[algogeeks] Re: BST Problem

2010-08-25 Thread Giri
@arvind:
had i knwn would hv posted it

On Aug 23, 8:59 am, "R.ARAVINDH"  wrote:
> @giri:
>
> can u post d correct answer??

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Re: [algogeeks] Re: BST Problem

2010-08-23 Thread TurksHead Education
I am not sure if I am repeating the answer:

The problem will reduce to find the pair of elements which will sum up to a
particular number. Then read the below article,

http://www.rawkam.com/?p=345



On Mon, Aug 23, 2010 at 9:29 AM, R.ARAVINDH  wrote:

> @giri:
>
> can u post d correct answer??
>
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Re: [algogeeks] Re: BST Problem

2010-08-23 Thread Raj N
Perform inorder traversal and store in an array.
low = 0, high = size-1

while(low<=high)
{
if ( a[low] + a[high] < sum)
low++;
else if (a[low] + a[high] > sum)
high--;
else
return a[high] and [low]
}
On Mon, Aug 23, 2010 at 9:29 AM, R.ARAVINDH  wrote:

> @giri:
>
> can u post d correct answer??
>
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[algogeeks] Re: BST Problem

2010-08-22 Thread R.ARAVINDH
@giri:

can u post d correct answer??

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[algogeeks] Re: BST Problem

2010-08-22 Thread R.ARAVINDH
@giri:

thnx frnd...sorry ppl . ignore my post :(

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[algogeeks] Re: BST Problem

2010-08-22 Thread Giri
frnd check ur code.. t contains much errors..
consider the following eg.:
k=5;3
   1 4
 2 5

On Aug 22, 2:30 pm, "R.ARAVINDH"  wrote:
> can v do like  this???
>
> findnodes(root,sum)
> {
> if(root==abs(sum-root->data))
> print (the data is root->data, sum-(root->data));
> else
> if(rootdata))
> findnodes(root->right,sum-root->data)
> else if(root>abs(sum-root->data))
> findnodes(root->left,sum-root->data)
> else if(root->left==NULL || root->right==NULL) print(root,sum-root);
>
>
>
> }

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[algogeeks] Re: BST Problem

2010-08-22 Thread R.ARAVINDH
for the above post i have assumed that the two nodes whose sum is k is
present in the BST...

so correct me if m wrong

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[algogeeks] Re: BST Problem

2010-08-22 Thread R.ARAVINDH
can v do like  this???

findnodes(root,sum)
{
if(root==abs(sum-root->data))
print (the data is root->data, sum-(root->data));
else
if(rootdata))
findnodes(root->right,sum-root->data)
else if(root>abs(sum-root->data))
findnodes(root->left,sum-root->data)
else if(root->left==NULL || root->right==NULL) print(root,sum-root);
}

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[algogeeks] Re: BST Problem

2010-08-10 Thread Avik Mitra
@Sekin

Yes that's true..

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Re: [algogeeks] Re: BST Problem

2010-08-10 Thread Seçkin Can Şahin
Avik, yes the answer is obvious but your code doesn't find that.
that 2 pointer approach is the correct one.

On Tue, Aug 10, 2010 at 12:07 AM, Avik Mitra  wrote:

> @Sekin.
>
> Sort the elements (increasing order). This has already been mentioned.
> So, answer will be 1, 100.
>
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[algogeeks] Re: BST Problem

2010-08-10 Thread Avik Mitra
@Sekin.

Sort the elements (increasing order). This has already been mentioned.
So, answer will be 1, 100.

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Re: [algogeeks] Re: BST Problem

2010-08-06 Thread Seçkin Can Şahin
Chonku, you can do that only when you have the links to parent nodes. I
couldn't come up with a way of doing what you said on a basic BST(nodes
having pointers only to their 2 children) that is why I suggested using an
array. It doesn't change the overall complexity but if you have an idea
about how implement your idea on a basic BST, I would like to hear it.

Thanks,
Seckin

On Fri, Aug 6, 2010 at 2:56 AM, sharad kumar wrote:

> do the inorder traversal of the bst ...this gives the sorted array..
> from that use
>
> int i=0,j=length(array)
> while(i {
> if(array[i]+array[j]>sum)
> --j;
> else if(array[i]+array[j] ++i;
> else if((array[i]+array[j])==sum)
> return i,j
> else
> ++i,--j;
> }
>
>
> On Fri, Aug 6, 2010 at 3:10 PM, Chonku  wrote:
>
>> Two inorders would achieve the same thing without using an array. One
>> pointer running inorder with LDR and other pointer running inorder with RDL.
>> Compare the sum at the two nodes and then adjust them accordingly.
>>
>> On Fri, Aug 6, 2010 at 2:11 PM, Manjunath Manohar <
>> manjunath.n...@gmail.com> wrote:
>>
>>> the solution elegant..but is there any on the fly method by just
>>> exploiting the BST propertyby using left and right pointers
>>>
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Re: [algogeeks] Re: BST Problem

