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# Level order traversal line by line | Set 3 (Using One Queue)

• Difficulty Level : Easy
• Last Updated : 21 Dec, 2022

Given a Binary Tree, print the nodes level-wise, each level on a new line. ```Output:
1
2 3
4 5```

We have discussed two solutions in the articles below.
Print level order traversal line by line | Set 1
Level order traversal line by line | Set 2 (Using Two Queues)
In this post, a different approach using one queue is discussed. First insert the root and a null element into the queue. This null element acts as a delimiter. Next, pop from the top of the queue and add its left and right nodes to the end of the queue and then print at the top of the queue. Continue this process till the queues become empty.

## C++

 `/* C++ program to print levels ` `line by line */` `#include ` `using` `namespace` `std;`   `// A Binary Tree Node` `struct` `node` `{` `    ``struct` `node *left;` `    ``int` `data;` `    ``struct` `node *right;` `};`   `// Function to do level order` `// traversal line by line` `void` `levelOrder(node *root)` `{` `    ``if` `(root == NULL) ``return``;`   `    ``// Create an empty queue for` `    ``// level order traversal` `    ``queue q;` `    `  `    ``// to store front element of ` `    ``// queue.` `    ``node *curr;`   `    ``// Enqueue Root and NULL node.` `    ``q.push(root);` `    ``q.push(NULL);`   `    ``while` `(q.size() > 1)` `    ``{` `        ``curr = q.front();` `        ``q.pop();` `        `  `        ``// condition to check ` `        ``// occurrence of next ` `        ``// level.` `        ``if` `(curr == NULL)` `        ``{` `           ``q.push(NULL);` `           ``cout << ``"\n"``;` `        ``}` `        `  `        ``else` `{` `            `  `            ``// pushing left child of ` `            ``// current node.` `            ``if``(curr->left)` `            ``q.push(curr->left);` `            `  `            ``// pushing right child of` `            ``// current node.` `            ``if``(curr->right)` `            ``q.push(curr->right);` `            `  `            ``cout << curr->data << ``" "``;` `        ``}` `    ``}` `}`   `// Utility function to create a` `// new tree node` `node* newNode(``int` `data)` `{` `    ``node *temp = ``new` `node;` `    ``temp->data = data;` `    ``temp->left = NULL;` `    ``temp->right = NULL;` `    ``return` `temp;` `}`   `// Driver program to test above` `// functions` `int` `main()` `{` `    `  `    ``// Let us create binary tree` `    ``// shown above` `    ``node *root = newNode(1);` `    ``root->left = newNode(2);` `    ``root->right = newNode(3);` `    ``root->left->left = newNode(4);` `    ``root->left->right = newNode(5);` `    ``root->right->right = newNode(6);`   `    ``levelOrder(root);` `    ``return` `0;` `}`   `// This code is contributed by` `// Nikhil Jindal.`

## Java

 `// Java program to do level order` `// traversal line by line` `import` `java.util.LinkedList;` `import` `java.util.Queue;`   `public` `class` `GFG {` `  ``static` `class` `Node {` `    ``int` `data;` `    ``Node left;` `    ``Node right;`   `    ``Node(``int` `data) {` `      ``this``.data = data;` `      ``left = ``null``;` `      ``right = ``null``;` `    ``}` `  ``}`   `  ``// Prints level order traversal line` `  ``// by line using two queues.` `  ``static` `void` `levelOrder(Node root) {` `    ``if` `(root == ``null``)` `      ``return``;`   `    ``Queue q = ``new` `LinkedList<>();`   `    ``// Pushing root node into the queue.` `    ``q.add(root);`   `    ``// Pushing delimiter into the queue.` `    ``q.add(``null``);`   `    ``// Executing loop till queue becomes` `    ``// empty` `    ``while` `(!q.isEmpty()) {`   `      ``Node curr = q.poll();`   `      ``// condition to check the` `      ``// occurrence of next level` `      ``if` `(curr == ``null``) {` `        ``if` `(!q.isEmpty()) {` `          ``q.add(``null``);` `          ``System.out.println();` `        ``}` `      ``} ``else` `{` `        ``// Pushing left child current node` `        ``if` `(curr.left != ``null``)` `          ``q.add(curr.left);`   `        ``// Pushing right child current node` `        ``if` `(curr.right != ``null``)` `          ``q.add(curr.right);`   `        ``System.out.print(curr.data + ``" "``);` `      ``}` `    ``}` `  ``}`   `  ``// Driver function` `  ``public` `static` `void` `main(String[] args) {`   `    ``Node root = ``new` `Node(``1``);` `    ``root.left = ``new` `Node(``2``);` `    ``root.right = ``new` `Node(``3``);` `    ``root.left.left = ``new` `Node(``4``);` `    ``root.left.right = ``new` `Node(``5``);` `    ``root.right.right = ``new` `Node(``6``);`   `    ``levelOrder(root);` `  ``}` `}`   `// This code is Contributed by Rishabh Jindal`

