@Surender: sorry, i had missed a case. We need a 3 way comparison.
heres the correct version

btree* max_tree(btree *root)
{
        if(root == NULL)
                return root;
        btree * current = root;
        while(current->right != NULL)
        {
                current = current->right;
        }
        return current;
}

btree * min_tree(btree *root)
{
        if(root == NULL)
                return root;
        btree * current = root;
        while(current->left != NULL)
        {
                current = current->left;
        }
        return current;
}

void binarytreetobst(btree *root)
{
//base-case tree is empty
        if(root == NULL)
                return;
        else if(root->left == NULL && root->right == NULL) //base-case
tree
of size 1
                return;
        else
        {
                binarytreetobst(root->left);
                binarytreetobst(root->right);
                btree* max = max_tree(root->left);
                if(max && max->data > root->data)
                {
                        int temp = root->data;
                        root->data = max->data;
                        max->data = temp;
                        binarytreetobst(root->left);
                }
                btree* min = min_tree(root->right);
                if(min && min->data  < root->data)
                {
                        int temp = root->data;
                        root->data = min->data;
                        min->data = temp;
                        binarytreetobst(root->right);
                        max = max_tree(root->left);
                        if(max && max->data > root->data)
                        {
                             int temp = root->data;
                             root->data = max->data;
                             max->data = temp;
                             binarytreetobst(root->left);
                        }
                }
        }
}

On Jul 19, 8:38 am, surender sanke <[email protected]> wrote:
> @Damanshu for
>         1
>       /   \
>      2     3
>     /  \
>    4   5
>   /  \
> 6    7
>
> im ending up at some non BST
>
> surender
>
>
>
>
>
>
>
> On Tue, Jul 19, 2011 at 4:06 AM, Dumanshu <[email protected]> wrote:
> > @Gaurav: The best solution would be to manipulate the given BTree in
> > place and get the BST. We don't need a separate tree but convert the
> > existing one using the same space occupied by nodes of BT already in
> > BST.
>
> > On Jul 19, 2:06 am, Gaurav Popli <[email protected]> wrote:
> > > cant we just do....traverse using recursion and instead of printing it
> > > just pass to function which is making BST....??
> > > and is this right as someone above said first sort it and then make
> > BST...
> > > dont we want that root of both Tree to be same or something like
> > that...??
>
> > > On Tue, Jul 19, 2011 at 2:17 AM, Dumanshu <[email protected]> wrote:
> > > > @Balaji: for third question, were u asked to write the code?
>
> > > > On Jul 18, 10:04 pm, Balaji S <[email protected]> wrote:
> > > >> wats the mistake..
>
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