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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