Problem description:

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:

The length of the given string is ≤ 10000.
Each number will contain only one digit.
The conditional expressions group right-to-left (as usual in most languages).
The condition will always be either T or F. That is, the condition will never be a digit.
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:

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   "(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:

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

Solution:

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class Solution {
public:
    string parseTernary(string expression) {
        stack<char> stk;
        for(int i= expression.length()-1; i>= 0; i--){
            char c= expression[i]; //current condition
            if(!stk.empty() && stk.top() == '?'){
                stk.pop(); //pop out the '?'
                char first= stk.top();
                stk.pop(); //pop out first digit
                stk.pop(); //pop out ':'
                char second= stk.top();
                stk.top()= (c== 'T')? first: second;
            }
            else
                stk.push(c);
        }
        
        return string(1, stk.top());
    }
};

reference:
https://goo.gl/jKfet6