This can be done using single array too... :) :) Do anybody wants the code?
On Sun, Jul 17, 2011 at 3:04 PM, sagar pareek <sagarpar...@gmail.com> wrote: > This can be done like this > > 1. find out the height of the tree > 2. make the number of arrays(node* pointers)=height of tree > 3. traverse the tree from root as > arr0[0]=root; > arr1[0]=root->left; > arr1[1]=root->right > arr2[0]=arr1[0]->left > arr2[1]=arr1[1]->right > . > . > . and so on > note:- if any arr[n]==NULL then make corresponding left and right entries > NULL too > now make the tree entries as :- > arr[n]->right=arr[n+1] > if arr[n] is last entry of tree make its right node NULL > > we are done :) > > > > On Sun, Jul 17, 2011 at 11:22 AM, naveen ms <naveenms...@gmail.com> wrote: > >> in this recursive code...the right link node will point to its sibling >> to the right (if it has) or else it will be null. >> the left link of the node will point to its child(if it has) or else >> it will be null. >> >> regards, >> Naveen >> CSE >> R.V.C.E, Bangalore. >> >> -- >> You received this message because you are subscribed to the Google Groups >> "Algorithm Geeks" group. >> To post to this group, send email to algogeeks@googlegroups.com. >> To unsubscribe from this group, send email to >> algogeeks+unsubscr...@googlegroups.com. >> For more options, visit this group at >> http://groups.google.com/group/algogeeks?hl=en. >> >> > > > -- > **Regards > SAGAR PAREEK > COMPUTER SCIENCE AND ENGINEERING > NIT ALLAHABAD > > -- **Regards SAGAR PAREEK COMPUTER SCIENCE AND ENGINEERING NIT ALLAHABAD -- You received this message because you are subscribed to the Google Groups "Algorithm Geeks" group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en.
