Surrounded Regions

Description：

Given a 2D board containing 'X' and 'O', capture all regions surrounded by 'X'.

A region is captured by flipping all 'O's into 'X's in that surrounded region.

For example,

X X X X
X O O X
X X O X
X O X X

After running your function, the board should be:

X X X X
X X X X
X X X X
X O X X

Code：

struct vertex
{
    int row;
    int col;
    vertex (int x, int y):row(x),col(y)
    {
        
    }
};

  void BFSTraverse (vector< vector<char> >& board, int i, int j)
  {//因为值被修改为'S'就说明该元素已被访问，所以本题中可以不用另外的变量记录访问状态
            board[i][j] = 'S';
            deque<vertex>m;
            m.push_back( vertex(i,j) );
            while ( !m.empty() )
            {
                   vertex v = m.front();
                   m.pop_front();

                   int mRow = v.row-1;
                   if ( mRow >=0 && board[mRow][v.col] == 'O')
                    {//上面的元素
                           board[mRow][v.col] = 'S';
                           m.push_back( vertex(mRow, v.col) );
                   }
                   mRow = v.row+1;
                   if ( mRow < board.size() && board[mRow][v.col] == 'O')
                   {//下面的元素
                           board[mRow][v.col] = 'S';
                           m.push_back( vertex(mRow, v.col) );
                   }
                   int mCol = v.col - 1;
                   if ( mCol >=0 && board[v.row][mCol] == 'O')
                   {//左面的元素
                           board[v.row][mCol] = 'S';
                           m.push_back( vertex(v.row, mCol));
                   }
                    mCol = v.col + 1;
                   if ( mCol < board[0].size()  && board[v.row][mCol] == 'O' )
                   {//右面的元素
                           board[v.row][mCol] = 'S';
                           m.push_back( vertex(v.row, mCol));
                   }
           }
  }
    
    void solve(vector< vector<char> >& board)
    {
        int row = board.size();
        if ( row!=0 )
        {
            int col = board[0].size();
            for (int j = 0; j < col; ++j)
           {
                if ( board[0][j] == 'O')
                    BFSTraverse (board, 0, j);
                if ( board[row-1][j] == 'O')
                    BFSTraverse (board, row-1, j);
           }
           for (int i = 1;  i < row-1; ++i)
           {
               if ( board[i][0] == 'O')
                    BFSTraverse (board, i, 0);
                if ( board[i][col-1] == 'O' )
                    BFSTraverse (board, i, col-1);
           }
           
           for (int  i = 0; i < row; ++i )
           {
               for ( int j = 0; j < col; ++j )
               {
                   if (board[i][j] == 'S')
                        board[i][j] = 'O';
                   else if (board[i][j] == 'O')
                        board[i][j] = 'X';
                   else
                   ;
               }
           }//for
        }//if
    }