Sorry i typed wrongly
  tree is
       2
1              3

preorder traversal is 123 and same for other tree as well. Please check !

On Wed, Jul 4, 2012 at 5:24 PM, a g <[email protected]> wrote:

>
>          1
> 2             3   is not a BST and its pre-order traversal is 1 2 3, pre
> order of other is 3 2 1 .
>
>
>
> On Wed, Jul 4, 2012 at 5:17 PM, Bhupendra Dubey 
> <[email protected]>wrote:
>
>> Consider trees:
>>
>>               1                              3
>>        2           3                                 2
>>                                               1
>> pre order traversal  is same still (1 , 2 ,3) even though ist tree
>> doesn't satisfy the criteria . So I don't think it can be determined.
>>
>> On Wed, Jul 4, 2012 at 4:16 PM, Ashish Goel <[email protected]> wrote:
>>
>>> Q4
>>>
>>>
>>> vector<String> prefix;
>>> prefix[0]=NULL;
>>> prefixCount =1;
>>> for (int i=0;i<n;i++)
>>>   for (int j=0;j<n;j++)
>>>     for (int k=0; k<prefixCount;k++)
>>>     {
>>>          if (visited[i][j]) continue;
>>>          visited[i][j] = true;
>>>          String s=prefix[k]+a[i][j];
>>>          if (isWord(s) { printWord(s); prefix[k]=s; continue;}
>>>          else if isPrefix(s) prefix[prefixCount++] = s;
>>>          else removePrefix(prefix[k], prefixCount);
>>>          prefix[prefixCount++] = String(a[i][j];
>>>      }
>>> Best Regards
>>> Ashish Goel
>>> "Think positive and find fuel in failure"
>>> +919985813081
>>> +919966006652
>>>
>>>
>>> On Wed, Jul 4, 2012 at 4:13 PM, Ashish Goel <[email protected]> wrote:
>>>
>>>> 1. inverted hasp map
>>>> 2. not clear
>>>> 3. VLR, how do you identify end of L and start of R, question incomplete
>>>> 4. One problem: consider
>>>>
>>>> .......
>>>> a b...
>>>> c d...
>>>> .......
>>>>
>>>> if ab is a prefix, can aba be another prefix, i would assume so. But if
>>>> that is true, i am not sure if this program will come to an end.
>>>> vector<String> prefix;
>>>> prefix[0]=NULL;
>>>> prefixCount =1;
>>>> for (int i=0;i<n;i++)
>>>>   for (int j=0;j<n;j++)
>>>>     for (int k=0; k<prefixCount;k++)
>>>>     {
>>>>          String s=prefix[k]+a[i][j];
>>>>          if (isWord(s) { printWord(s); prefix[k]=s; continue;}
>>>>          else if isPrefix(s) prefix[prefixCount++] = s;
>>>>          else removePrefix(prefix[k], prefixCount);
>>>>          prefix[prefixCount++] = String(a[i][j];
>>>>      }
>>>> Best Regards
>>>> Ashish Goel
>>>> "Think positive and find fuel in failure"
>>>> +919985813081
>>>> +919966006652
>>>>
>>>>
>>>>
>>>> On Wed, Jul 4, 2012 at 12:22 PM, Decipher <[email protected]>wrote:
>>>>
>>>>> Find the next higher number in set of permutations of a given number
>>>>
>>>>
>>>>
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>>
>>
>>
>> --
>> Thanks & regards
>> Bhupendra
>>
>>
>>
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-- 
Thanks & regards
Bhupendra

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