Any help regarding the preprocessing required for O(1) operation for
LCA??? I read the wiki's article involving eulers traversal.. but its
nt clear.
On Aug 24, 1:12 am, DK wrote:
> Hmm.. Interesting question.
>
> *Here's a solution:*
>
> There are n elements in the tree and therefore n^2 paths in
Hmm.. Interesting question.
*Here's a solution:*
There are n elements in the tree and therefore n^2 paths in the tree
(duplicate endpoints allowed and 2 endpoints correspond to exactly 1 simple
path in the tree).
1. For each node in the BST, store the sum of node values on the path to the
root
I'm very sry dude..It was not me as any intelligent person can
understand.
On Tue, Aug 23, 2011 at 3:49 AM, Anil Arya wrote:
> fuck
>
>
> On Mon, Aug 22, 2011 at 12:52 AM, Dhriti Khanna wrote:
>
>> void search_path( root , int i )
>> {
>>static char str[100];
>>if(root == nul
fuck
On Mon, Aug 22, 2011 at 12:52 AM, Dhriti Khanna wrote:
> void search_path( root , int i )
> {
>static char str[100];
>if(root == null)
> return;
>
>if( i == 0 )
> str[i] = root->data;
>
>else str[i] = str[i-1] + root->data; // Maintaining t
void search_path( root , int i )
{
static char str[100];
if(root == null)
return;
if( i == 0 )
str[i] = root->data;
else str[i] = str[i-1] + root->data; // Maintaining the cumulative
sum of all the sums till now.
if ( str[i] == value )
10
20 30
4 66 7
now how do u make 60 => 20 +10+30
how do u make 36 :: 10 + 20 +6 30+6
On Fri, Aug 19, 2011 at 8:37 PM, Arun Vishwanathan
wrote:
> @mac: just to clear my understanding, u need to print paths that sum up to
> a Val
@mac: just to clear my understanding, u need to print paths that sum up to a
Value which means the value of the nodes in a path is added right to see if
it satisfies this Value? does the value at each node have any relevance as
such? I mean can it be any value at any node or does value at a node sa
start from leaves.(leaves have possible sum only its value)
go up step by step
save all the possible sums on that node.
for example
if left node has possible sums( 4, 6, 7 ,13) and right node has
possible sums (3, 5, 9); and node itself value has 5 than
this node all possible sums will be (8, 10,
any suggestion frds ?? my thinking is eitther
a) root + left formsa path
b) oot +right form a path
c) left +root +right form a path
d) path only in left
e) path only in right
this thought process seems to be worng to me as its highly expensive and i
am myself not convinced of what to do next . An
