Create tree using inorder and preorder traversals

Why is my print function wrong?

#include<iostream>
#include<queue>
using namespace std;

class node{
public:
	int data;
	node *left;
	node *right;
	// constructor
	node(int d){
		data = d;
		left = NULL;
		right = NULL;
	}
};

node *build_tree(int pre[], int i, int in[], int s, int e){
	// base case
	if(s>e){
		return NULL;
	}


	// rec case
	node *root = new node(pre[i]);
	int index = -1;
	for(int j=s; j<=e; j++){
		if(in[j] == pre[i]){
			index = j;
			break;
		}
	}
	i++;
	root->left = build_tree(pre,i,in,s,index-1);
	root->right = build_tree(pre,i,in,index+1,e);

	return root;
}

void print(node *root){
    queue<node *> q;
    q.push(root);
    while(!q.empty()){
        // if front of the queue == -1, pop it
        while(q.front()->data == -1){
            q.pop();
        }
        if(q.empty()){
            break;
        }


        // push left and right children
        if(q.front()->left == NULL){
            q.push(new node(-1));
        }
        else{
            q.push(q.front()->left);
        }
    
        if(q.front()->right == NULL){
            q.push(new node(-1));
        }
        else{
            q.push(q.front()->right);
        }


        // print leftChild
        if(q.front()->left != NULL){
            cout<<q.front()->left->data<<" => ";
        }
        else{
            cout<<"END => ";
        }

        // print root
        cout<<q.front()->data;

        // print rightChild
        if(q.front()->right != NULL){
            cout<<" <= "<<q.front()->right->data;
        }
        else{
            cout<<" <= END";
        }
        cout<<endl;
        q.pop();
    }
}


int main() {

	int n;
	cin>>n;
	int pre[10000];
	for(int i=0; i<n; i++){
		cin>>pre[i];
	}

	int m;
	cin>>m;
	int in[10000];
	for(int i=0; i<m; i++){
		cin>>in[i];
	}

	node *root = build_tree(pre,0,in,0,m-1);

    print(root);



	return 0;
}

hi Anurag
refer this --> https://ide.codingblocks.com/s/619328
i dont understand what u are trying to do by pushing -1 and all

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