SPOJ #4 Transform the Expression

Not hard to know it is simply transform from in-order to post-order.
My first idea is to build a tree from in-order string and then traverse the tree by post-order - naive so slow.
Subtle stack manipulation solves it - stack only for operators:
http://cs.nyu.edu/courses/Fall12/CSCI-GA.1133-002/notes/InfixToPostfixExamples.pdf

BTW, I love Ruby more.

 # SPOJ #4 QNR

 cnt = gets
 cnt = cnt.to_i

 #
 def process(str)
     stk = []
     i = 0
     while i < str.length do
         c = str[i]
         case c
         when 'a'..'z'
             print c
         when '+', '-'
             stk.push(c)
         when '*', '/'
             while !stk.empty? && (stk.last != '+' or stk.last != '-') do
                 if(stk.last == '(')
                     break
                 end
                 print stk.pop()
             end
             stk.push(c)
         when '^'
             while !stk.empty? && stk.last == '^' do
                 if(stk.last == '(')
                     break
                 end
                 print stk.pop()
             end
             stk.push(c)
         when '('
             stk.push(c)
         when ')'
             while !stk.empty? && stk.last != '(' do
                 if(stk.last == '(')
                     break
                 end
                 print stk.pop()
             end
             if(stk.last == '(')
                 stk.pop()
             end
         end
         i += 1
     end
     while !stk.empty? do
         print stk.pop()
     end
     puts
 end # def process

 i = 0
 while i < cnt do
     str = gets.to_s.downcase
     process(str)
     i += 1
 end