For solution to this problem see
http://groups.google.co.in/group/algogeeks/browse_thread/thread/adc95dcf96020a7/2943fd3c5a116fee?hl=en&lnk=gst&q=binary+tree+to+binary+serach+tree#
In that thread, I have a doubt regarding solution posted by @"Algoose
Chase". The code posted is good as I feel except for the condition
if( NodeL->Data > root->data)
{
....
}
if( NodeR->Data <= root->data)
{
....
}
Here If you find that the present node's (say 'n1') value if less than
the MAX (say 'n2' ) of so far constructed BST in the left tree, then
if is for sure that that MAX ('n2') of the left tree will replace the
present node 'n1'. This will make sure that all nodes to the left are
less than the new root 'n2', but we are not sure that the remaining
nodes n the left tree are all less than 'n1'.
So in "if( NodeL->Data > root->data)" condition, "temp = root-data;
root->data = NodeL->data" is correct but instead of doing
Nodel->data = root->data;
BinarytoBST(NodeL)
we should simply say
deleteNode(tree->left,NodeL);//This will delete NodeL from a BST
rooted at tree->left, taking into account delete conditions for
deleting right most child of BST //(because NodeL was the
right most child)
InsertNode(tree->left,temp);
Do share your views.
Thanks,
Sourav
On Sep 27, 7:58 am, prodigy <1abhishekshu...@gmail.com> wrote:
> 15
> / \
> 8 25
> / \
> 20 22
>
> On Sep 26, 10:45 am, Chonku <cho...@gmail.com> wrote:
>
> > This can also be done if we do an inorder traversal of the binary tree and
> > look for the longest continuous sequence of numbers in ascending order.
>
> Your idea will fail for above case.
>
> In Order => 8 15 20 25 22
> longest continuous sequence of numbers in ascending order => 8 15 20
> 25
>
> But that's not the answer (I hope you realize what correct output
> would be )
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