Ternary Expression Parser

Given a string representing arbitrarily nested ternary expressions, calculate the result of the expression. You can always assume that the given expression is valid and only consists of digits 0-9, ?, :, T and F (T and F represent True and False respectively).

Note:

1. The length of the given string is ≤ 10000.
2. Each number will contain only one digit.
3. The conditional expressions group right-to-left (as usual in most languages).
4. The condition will always be either T or F. That is, the condition will never be a digit.
5. The result of the expression will always evaluate to either a digit 0-9, T or F.

Example 1:

Input: "T?2:3"

Output: "2"

Explanation: If true, then result is 2; otherwise result is 3.

Example 2:

Input: "F?1:T?4:5"

Output: "4"

Explanation: The conditional expressions group right-to-left. Using parenthesis, it is read/evaluated as:

             "(F ? 1 : (T ? 4 : 5))"                   "(F ? 1 : (T ? 4 : 5))"
          -> "(F ? 1 : 4)"                 or       -> "(T ? 4 : 5)"
          -> "4"                                    -> "4"

Example 3:

Input: "T?T?F:5:3"

Output: "F"

Explanation: The conditional expressions group right-to-left. Using parenthesis, it is read/evaluated as:

             "(T ? (T ? F : 5) : 3)"                   "(T ? (T ? F : 5) : 3)"
          -> "(T ? F : 3)"                 or       -> "(T ? F : 5)"
          -> "F"                                    -> "F"

 public class Solution {
     public String parseTernary(String expression) {
         if (expression == null || expression.length() == )
             return "";
         Stack<Character> stack = new Stack<>();

         for (int i = expression.length() - ; i >= ; i--) {
             char c = expression.charAt(i);
             if (!stack.isEmpty() && stack.peek() == '?') {

                 stack.pop(); // pop '?'
                 char first = stack.pop();
                 stack.pop(); // pop ':'
                 char second = stack.pop();

                 if (c == 'T')
                     stack.push(first);
                 else
                     stack.push(second);
             } else {
                 stack.push(c);
             }
         }
         return String.valueOf(stack.peek());
     }
 }