@ 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 <anika.jai...@gmail.com> 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 <mohit89m...@gmail.com> 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 <
> > ankit.agarwal.n...@gmail.com> 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 <ankuj2...@gmail.com> 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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>
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