@Dave I Think So We Can Construct In the same we construct singly
linked list ton BST .
Shashank
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struct bst * BSTsearch(struct bst ** MAIN_ROOT,key)
{
if ( MAIN_ROOT==NULL )
return NULL;
if(MAIN_ROOT-data==key)
return MAIN_ROOT;
@Muthuraj
DLL to BST back is not possible
In the first step while converting to DLLwe will create a sorted DLL
using inorder walk of the tree so DLL will represent the inorder sequence of
nodes of BST
and we know there can be many BST's for the given inorder
so we can not construct the same
Anyone able to code the dave's proposal to do inoder and reverse of inorder
at the same time?
On Mon, Jun 27, 2011 at 11:32 PM, Swathi chukka.swa...@gmail.com wrote:
Dave,
I am unable to write code for this so i am asking your help.
Thanks,
Swathi
On Mon, Jun 27, 2011 at 11:28 PM, Dave
@Bittu: When you are finishedm can you change the DLL back into the
original BST?
Dave
On Jun 29, 5:54 pm, bittu shashank7andr...@gmail.com wrote:
Algorithm:
1.Convert BST into sorted DLL which Will Take O(N) Time(Check previous
Posts Already Coded) you can see here cslibrary.stanford.edu/109
@Nishant: No need to store the data in an array. Do two inorder
traversals simultaneously. Let u and v be the current nodes of the two
traversals, respectively. If u + v x, then advance the v
traversal. If u + v x, advance the u traversal.
Dave
On Jun 27, 3:40 am, Nishant Mittal
@Dave
i think your solution won't work
consider inorder traversal of a BST is 1 6 7 8 15 and x = 14
initially both u,v (1,1)
according to u your algorithm will proceed like
(1,1) - (1,6) - (1,7) - (1,8) - (1,15) - (6,15) - (15,15)
and clearly in second step of your solution if (u+v)
It doesn't work. In your idea values of nodes represented by u and v always
increased, cause you are doing inorder traversal.
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@Sunny. Mea culpa. You are correct. Revised (and correct) algorithm.
Do two inorder traversals, one in the usual (descend to the left
before descendung to the right) direction and the other in the
reversed(descend to the right before descending to the left)
direction. Let u and r be the current
Yes, now it is fine enough :)
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Dave,
Can you provide the psuedo code for this..
Thanks,
Swathi
On Mon, Jun 27, 2011 at 7:30 PM, Dave dave_and_da...@juno.com wrote:
@Sunny. Mea culpa. You are correct. Revised (and correct) algorithm.
Do two inorder traversals, one in the usual (descend to the left
before descendung to the
@ankit, sorry i was mistaken its O(nlogn) for searching the two elements...
On Mon, Jun 27, 2011 at 8:04 PM, Swathi chukka.swa...@gmail.com wrote:
Dave,
Can you provide the psuedo code for this..
Thanks,
Swathi
On Mon, Jun 27, 2011 at 7:30 PM, Dave dave_and_da...@juno.com wrote:
suppose k is the sum to be found.
@vaibhav: yes it will stop when crossing is there means (j must be greater
than i).initially sum = a[i] + a[j] (where i = 0 n j = n- 1) n we will
increase i when sum is less than k and decrease j when sum k.stop if sum
== k. and if no i , j found till j i. then
@varun: Thanks...
On Mon, Jun 27, 2011 at 9:18 PM, varun pahwa varunpahwa2...@gmail.comwrote:
suppose k is the sum to be found.
@vaibhav: yes it will stop when crossing is there means (j must be greater
than i).initially sum = a[i] + a[j] (where i = 0 n j = n- 1) n we will
increase i when
@Swathi: No. I think the high level description should be adequate for
you to write your own code or pseudocode, albeit recognizing that you
may have to look up how to do an inorder traversal using a stack
instead of recursion.
Dave
On Jun 27, 9:34 am, Swathi chukka.swa...@gmail.com wrote:
Dave,
I am unable to write code for this so i am asking your help.
Thanks,
Swathi
On Mon, Jun 27, 2011 at 11:28 PM, Dave dave_and_da...@juno.com wrote:
@Swathi: No. I think the high level description should be adequate for
you to write your own code or pseudocode, albeit recognizing that you
how about converting the BST into a doubly linked list...but this will
definitely modify the BST..
On Mon, Jun 27, 2011 at 11:32 PM, Swathi chukka.swa...@gmail.com wrote:
Dave,
I am unable to write code for this so i am asking your help.
Thanks,
Swathi
On Mon, Jun 27, 2011 at 11:28 PM,
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