Made some changes.
flattenTree(TreeNode node,TreeNode previous) {
if (node is leaf) {
previous = node
return;
}
flattenTree(node->left,previous);
if (previous != null) {
previous->next = node;
previous=node;
}
flattenTree(node->right,previous);
}//End-flattenTree
On Sat, Aug 28, 2010 at 9:03 PM, Giri <giri.pe...@gmail.com> wrote:
> @Chonku:
> yours is wrong. consider the given ex,.
> 50
> / \
> 25 60
> / \ / \
> 5 30 55 75
>
> 5 will become head. 5->next=25. 25->next=30. then 25 will be returned
> up.
> so 25->next=50. which is wrong
>
> On Aug 26, 11:36 pm, Chonku <cho...@gmail.com> wrote:
> > At first, store the pointer to the first node in inorder traversal (in
> this
> > case 5) because its going to be the head of the list.
> > Then use the following logic.
> >
> > flattenTree(TreeNode node) {
> > if (node is leaf node)
> > return node;
> >
> > if (node.left exists) then
> > flattenTree(node.left)->next = node;
> >
> > if (node.right exists) then
> > node->next = flattenTree(node.right);
> >
> > return node;
> >
> > }
> >
> > On Thu, Aug 26, 2010 at 11:07 PM, Yan Wang <wangyanadam1...@gmail.com
> >wrote:
> >
> >
> >
> > > I can only figure out the inorder traversal...
> >
> > > On Thu, Aug 26, 2010 at 9:59 AM, krazee koder <aravindhr...@gmail.com>
> > > wrote:
> > > > Give all possible methods to flatten a binary tree to a linked list.
> >
> > > > for eg.
> >
> > > > 50
> > > > / \
> > > > 25 60
> > > > / \ / \
> > > > 5 30 55 75
> >
> > > > should be flattened to 5->25->30->50->55->60->75
> >
> > > > PS: note that the tree should be converted to the LL and no separate
> > > > LL should be formed.
> >
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