"We dont have to do anything more for BST2 as all children of BST2 are
already at their proper place "
how can we insert BST2 into BST1 only checking the root value?? say
BST1's root value is 25 and BST2's root value is 35, then BST2 will be
inserted in right to BST1's root. If BST2 has minimum value 1
(leftmost child), BST1 has minimum 2, then it violates the BST
conditions. or even left child of BST2 is 24 it again violates the
condition :P

On Jan 29, 5:10 am, Ashish Meena <ashishmee...@gmail.com> wrote:
> Hi all,
>
> Lets say we do following steps to merge two BSTS's. Lets call them BST1 and
> BST2.
>
> 1. Check the value of root of BST2.
> 2. In BST1 find a place where root of BST2 can be put.
> 3. Remove the children from this place and store them as BST3 and BST4.
> 3. Attach BST2's root pointer at this place.
> We dont have to do anything more for BST2 as all children of BST2 are
> already at their proper place
> Now our BST1 has BST2 merged to it but it doesnt have BST3 and BST4.
> 4. Do inorder traversal of BST3 and BST4 and insert it to BST1.
>
> If we know the approximate size of BST1 and BST2 we can choose tree for our
> step 1 accordingly.
>
> This algorithm doesnt need any extra memory and can be almost O(n), if BST3
> and BST4 are very small trees.
>
> Does anybody see any issues with this approach?
>
> Regards,
> Ashish
>
>
>
> On Fri, Jan 29, 2010 at 2:52 PM, vivek bijlwan <viv...@gmail.com> wrote:
> > @ shingray ...  cartesian trees are not BSTs .  I guess nirmal's question
> > asks for a BST.
> > also cartesian trees have extra space requirements.
>
> > On Fri, Jan 29, 2010 at 12:07 AM, naga vinod kumar <
> > vinodkumark...@gmail.com> wrote:
>
> >> hi varun ,it cant be in O(n) time ,it can be merged in O(nlogn) time.
>
> >> On Thu, Jan 28, 2010 at 10:37 PM, Varun S V <varun...@gmail.com> wrote:
>
> >>> Delete the nodes in the second BST in postorder. As and when you delete
> >>> this node, insert it into the first BST.
>
> >>> On Thu, Jan 28, 2010 at 9:35 PM, Bijlwan <viv...@gmail.com> wrote:
>
> >>>> hey nirmal  . i don't get that when you merge the two linked list ,
> >>>> how do you get the BST?
> >>>> making the BST would itself be a O(nlogn) process?
>
> >>>> On Jan 28, 5:03 am, Nirmal <nirm...@gmail.com> wrote:
> >>>> > Given two binary search trees, how to merge them in O(n) time and O(1)
> >>>> > space?
>
> >>>> > It can be done using O(n) space as below,
>
> >>>> > 1. covert BST #1 into linked list or sorted array
> >>>> > 2. covert BST #2 into linked list or sorted array
> >>>> > 3. merge them...
>
> >>>> > but how to do this in place?
>
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