Just check the stack.
    Though actually IIRC, you'd actually need a queue.

  • How to remember the path while traverse a binary tree?

    Hi, again, I have difficulty handling tree search problems.
    The quesion is How to search for a node from a binary tree A, return true if found meanwhile generate the path which can be used to locate the node.
    I think the signature should be:
    // The path contains only 0s and 1s. 0 means go left and 1 means go right.
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    I tried to use recursion but havn't got a correct solution. I am thinking of usinga non-recursion algo.
    Anyone wants to help?

    Hi, JosAh,
    That's mind provoking. However, I do think it works. It does not pop some stuff it should pop. I tested it over a very simple tree and it failed. Did you test it? I might be wrong...
    The tree I am working on does not have null pointers, the condition to test if a node is a leaf is if(node.right == right). Namly, all the right pointer of leaves points to itself.
    So I changed your code to be:
    Stack search(Node root, Node node, Stack<Integer> path) {
         if (root == null || root.right ==right) return null;
         if (root.equals(node)) return path;
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    path.pop();
    path.push(1);
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    }I simply tested it with
    Stack<Integer> path = new Stack<Integer>();
    System.out.println( root, root.right.right, path);
    root is the root of a complete binary tree with 7 nodes(4 leaves).
    Apparenly, if the path is built correctly search(root, root.right.right, path) would return [1,1] whereas this seach returns [ 0 , 1, 1].
    Considerring , the right branch never pops, I changed it into
    Then I changed it to :
    Stack search(Node root, Node node, Stack<Integer> path) {
         if (root == null || root.right ==right ) return null;
         if (root.equals(node)) return path;
         path.push(0);
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    path.pop();
    path.push(1);
    if (search(root.right, node, path) != null) return path;
    path.pop();
    return path;
    With the same test case, it returns [].
    I will keep working on it.
    Cheers,
    Message was edited by:
    since81
    Message was edited by:
    since81

  • Traverse a binary tree from root to every branch

    I have a couple of other questions. I need to get all the different combinations of a binary tree and store them into a data structure. For the example in the code below, the combinations would be:
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    2) Start, A1, A2, B1, A3, B2, B3
    3) Start, A1, A2, B1, B2, A3, B3
    4) Start, A1, A2, B1, B2, B3, A3
    5) Start, A1, B1, A2, A3, B2, B3
    etc.
    I understand that this is very similar to the preorder traversal, but preorder does not output the parent nodes another time when the node splits into a left and right node. Any suggestions?
    * To change this template, choose Tools | Templates
    * and open the template in the editor.
    package binarytreetest;
    import java.util.ArrayList;
    import java.util.Iterator;
    * @author vluong
    public class BinaryTreeTest {
         * @param args the command line arguments
        public static void main(String[] args) {
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            int countB = 0;
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            ArrayList listB = new ArrayList();
            listA.add("A1");
            listA.add("A2");
            listA.add("A3");
            listB.add("B1");
            listB.add("B2");
            listB.add("B3");
            //listB.add("B1");
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            //printFromRoot(root);
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            private Node left;
            private Node right;
            private String value;
            public Node(String value){
                this.value = value;
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                    System.out.println("There is no left node. CountA is " + countA);
                    System.out.println("Created new node with value: " + listA.get(countA).toString() + " with parent, "
                            + node.value);
                    System.out.println();
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                    System.out.println("There is a left node. CountA + 1 is " + countA+1);
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                    System.out.println("There is no right node. CountB is " + countB);
                    System.out.println("Created new node with value: " + listB.get(countB).toString() + " with parent, "
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                    System.out.println();
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                    System.out.println("There is a right node. CountB + 1 is " + countB+1);
                    constructTree(node.right, countA, countB+1, listA, listB);
        }My second question is, if I need to add another list (listC) and find all the combinations of List A, listB and list C, is it correct to define the node class as
    public static class Node{
            private Node left;
            private Node mid;
            private Node right;
            private String value;
            public Node(String value){
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        }Node left = listA, Node mid = listB, Node right = listC
    The code for the 3 lists is below.
    3 lists (A, B, C):
    * To change this template, choose Tools | Templates
    * and open the template in the editor.
    package binarytreetest;
    import java.util.ArrayList;
    * @author vluong
    public class BinaryTreeTest {
         * @param args the command line arguments
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            insert(root, "A2");
            insert(root, "B1");
            insert(root, "B2"); 
            insert(root, "A2");
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            int countB = 0;
            int countC = 0;
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            ArrayList listB = new ArrayList();
            ArrayList listC = new ArrayList();
            listA.add("A1");
            listA.add("A2");
            //listA.add("A3");
            listB.add("B1");
            listB.add("B2");
            //listB.add("B3");
            //listB.add("B1");
            listC.add("C1");
            listC.add("C2");
            Node root = new Node("START");
            constructTree(root, countA, countB, countC, listA, listB, listC);
            //ConstructTree(root, countA, countB, listA, listB);
            //ConstructTree(root, countA, countB, listA, listB);
            printInOrder(root);
            //printFromRoot(root);
        public static class Node{
            private Node left;
            private Node mid;
            private Node right;
            private String value;
            public Node(String value){
                this.value = value;
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            if(countA < listA.size()){
                if(node.left == null){
                    System.out.println("There is no left node. CountA is " + countA);
                    System.out.println("Created new node with value: " + listA.get(countA).toString() + " with parent, "
                            + node.value);
                    System.out.println();
                    node.left = new Node(listA.get(countA).toString()); 
                    constructTree(node.left, countA+1, countB, countC, listA, listB, listC);   
                }else{
                    System.out.println("There is a left node. CountA + 1 is " + countA+1);
                    constructTree(node.left, countA+1, countB, countC, listA, listB, listC);   
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                if(node.mid == null){
                    System.out.println("There is no mid node. CountB is " + countB);
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                    System.out.println();
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                }else{
                    System.out.println("There is a right node. CountB + 1 is " + countB+1);
                    constructTree(node.mid, countA, countB+1, countC, listA, listB, listC);
            if(countC < listC.size()){
                if(node.right == null){
                    System.out.println("There is no right node. CountC is " + countC);
                    System.out.println("Created new node with value: " + listC.get(countC).toString() + " with parent, "
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                    System.out.println();
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                    constructTree(node.right, countA, countB, countC+1, listA, listB, listC);
                }else{
                    System.out.println("There is a right node. CountC + 1 is " + countC+1);
                    constructTree(node.mid, countA, countB, countC+1, listA, listB, listC);
        }Thank you in advance!

