Binary Tree Insert Algorithm

Question

I recently finished implementing a Binary search tree for a project I was working on. It went well and I learned a lot. However, now I need to implement a regular Binary Tree... which for some reason has me stumped.

I'm looking for a way to do my InsertNode function..

normally in a BST you just check if data < root then insert left and vice versa. However, In a normal Binary tree, it is just filled from left to right, one level at a time..

could anyone help me implement a function that just adds a new Node to the Binary tree from left to right in no specific order?

Here's my Insert for a BST:

void Insert(Node *& root, int data)
{
  if(root == nullptr)
  {
    Node * NN = new Node;
    root = NN;
  }
  else
  {
    if(data < root->data)
    { 
      Insert(root->left, data);
    }
    else
    {
      Insert(root->right, data);
    }
  }
}

Answer 1

I am aware of the fact that this is a question posted some time ago, but I still wanted to share my thoughts on it.

What I would do (since this indeed is not very well documented) is use a Breadth-First-Search (using a queue) and insert the child into the first null I encounter. This will ensure that your tree will fill up the levels first before it goes to another level. With the right number of nodes, it will always be complete.

I haven't worked that much with c++, so to be sure it was correct I did it in Java, but you get the idea:

public void insert(Node node) {
    if(root == null) {
        root = node;
        return;
    }

    /* insert using Breadth-first-search (queue to the rescue!) */
    Queue<Node> queue = new LinkedList<Node>();
    queue.offer(root);

    while(true) {
        Node n = queue.remove();
        if(!n.visited) System.out.println(n.data);
        n.visited = true;

        if(n.left == null) {
            n.left = node;
            break;
        } else {
            queue.offer(n.left);
        }           

        if(n.right == null) {
            n.right = node;
            break;
        } else {
            queue.offer(n.right);
        }
    }
}