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Asked: 2026-06-01T11:56:50+00:00

I’m trying to implement AVL. Here’s my insert, balance_tree, check_bf (balance factor), and single

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I’m trying to implement AVL. Here’s my insert, balance_tree, check_bf (balance factor), and single left rotate functions in order:

BinaryNode *BinarySearchTree::insert(int x,BinaryNode *t, int dpt) throw(DuplicateItem)
{
    if (t == NULL) t = new BinaryNode(x,NULL,NULL,dpt+1);
    else if (x < t->element) t->left = insert(x, t->left, dpt+1);
    else if (x > t->element) t->right = insert(x, t->right, dpt+1);
    else
        throw DuplicateItem();
    balance_tree(t);
    return t;
}
BinaryNode* BinarySearchTree::balance_tree(BinaryNode *t)
{
    double debug = check_BF(t);
    while(check_BF(t)>1 || check_BF(t)<-1)
    {
        if(check_BF(t)>1)
        {
            if(check_BF(t->right)<-1) return doubleLeft(t);
            else return singleLeft(t);
        }
        else if(check_BF(t)<-1)
        {
            if(check_BF(t->left)>1) return doubleRight(t);
            else return singleRight(t);
        }
    }
}
double BinarySearchTree::check_BF(BinaryNode *t)
{
    double l, r;
    if(t->left!=NULL) l = t->height(t->left)+1;
    else l=0;

    if(t->right!=NULL) r = t->height(t->right)+1;
    else r=0;

    return r-l;
}
BinaryNode* BinarySearchTree::singleLeft(BinaryNode *t)
{
    BinaryNode* Y = t;
    if(Y!=NULL)
    {
        t = t->right;
        Y->right = t->left;
        t->left=Y;
    }
    return t;
}

I tried it out with a small tree that requires a single left rotation:

1 <----t
 \
  2
   \
    3

At the end of the single left rotation function, t points to 2 with 1 as the left node and 3 as the right node, so the function works. The tree looks like this:

  2<----t
 / \
1   3

However, when it exits the function, t points to 1 with no left or right node. I don’t understand what happens between that return t and the right bracket } that ends the function that changes t. Can anyone help?

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1 Answer

  1. Editorial Team

    What is missing here is the line in which you call the function in your testing. I think I hear you saying that t is unchanged, but the node to which t points is changed. That the node to which t points (the 1) is changed is the expected behaviour. That t is unchanged is not what you’re expecting. Your routine returns a value. Are you assigning that value to t, or just expecting t to be altered by the routine?

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