Failing for specific case - c++

Question

This is for finding the top view of a binary tree. My logic has been line by line traversal of the tree. I have used two maps here, m2 for storing node and horizontal distance, the other(m1) with horizontal distance and node data. It seems to fail for a specific case(I cannot copy the test case here because the test case is too large and is not shown completely) in GFG : Link to question
Here is the partial test case I can get :
1
338
5 4 L 5 4 R 4 8 L 4 -1 N 4 2 L 4 2 R 8 8 L 8 -1 N 2 4 L 2 7 R 2 5 L 2 -1 N 8 9 L 8 3 R 4 9 L 4 -1 N 7 5 L 7 3 R 5 8 L 5 -1 N 9 1 L 9 8 R 3 1 L 3 -1 N 9 10 L 9 10 R 5 5 L 5 -1 N 3 7 L 3 6 R 8 9 L 8 -1 N 1 3 L 1 10 R 8 6 L 8 -1 N 1 7 L 1 10 R 10 4 L 10 -1 N 10 7 L 10 10 R 5 8 L 5 -1 N 7 1 L 7 9 R 6 10 L 6 -1 N 9 3 L 9 10 R 3 10 L 3 -1 N 10 2 L 10 4 R 6 2 L 6 -1 N 7 7 L 7 7 R 10 3 L 10 -1 N 4 5 L 4 3 R 7 1 L 7 -1 N 10 3 L 10 2 R 8 10 L 8 -1 N 1 2 L 1 1 R 9 10 L 9 -1 N 10 1 L 10 4 R 3 8 L 3 -1 N 10 8 L 10 7 R 10 9 L 10 -1 N 2 4 L 2 7 R 4 2 L 4 -1 N 2 6 L 2 1 R 7 10 L 7 -1 N 7 6 L 7 10 R 3 3 L 3 -1 N 5 8 L 5 5 R 3 9 L 3 -1 N 1 9 L 1 4 R 3 6 L 3 -1 N 2 3 L 2 3 R 10 7 L 10 -1 N 2 4 L 2 7 R 1 7 L 1 -1 N 10 6 L 10 6 R 1 8 L 1 -1 N 4 6 L 4 8 R 8 7 L 8 -1 N 8 9 L 8 1 R 7 4 L 7 -1 N 9 10 L 9 2 R 4 6 L 4 -1 N 7 1 L 7 1 R 2 9 L 2 -1 N 6 4 L 6 2 R 1 6 L 1 -1 N 10 9 L 10 4 R 6 9 L 6 -1 N 10 2 L 10 6 R 3 1 L 3 -1 N 8 2 L 8 8 R 5 1 L 5 -1 N 9 3 L 9 6 R 9 7 L 9 -1 N 4 3 L 4 2 R 6 2 L 6 -1 N 3 10 L 3 .................
Expected Output :
5 3 10 9 10 3 1 9 8 8 4 5 4 2 6 8 3
My Output:
5 3 10 9 10 3 1 9 8 8 4 5 4 2
My Code :
// { Driver Code Starts
#include <bits/stdc++.h>
using namespace std;
struct Node
{
int data;
struct Node *left;
struct Node *right;
Node(int x){
data = x;
left = NULL;
right = NULL;
}
};
void topView(struct Node *root);
int main()
{
int t;cin>>t;
while(t--)
{
int n,i,k;
cin>>n;
i=n;
Node* root=NULL;
queue <Node*> q;
while(i>0)
{
int a,b;
char c;
cin>> a >> b >> c;
if(!root){
root = new Node(a);
q.push(root);
}
Node* pick = q.front();
q.pop();
if(c == 'L'){
pick->left = new Node(b);
q.push( pick->left );
}
cin>> a >> b >> c;
if(c == 'R'){
pick->right = new Node(b);
q.push( pick->right );
}
i-=2;
}
// inorder(root);
// cout<<endl;
topView(root);
cout << endl;
}
return 0;
}
// } Driver Code Ends
//Structure of binary tree
/*struct Node
struct Node
{
int data;
struct Node* left;
struct Node* right;
Node(int x){
data = x;
left = right = NULL;
}
};*/
// function should print the topView of the binary tree
void topView(struct Node *root)
{
if(root==NULL)
return;
queue<Node *> q;
map<Node *,int> m2;
map<int,int> m1;
m1[0]=root->data;
m2[root]=0;
q.push(root);
while(!q.empty())
{
if(q.front()->left)
{
q.push(root->left);
if(m1.find(m2[root]-1)==m1.end())
{
m2[root->left]=m2[root]-1;
m1[m2[root]-1]=root->left->data;
}
}
if(q.front()->right)
{
q.push(root->right);
if(m1.find(m2[root]+1)==m1.end())
{
m2[root->right]=m2[root]+1;
m1[m2[root]+1]=root->right->data;
}
}
q.pop();
root=q.front();
}
for(auto i:m1)
cout<<i.second<<" ";
}

Answer 1

The problem is that you add new members to m2 when the root is seen from the top of the tree:
if(m1.find(m2[root]+1)==m1.end())
{
m2[root->right]=m2[root]+1;
This will not work for a test case like this
1
10
1 2 L 1 5 R 2 -1 N 2 3 R 5 -1 N 5 -1 N 3 -1 N 3 4 R 4 -1 N 4 6 R
which has the following tree:
1
/ \
2 5
\
3
\
4
\
6
The expected output would be
2 1 5 6
but your code will result in
2 1 5
Here, when the root is node 2, node 3 will be added to m2. But when the root node 3, it will not add node 4 to m2 as node 3 cannot be seen from the top and thus not initialize it correctly to position 1.
Afterwards when root is set to node 4 it will look for it in m2. But as that key doesn't exist yet, it will create a new value with the standard constructor of int (0). Because of that your code suggests that node 6 is at position 1 which is overshadowed by node 5 already.
The fix is simple though. Just move the assignments of m2 out of the if-clauses:
m2[root->left]=m2[root]-1;
if(m1.find(m2[root]-1)==m1.end())
{
m1[m2[root]-1]=root->left->data;
}
Same for the right case.