Let No of flips reqd = solve(N, V) L=left subtree. R = right subtree N = root node.
If V=0, then only problem is when L = 1 and R=1. (otherwise atmax changing the root node will do) then ans = min(solve(L, 0), solve(R,0)) + (R=AND)?0:1 If V=1 then only problem is when L=0 and R=0 similarly dealt. If V=something else output -1 -- 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.
