Printing BFS (Binary Tree) in Level Order with Specific Formatting

Question

To begin with, this question is not a dup of this one, but builds on it.

Taking the tree in that question as an example,

    1 
   / \\
  2   3
 /            


        

Answer 1

Here my code prints the tree level by level as well as upside down

int counter=0;// to count the toatl no. of elments in the tree

void tree::print_treeupsidedown_levelbylevel(int *array)
{
    int j=2;  
    int next=j;
    int temp=0;
    while(j<2*counter)
    {
        if(array[j]==0)
        break;

        while(array[j]!=-1)
        {
            j++;
        }

        for(int i=next,k=j-1 ;i=0;i--)
    {
        if(array[i]>0)
        printf("%d ",array[i]);

        if(array[i]==-1)
        printf("\n");
    }
}

void tree::BFS()
{
    queuep;

    node *leaf=root;

    int array[2*counter];
    for(int i=0;i<2*counter;i++)
    array[i]=0;

    int count=0;

    node *newline=new node; //this node helps to print a tree level by level
    newline->val=0;
    newline->left=NULL;
    newline->right=NULL;
    newline->parent=NULL;

    p.push(leaf);
    p.push(newline);

    while(!p.empty())
    {
        leaf=p.front();
        if(leaf==newline)
        {
            printf("\n");
            p.pop();
            if(!p.empty())
            p.push(newline);
            array[count++]=-1;
        }
        else
        {
            cout<val<<" ";
            array[count++]=leaf->val;

            if(leaf->left!=NULL)
            {
                p.push(leaf->left);
            }
            if(leaf->right!=NULL)
            {
                p.push(leaf->right);
            }
            p.pop();
        }
    }
    delete newline;

    print_treeupsidedown_levelbylevel(array);
}

Here in my code the function BFS prints the tree level by level, which also fills the data in an int array for printing the tree upside down. (note there is a bit of swapping is used while printing the tree upside down which helps to achieve our goal). If the swapping is not performed then for a tree like

                    8
                   /  \
                  1    12
                  \     /
                   5   9
                 /   \
                4     7
                     /
                    6

o/p will be

  6
  7 4
  9 5
  12 1
  8

but the o/p has to be

  6
  4 7
  5 9
  1 12
  8

this the reason why swapping part was needed in that array.