## Python3

 `# Python3 program to print levels` `# line by line` `from` `collections ``import` `deque as queue`   `# A Binary Tree Node`     `class` `Node:`   `    ``def` `__init__(``self``, key):`   `        ``self``.data ``=` `key` `        ``self``.left ``=` `None` `        ``self``.right ``=` `None`   `# Function to do level order` `# traversal line by line`     `def` `levelOrder(root):`   `    ``if` `(root ``=``=` `None``):` `        ``return`   `    ``# Create an empty queue for` `    ``# level order traversal` `    ``q ``=` `queue()`   `    ``# To store front element of` `    ``# queue.` `    ``#node *curr`   `    ``# Enqueue Root and None node.` `    ``q.append(root)` `    ``q.append(``None``)`   `    ``while` `(``len``(q) > ``1``):` `        ``curr ``=` `q.popleft()` `        ``# q.pop()`   `        ``# Condition to check` `        ``# occurrence of next` `        ``# level.` `        ``if` `(curr ``=``=` `None``):` `            ``q.append(``None``)` `            ``print``()`   `        ``else``:`   `            ``# Pushing left child of` `            ``# current node.` `            ``if` `(curr.left):` `                ``q.append(curr.left)`   `            ``# Pushing right child of` `            ``# current node.` `            ``if` `(curr.right):` `                ``q.append(curr.right)`   `            ``print``(curr.data, end``=``" "``)`     `# Driver code` `if` `__name__ ``=``=` `'__main__'``:`   `    ``# Let us create binary tree` `    ``# shown above` `    ``root ``=` `Node(``1``)` `    ``root.left ``=` `Node(``2``)` `    ``root.right ``=` `Node(``3``)` `    ``root.left.left ``=` `Node(``4``)` `    ``root.left.right ``=` `Node(``5``)` `    ``root.right.right ``=` `Node(``6``)`   `    ``levelOrder(root)`   `# This code is contributed by mohit kumar 29`

## C#

 `// C# program to do level order` `// traversal line by line` `using` `System;` `using` `System.Collections;`   `class` `GFG ` `{` `public` `class` `Node ` `{` `    ``public` `int` `data;` `    ``public` `Node left;` `    ``public` `Node right;`   `    ``public` `Node(``int` `data) ` `    ``{` `        ``this``.data = data;` `        ``left = ``null``;` `        ``right = ``null``;` `    ``}` `}`   `// Prints level order traversal line` `// by line using two queues.` `static` `void` `levelOrder(Node root) ` `{` `    ``if` `(root == ``null``)` `    ``return``;`   `    ``Queue q = ``new` `Queue();`   `    ``// Pushing root node into the queue.` `    ``q.Enqueue(root);`   `    ``// Pushing delimiter into the queue.` `    ``q.Enqueue(``null``);`   `    ``// Executing loop till queue becomes` `    ``// empty` `    ``while` `(q.Count>0)` `    ``{`   `        ``Node curr = (Node)q.Peek();` `        ``q.Dequeue();`   `        ``// condition to check the` `        ``// occurrence of next level` `        ``if` `(curr == ``null``) ` `        ``{` `            ``if` `(q.Count > 0)` `            ``{` `                ``q.Enqueue(``null``);` `                 ``Console.WriteLine();` `            ``}` `        ``} ` `        ``else` `        ``{` `            ``// Pushing left child current node` `            ``if` `(curr.left != ``null``)` `            ``q.Enqueue(curr.left);` `    `  `            ``// Pushing right child current node` `            ``if` `(curr.right != ``null``)` `            ``q.Enqueue(curr.right);` `    `  `            ``Console.Write(curr.data + ``" "``);` `        ``}` `    ``}` `}`   `// Driver code` `static` `public` `void` `Main(String []args) ` `{`   `    ``Node root = ``new` `Node(1);` `    ``root.left = ``new` `Node(2);` `    ``root.right = ``new` `Node(3);` `    ``root.left.left = ``new` `Node(4);` `    ``root.left.right = ``new` `Node(5);` `    ``root.right.right = ``new` `Node(6);`   `    ``levelOrder(root);` `}` `}`   `// This code is Contributed by Arnab Kundu`

## Javascript

 ``

Output

```1
2 3
4 5 6 ```

Time Complexity: O(n)
Auxiliary Space: O(n) for queue, where n is no of nodes of binary tree

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