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