2010-08-06 Thread sharad kumar
do the inorder traversal of the bst ...this gives the sorted array..
from that use

int i=0,j=length(array)
while(isum)
--j;
else if(array[i]+array[j] wrote:

> Two inorders would achieve the same thing without using an array. One
> pointer running inorder with LDR and other pointer running inorder with RDL.
> Compare the sum at the two nodes and then adjust them accordingly.
>
> On Fri, Aug 6, 2010 at 2:11 PM, Manjunath Manohar <
> manjunath.n...@gmail.com> wrote:
>
>> the solution elegant..but is there any on the fly method by just
>> exploiting the BST propertyby using left and right pointers
>>
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Re: [algogeeks] Re: BST Problem

2010-08-06 Thread Chonku
Two inorders would achieve the same thing without using an array. One
pointer running inorder with LDR and other pointer running inorder with RDL.
Compare the sum at the two nodes and then adjust them accordingly.

On Fri, Aug 6, 2010 at 2:11 PM, Manjunath Manohar
wrote:

> the solution elegant..but is there any on the fly method by just exploiting
> the BST propertyby using left and right pointers
>
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Re: [algogeeks] Re: BST Problem

2010-08-06 Thread Manjunath Manohar
the solution elegant..but is there any on the fly method by just exploiting
the BST propertyby using left and right pointers

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Re: [algogeeks] Re: BST Problem

2010-08-05 Thread Seçkin Can Şahin
as a hint, convert the BST to a sorted array and take two pointers one
pointing to the first number and the other pointing to the last. Then, move
pointers appropriately to find the two numbers summing up to k.

complexity: O(n)

2010/8/5 Seçkin Can Şahin 

> what about the case:
> array : 1 3 10 100 and k = 101. Your code doesn't find it I suppose.
>
>
> On Thu, Aug 5, 2010 at 11:15 PM, Avik Mitra  wrote:
>
>>
>> Inorder traversal of the BST will give elements in sorted way. Let us
>> assume that the sorted elements are in an array A of length N.
>> set i=1;
>> while i > {
>>  if a[i] > k, then output: "No such node"
>>  else if(a[i]==k)
>>  {
>>if (a[i+1] ==0)
>> output: "Two nodes found" BREAK;
>>else
>>   output: "No such node."  BREAK.
>>
>>  }
>>
>>  else if(a[i] > {
>>   if(a[i]+a[i+1]==k)
>>output: "Two nodes found" BREAK.
>>  else if(a[i]+a[i+1] >k)
>>output: "No such node" BREAK
>>  else if(a[i] +a[i+1] < k)
>>  i++ ;
>>  }
>> }//End of while-loop.
>>
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Re: [algogeeks] Re: BST Problem

2010-08-05 Thread Seçkin Can Şahin
what about the case:
array : 1 3 10 100 and k = 101. Your code doesn't find it I suppose.

On Thu, Aug 5, 2010 at 11:15 PM, Avik Mitra  wrote:

>
> Inorder traversal of the BST will give elements in sorted way. Let us
> assume that the sorted elements are in an array A of length N.
> set i=1;
> while i  {
>  if a[i] > k, then output: "No such node"
>  else if(a[i]==k)
>  {
>if (a[i+1] ==0)
> output: "Two nodes found" BREAK;
>else
>   output: "No such node."  BREAK.
>
>  }
>
>  else if(a[i]  {
>   if(a[i]+a[i+1]==k)
>output: "Two nodes found" BREAK.
>  else if(a[i]+a[i+1] >k)
>output: "No such node" BREAK
>  else if(a[i] +a[i+1] < k)
>  i++ ;
>  }
> }//End of while-loop.
>
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[algogeeks] Re: BST Problem

2010-08-05 Thread Avik Mitra

Inorder traversal of the BST will give elements in sorted way. Let us
assume that the sorted elements are in an array A of length N.
set i=1;
while i  k, then output: "No such node"
  else if(a[i]==k)
  {
if (a[i+1] ==0)
 output: "Two nodes found" BREAK;
else
   output: "No such node."  BREAK.

  }

 else if(a[i] k)
output: "No such node" BREAK
  else if(a[i] +a[i+1] < k)
  i++ ;
 }
}//End of while-loop.

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