LintCode "Expression Tree Build"

Lesson learnt: for any calculator problems, keep 2 stacks: 1 for operators and 1 for operands.

class Solution
{
    stack<ExpressionTreeNode*> op;
    stack<ExpressionTreeNode*> data;

    void eval()
    {
    ExpressionTreeNode *opnode = op.top(); op.pop();
    ExpressionTreeNode *data1 = data.top(); data.pop();
    ExpressionTreeNode *data2 = data.top(); data.pop();
    opnode->left = data2;
    opnode->right = data1;
    data.push(opnode);
    }
public:
    ExpressionTreeNode* build(vector<string> &expression)
    {
    for(auto &tmp : expression)
    {
        ExpressionTreeNode *node = new ExpressionTreeNode(tmp);

        switch(tmp[])
        {
        case '(':
            op.push(node);
            break;
        case '+':
        case '-':
            while(!op.empty()&&op.top()->symbol[]!='(')
            {
            eval();
            }
            op.push(node);
            break;
        case '*':
        case '/':
            while(!op.empty()&&(op.top()->symbol[]=='*'||op.top()->symbol[]=='/'))
            {
            eval();
            }
            op.push(node);
            break;
        case ')':
            while(op.top()->symbol[]!='(')
            {
            eval();
            }
            op.pop();
            break;
        default:
        data.push(node);
        break;
        }
         }
         while(!op.empty())
     {
             eval();
         }

         if(data.empty()) return nullptr;
         return data.top();
   }
};