Left-view of a binary tree

using namespace std;
/* A binary tree node has data, pointer to left child
   and a pointer to right child */
struct Node
{
    int data;
    struct Node* left;
    struct Node* right;

    Node(int x){
        data = x;
        left = right = NULL;
    }
};

// A wrapper over leftViewUtil()
void leftView(struct Node *root);


/* Driver program to test size function*/
int main()
{
  int t;
  struct Node *child;
  scanf("%d\n", &t);
  while (t--)
  {
     map<int, Node*> m;
     int n;
     scanf("%d\n",&n);
     struct Node *root = NULL;
     while (n--)
     {
        Node *parent;
        char lr;
        int n1, n2;
        scanf("%d %d %c", &n1, &n2, &lr);

        if (m.find(n1) == m.end())
        {
           parent = new Node(n1);
           m[n1] = parent;
           if (root == NULL)
             root = parent;
        }
        else
           parent = m[n1];

        child = new Node(n2);
        if (lr == 'L')
          parent->left = child;
        else
          parent->right = child;
        m[n2]  = child;
     }

     leftView(root);
     cout << endl;
  }
  return 0;
}
/*This is a function problem.You only need to complete the function given below*/
/* A binary tree node
struct Node
{
    int data;
    struct Node* left;
    struct Node* right;

    Node(int x){
        data = x;
        left = right = NULL;
    }
};
 */
// A wrapper over leftViewUtil()
void leftView(Node *root)
{
    if(!root)
        return;
    map<int, int> m;
    queue<pair<Node *, int> > q;
    q.push({root, 0});
    while(!q.empty()){
        Node *n = q.front().first;
        int l = q.front().second;
        q.pop();
        m.insert({l, n->data});
        if(n->left)
            q.push({n->left, l+1});
        if(n->right)
            q.push({n->right, l+1});
    }
    for(auto it=m.begin(); it!=m.end(); ++it)
        cout<<(it->second)<<" ";
}