HackerEarth Gandhi Tree March problem solution

In this HackerEarth Gandhi Tree March problem solution, Gandhijee is interested in building a human tree. He defined a human node as follows :

Person_Id = English alphabet {a…z} .

Person_Chain_Definition = Person_Id ( Person_Chain_Definition Person_Chain_Definition )

For example :

a( b(..) c( d( . e(..) ) f( . . ) ) ) refers to a human tree having a as the root human, whose chain links are are b and c, (spaces here are for clarity, input file has no spaces).

Note: After every Person_Id there is necessarily an opening parenthesis ( that starts the definition of the sub-tree under the person. For empty(NULL) human nodes, there is a ‘.’ character.

Now, Gandhijee figures out the tree as follows :

Step 1: He writes root human root id in column 0.

Step 2: Then he writes the left link’s person id in 1 column to the left and the right link’s person id in 1 column to the right.

He continues writing the subtree, in a way that, for an id written in column C, its left link’s id is written in column C-1 and the right link’s id is written in column C+1.

Now, Gandhijee wants to know the ids of people standing in a particular column C in lexicographically sorted order.

You are provided with column number C and the Tree definition string S. Gandhijee is very busy and hence he wants you to help him.

#include<string>
#include<vector>
#include<cstdio>
#include<sstream>
#include<cassert>
#include<iostream>
#include<algorithm>
using namespace std;

int pos;
string S;
vector<char> L[10001], R[10001];

void buildTree(int C)
{
    if(S[pos]!='.')
    {
        if(abs(C)<=10000)
        {
            if(C>=0)  L[C].push_back(S[pos]);
            else  R[-C].push_back(S[pos]);
        }
        pos+=2;
        buildTree(C-1);
        buildTree(C+1);
    }
    pos++;
}

int main()
{
    int i, T, C, Size;
    cin>>T;
    while(T--)
    {
        pos=0;
        cin>>C>>S;
        for(i=0; i<=10000; i++)  L[i].clear(), R[i].clear();

        buildTree(0);

        if(C>=0)
        {
            if(Size=L[C].size())
            {
                sort(L[C].begin(), L[C].end());
                for(i=0; i<Size; i++) cout<<L[C][i];
                cout<<endl;
            }
            else cout<<"Common Gandhijee!n";
        }
        else
        {
            if(Size=R[-C].size())
            {
                sort(R[-C].begin(),R[-C].end());
                for(i=0; i<Size; i++)cout<<R[-C][i];
                cout<<endl;
            }
            else cout<<"Common Gandhijee!n";
        }
    }
    return 0;
}