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find the inorder traversal of the tree and print them all by negating every alternate number


find the inorder traversal of the tree and print them all by negating every alternate number

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find the inorder traversal of the tree and print them all by negating every alternate number
Tag : cpp , By : Mariamario
Date : January 12 2021, 08:33 AM

it helps some times All you might have to do , add a additional argument to keep track of alternate sign, something like following :
Node * func(struct Node * root, int& signV ){
  if(root!=NULL)
  {
    func(root->lChild, signV);
    cout<<root->nodeValue * signV <<" "; signV *= -1 ; // Change sign here
    func(root->rChild, signV);
  }
  return NULL;
}

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How to print tree in an inorder traversal (java)?


Tag : java , By : acolomba
Date : March 29 2020, 07:55 AM
I hope this helps you . I am trying to print my AVLtree as following, , Here's the code:
void inOrder(AVLNode node, int depth, StringBuilder sb)
{
  if(node == null) return;

  inOrder(node.left, depth+1, sb);

  for(int a=0;a<depth;a++) sb.append(".");
  sb.append(node.data + "\n");

  inOrder(node.right, depth+1, sb);

}

Find the inorder traversal from its preorder traversal sequence in binary tree


Tag : development , By : Dominique Vocat
Date : March 29 2020, 07:55 AM
this will help I don't think you can find out the inorder traversal based on just the preorder traversal for a binary tree. As you said for binary search tree, sorting will give you the inorder traversal.

How to print a parenthesized inorder AVL tree traversal in java


Tag : java , By : markku
Date : March 29 2020, 07:55 AM
To fix the issue you can do I am trying to print my AVL tree in the following way: , As Rohit Jain commented, just add them in the printTree method:
private void printTree(Node t){
    if(t != null){
        System.out.print("(");
        printTree(t.left);
        System.out.print(t.element + " ")
        printTree(t.right);
        System.out.print(")");
    }
}

Binary tree - find position in inorder traversal


Tag : java , By : Anand
Date : March 29 2020, 07:55 AM
wish help you to fix your issue You can also use a counter in the recursive approach. However, you can't simply pass an int counter argument - you need all calls to see the "same" counter, so you will have to wrap it in a class (or, as in this case, an inner class):
public static class Counter {
   private int value;
   public Counter(int initialValue) { value = initialValue; }
   public boolean decrement() { value--; return value == 0; }
   public boolean expired() { return value <= 0; }
}

public Node inOrderTraverseTree(Node root, Counter counter){
   if  (root != null && ! counter.expired()) {
       Node left = inOrderTraverseTree(root.leftChild, counter);
       if (left != null) {
            return left;
       } else if (counter.decrement()) {
            return root;
       } else {
            return inOrderTraverseTree(root.rightChild, counter); 
       }
   } else {
       return null;
   }
}
Node the9th = inOrderTraverseTree(root, new Counter(9));

Binary Search Tree in java, Inorder traversal is only recognizing a single number in the tree


Tag : java , By : new Blackberry devel
Date : March 29 2020, 07:55 AM
I wish this helpful for you In your inorder method, you are always printing the root node, instead of the node you are currently at.
Instead of:
System.out.println(root.data + " ");
System.out.println(subtree.data + " ");
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