    It looks to me like you are interleaving two lists. It looks like you are doing this while leaving the two subsequences in their original order.
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    List path = new List();
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    In order to interleave 3 lists, you would add C and c arguments to the function and you would add one more line in the else block.

  • Traversing a binary tree

    Hi
    I've written the following code for converting Postfix to Infix,it reads a string then reads it letter by letter ,a stack and a Tree is used,and I want to traverse the last object that has poped from the stack,in the following code I used this line:
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    Thax
    Bita

    Well, two things. First, you're trying to assign to a method call, which isn't valid. The compiler probably told you this.
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    Could you please learn [How To Ask Questions The Smart Way|http://www.catb.org/~esr/faqs/smart-questions.html]?
    Could you please ask a specific question about a concrete problem that you find while doing your own homework?
    thank you :)thank you :)

  • Need Help with a String Binary Tree

    Hi, I need the code to build a binary tree with string values as the nodes....i also need the code to insert, find, delete, print the nodes in the binarry tree
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    // demonstrates binary tree
    // to run this program: C>java TreeApp
    import java.io.*; // for I/O
    import java.util.*; // for Stack class
    import java.lang.Integer; // for parseInt()
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         //public int iData; // data item (key)
         public String iData;
         public double dData; // data item
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         public Node rightChild; // this node's right child
         public void displayNode() // display ourself
              System.out.print('{');
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              System.out.print(", ");
              System.out.print(dData);
              System.out.print("} ");
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    class Tree
         private Node root; // first node of tree
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         {                           // (assumes non-empty tree)
              Node current = root; // start at root
              while(current.iData != key) // while no match,
                   if(key < current.iData) // go left?
                        current = current.leftChild;
                   else // or go right?
                        current = current.rightChild;
                   if(current == null) // if no child,
                        return null; // didn't find it
              return current; // found it
         } // end find()
         public Node recfind(int key, Node cur)
              if (cur == null) return null;
              else if (key < cur.iData) return(recfind(key, cur.leftChild));
              else if (key > cur.iData) return (recfind(key, cur.rightChild));
              else return(cur);
         public Node find2(int key)
              return recfind(key, root);
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    newNode.iData = id; // insert data
    newNode.dData = dd;
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    root = newNode;
    else // root occupied
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    while(true) // (exits internally)
    parent = current;
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    current = current.leftChild;
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    {                 // insert on left
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    return;
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    current = current.rightChild;
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    {                 // insert on right
    parent.rightChild = newNode;
    return;
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    } // end while
    } // end else not root
    } // end insert()
    public void insert(String id, double dd)
         Node newNode = new Node(); // make new node
         newNode.iData = id; // insert data
         newNode.dData = dd;
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              root = newNode;
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              Node current = root; // start at root
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                   parent = current;
                   //if(id < current.iData) // go left?
                   if(id.compareTo(current.iData)>0)
                        current = current.leftChild;
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                             return;
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                        current = current.rightChild;
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                        {                 // insert on right
                             parent.rightChild = newNode;
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              } // end while
         } // end else not root
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                   newNode.dData = dd;
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                   Node parent;
                   while(current != null)
                        parent = current;
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                             current = current.leftChild;
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                                  newNode.iData = id; // insert data
                                  newNode.dData = dd;
                                  return_val = newNode;
                                  parent.leftChild = newNode;
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                             {                 // insert on right
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                                  newNode.dData = dd;
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                                  parent.rightChild = newNode;
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              } // end else not root
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              Node current = root;
              Node parent = root;
              boolean isLeftChild = true;
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                   parent = current;
                   if(key < current.iData) // go left?
                        isLeftChild = true;
                        current = current.leftChild;
                   else // or go right?
                        isLeftChild = false;
                        current = current.rightChild;
                   if(current == null)
                        return false; // didn't find it
              } // end while
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              if(current.leftChild==null &&
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                        parent.rightChild = null;
