You could do any order traversal to get the list of values. - O(n)
Apply the kth order statistic (where k = numNodes/2) to get the value - O(n)

Adjust the median depending on whether numNode is odd or even

On Sun, Mar 27, 2011 at 8:03 PM, Balaji Ramani
<rbalaji.psgt...@gmail.com> wrote:
> Hi,
>
> This is one approach to it:
>
> 1) Go to the first node in inorder traversal. Let the pointer be LP.
> (Push the intermediate nodes to Lstack while doing this )
> 2) Go to the last node in inorder traversal. Let the pointer be RP.
> (Push the intermediate nodes to Rstack while doing this )
>
> while(LP->data < RP->data){
>   LP = inordersuccessor(LP,Lstack)
>   // gets inorder successor and modifies Lstack accordingly
>   RP = inorderpredecessor(RP,Rstack)
>   // gets inorder predecessor and modifies Rstack accordingly
>   Lprev = LP
>   Rprev = RP;
> }
>
> if(LP->data = RP->data){ // even number of elements
>   retrun LP->data;
> }else{
>   return Lprev->data + Rprev->data / 2
> }
>
> Time: O(n)
> Memory: 2 * O ( log n) average
>
> Thanks,
> Balaji.
>
> On Sun, Mar 27, 2011 at 7:17 PM, bittu <shashank7andr...@gmail.com> wrote:
>>
>> @all
>>
>> wake up geeks
>>
>>
>> Thanks
>> Shashank
>>
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