Breadth First Traversal With Binary Search Tree C++

Question

Maybe fast/simple Question. I have a a Binary Tree Implemented already, Then I was hoping to convert binary search tree into an array or at least print it out as if in an array. Where I am having trouble with is how to get the NULL/flags in there '\0'.

for example lets say I have a tree like:

                10
               /  \
              6   12
             / \   \
            1  8   15
             \
              4   

And I want it to print how its supposed to print. Like:

        [10,6,12,1,8,\0,15,\0,4,\0,\0,\0,\0,\0,\0]
        ^Something Like this^ I don't know if I counted the NULL correctly.

Or Another Option on how i want to go about showing Visually my Tree is how to get the spacing correctly outputted like with the '/' and '\' pointing to the keys from the parents:

                10
               /  \
              6   12
             / \   \
            1  8   15
             \
              4   

Here is something that I tried elaborating on code wise but im stuck:

void BreadthFirstTravseral(struct node* root)
{
    queue<node*> q;

    if (!root) {
        return;
    }
    for (q.push(root); !q.empty(); q.pop()) {
        const node * const temp_node = q.front();
        cout<<temp_node->data << " ";

        if (temp_node->left) {
            q.push(temp_node->left);
        }
        if (temp_node->right) {
            q.push(temp_node->right);
        }
    }
}

Any Kind of Help or Link and or advice and or example code would be very much appreciated.

Solution 1

It will be very hard to get the spacing correctly as a key may have multiple digits and this should affect the spacing for all levels above the given node.

As for how to add NULL - simply add else clauses for your ifs where you print a NULL:

if (root) {
  q.push(root);
  cout << root->data << " ";  
} else {
  cout << "NULL ";
}
while (!q.empty()) {
    const node * const temp_node = q.front(); 
    q.pop();

    if (temp_node->left) {
      q.push(temp_node->left);
      cout << temp_node->left->data << " ";
    } else {
      cout << "NULL ";
    }


    if (temp_node->right) {
      q.push(temp_node->right);
      cout << temp_node->right->data << " ";
    } else {
      cout << "NULL ";
    }
}

Solution 2

void TreeBreadthFirst(Node*  treeRoot) 
{  
 Queue *queue  = new Queue();

      if (treeRoot == NULL) return;
       queue->insert(treeRoot); 
       while (!queue->IsEmpty())
          {          
          Node * traverse = queue->dequeue();
         cout<< traverse->data << “ “ ;
          if (traverse->left != NULL) 
            queue->insert( traverse->left); 
          if (traverse->right != NULL) 
            queue->insert(traverse->right); 
           } 
      delete queue;
       } 

Solution 3

I've made a program in c. This code will display somewhat like a tree.

struct node{
    int val;
    struct node *l,*r;
};

typedef struct node node;


int findDepth(node *t){
    if(!t) return 0;
    int l,r;
    l=findDepth(t->l);
    r=findDepth(t->r);

    return l>r?l+1:r+1;
}

void disp(node *t){
    if(!t)
        return;
    int l,r,i=0;
    node *a[100],*p;
    int front=0,rear=-1,d[100],dep,cur,h;
    a[++rear]=t;
    d[rear]=0;
    cur=-1;
    h=findDepth(t);
    printf("\nDepth : %d \n",h-1);

    while(rear>=front){
        dep = d[front];
        p=a[front++];
        if(dep>cur){
            cur=dep;
            printf("\n");
            for(i=0;i<h-cur;i++) printf("\t");
        }
        if(p){
            printf("%d\t\t",p->val);
            a[++rear]=p->l;
            d[rear]=dep+1;
            a[++rear]=p->r;
            d[rear]=dep+1;
        }
        else printf ("-\t\t");

    }
}