unlink each node in original tree in postorder, and insert these nodes in
new bst tree

surender

On Tue, Nov 8, 2011 at 4:48 AM, vikas <[email protected]> wrote:

> @ Above
> no need to have another array or nything
> binTreeToBST(node *root)
> {
>   if(!root )return;
>  node *newRoot;
>   binTreeToBSTConv(root, &newRoot);
> }
>   binTreeToBSTConv(node *old, node *new)
> {
>       if(!old ) return;
>          binTreeToBSTConv(old->left, new);
>          binTreeToBSTConv(old->rigth, new);
>          if(old){
>           if(!new) new =old
>           else{
>                 insertNode(new, old);
>              }
>           }
>
> }
>
> On Nov 6, 11:30 am, Anika Jain <[email protected]> wrote:
> > @mohit: your algo will add assurance that the tree is balanced..
> otherwise
> > ankit's approach is sufficient.
> >
> > On Sat, Nov 5, 2011 at 8:49 PM, mohit verma <[email protected]>
> wrote:
> > > another way is : convert binary tree to link list , sort the list and
> > > using divide  and conquer approach create the BST.
> >
> > > From link list to BST : find mid of sorted link list , make it root
> node
> > > and put left of it to recursive(list,start,mid->prev) and
> > > root->right=recursive(list,mid->next,last);
> >
> > > Let me know if something is wrong in this approach.
> >
> > > On Sat, Nov 5, 2011 at 3:48 PM, ankit agarwal <
> > > [email protected]> wrote:
> >
> > >> I think it's the only way as you need to traverse the entire binary
> > >> tree to do it.
> >
> > >> On Oct 31, 9:45 pm, Ankuj Gupta <[email protected]> wrote:
> > >> > How to convert a Binary tree to BST ? Naive way is to create each
> node
> > >> > of  Binary tree one by one and keep on creating the BST.
> >
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