Connecting all the nodes at the same level of a [binary tree](https://www.educative.io/shoteditor/6245974420750336/preview) involves linking the `right_neighbor` pointer of each node (in a tree) to its adjacent node on the same level.

# Implementation

**Level-Order traversal** refers to visiting all the nodes at the same level before iterating to the next level. In the following code, a level order traversal is done on the tree, and the `right_neighbor` is set to the immediate neighbor on the same level.

#include <iostream>
using namespace std;
#include <queue>
// A tree node
struct Tree
{
    char data;
    Tree *left;
    Tree *right;
 Tree *right_neighbor;
 Tree(char data)
   {
       this -> data = data;
       left = NULL;
       right = NULL;
       right_neighbor = NULL;
   }
};
// Function to print the right_neighbor of the nodes
void PreOrderTraversal(Tree *node)
{
    if(node == NULL){
      return;
 }
 else{
   if(node -> right_neighbor != NULL){
     cout << node -> data << "-> " << node -> right_neighbor -> data << endl;
   }
   else{
     cout << node -> data << "-> NULL" << endl;
   }
   PreOrderTraversal(node -> left);
   PreOrderTraversal(node -> right);
 }
}
int main() {
  // The tree
  Tree *root = new Tree('a');
  root -> left = new Tree('b');
  root -> right = new Tree('c');
  root -> left -> left = new Tree('d');
  root -> left -> right = new Tree('e');
  queue <Tree*> Q;
  // Enqueue the root node
  Q.push(root);
  // Enqueue NULL to serve a partition between levels
  Q.push(NULL);
  while (!Q.empty()){
    // Dequeue the element from the front of the queue
    Tree *node = Q.front();
    Q.pop();
    if(node == NULL){
      // This shows that a level has finished, so enqueue
      // NULL to denote that.
      if(!Q.empty()){
      Q.push(NULL);
      }
    }
    else{
      // Peek at the element at the front of the queue and
      // set the right pointer to the adjacent element in the queue
      node -> right_neighbor = Q.front();
      // Enqueue its left and right children if they exist
      if(node -> left != NULL){
        Q.push(node -> left);
      }
      if(node -> right != NULL){
        Q.push(node -> right);
      }
    }
  }
  PreOrderTraversal(root);
  return 0;
}

#  A tree node
class Tree: 
	def __init__(self, data): 
		self.data = data 
		self.left = None
		self.right = None
		self.right_neighbor = None
#  Function to print the right_neighbor of the nodes
def PreOrderTraversal(node):
	if node:
		if node.right_neighbor:
			print(node.data, "->", node.right_neighbor.data)
		else:
			print(node.data, "-> None")
		PreOrderTraversal(node.left)
		PreOrderTraversal(node.right)
#  The tree
root = Tree('a') 
root.left = Tree('b') 
root.right = Tree('c') 
root.left.left = Tree('d') 
root.left.right = Tree('e') 
# The queue
Q = [] 
#  Enqueue the root node
Q.append(root) 
#  Enqueue None to serve as a partition between levels
Q.append(None) 
while Q: 
	# Dequeue the element on the front of the queue
	node = Q.pop(0)
	if node == None:
		# This shows that a level has finished, so enqueue
    	# None to denote that.
		if Q:
			Q.append(None) 
	else: 
		# Peek at the element at the front of the queue and set
		# the right_neighbor of the node to it
		node.right_neighbor = Q[0] 
		# Enqueue the left and right children if they exist
		if node.left: 
			Q.append(node.left) 
		if node.right: 
			Q.append(node.right) 
PreOrderTraversal(root)

Python3

How to connect nodes at the same level in a binary tree

Connecting all the nodes at the same level of a binary tree involves linking the right_neighbor pointer of each node (in a tree) to its adjacent node on the same level. 