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              else if(current.rightChild==null)
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                        root = current.leftChild;
                   else if(isLeftChild)
                        parent.leftChild = current.leftChild;
                   else
                        parent.rightChild = current.leftChild;
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              else if(current.leftChild==null)
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                        root = current.rightChild;
                   else if(isLeftChild)
                        parent.leftChild = current.rightChild;
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                        parent.rightChild = current.rightChild;
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                        Node successor = getSuccessor(current);
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              {                                 // left children,
                   successorParent = successor;
                   successor = current;
                   current = current.leftChild; // go to left child
              // if successor not
              if(successor != delNode.rightChild) // right child,
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                   successorParent.leftChild = successor.rightChild;
                   successor.rightChild = delNode.rightChild;
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              switch(traverseType)
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                   preOrder(root);
                   break;
              case 2: System.out.print("\nInorder traversal: ");
                   inOrder(root);
                   break;
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                   postOrder(root);
                   break;
              System.out.println();
         private void preOrder(Node localRoot)
              if(localRoot != null)
                   localRoot.displayNode();
                   preOrder(localRoot.leftChild);
                   preOrder(localRoot.rightChild);
         private void inOrder(Node localRoot)
              if(localRoot != null)
                   inOrder(localRoot.leftChild);
                   localRoot.displayNode();
                   inOrder(localRoot.rightChild);
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              if(localRoot != null)
                   postOrder(localRoot.leftChild);
                   postOrder(localRoot.rightChild);
                   localRoot.displayNode();
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              globalStack.push(root);
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              boolean isRowEmpty = false;
              System.out.println(
              while(isRowEmpty==false)
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                   isRowEmpty = true;
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                        System.out.print(' ');
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                             System.out.print(temp.iData);
                             localStack.push(temp.leftChild);
                             localStack.push(temp.rightChild);
                             if(temp.leftChild != null ||
                                  temp.rightChild != null)
                                  isRowEmpty = false;
                        else
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                             localStack.push(null);
                             localStack.push(null);
                        for(int j=0; j<nBlanks*2-2; j++)
                             System.out.print(' ');
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                   nBlanks /= 2;
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         } // end displayTree()
    } // end class Tree
    class TreeApp
         public static void main(String[] args) throws IOException
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              input = new BufferedReader (new FileReader ("one.txt"));
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         val1=0.1;
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              //val1=Integer.parseInt{Term};
              val1=val1+1;
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              System.out.println(val1);          
    theTree.insert(50, 1.5);
    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);
              theTree.insert(50, 1.5);
              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);
              while(true)
                   putText("Enter first letter of ");
                   putText("show, insert, find, delete, or traverse: ");
                   int choice = getChar();
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                        theTree.displayTree();
                        break;
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                        value = getInt();
                        theTree.insert(value, value + 0.9);
                        break;
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                        putText("Enter value to find: ");
                        value = getInt();
                        Node found = theTree.find(value);
                        if(found != null)
                             putText("Found: ");
                             found.displayNode();
                             putText("\n");
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                        break;
                   case 'd':
                        putText("Enter value to delete: ");
                        value = getInt();
                        boolean didDelete = theTree.delete(value);
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                             putText("Deleted " + value + '\n');
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                        break;
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              System.out.print(s);
              System.out.flush();
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              InputStreamReader isr = new InputStreamReader(System.in);
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              return Integer.parseInt(s);
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    String str = "Hello";
              int index = 0, len = 0;
              len = str.length();
              while(index < len) {
                   System.out.println(str.charAt(index));
                   index++;
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  • A Binary Tree Implementation in ABAP

    Hi,
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    Manjula.

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    right type ref to data,
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    data: tree type stree.
    data: int type i.
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    pls reward if helps,
    regards.

  • Reconstructing a Binary Tree

    Hi guys ;
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