@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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