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 <bits/stdc++.h>using namespace std;// A Binary Tree Nodestruct node{ struct node *left; int data; struct node *right;};// Function to do level order// traversal line by linevoid levelOrder(node *root){ if (root == NULL) return; // Create an empty queue for // level order traversal queue<node *> 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 nodenode* newNode(int data){ node *temp = new node; temp->data = data; temp->left = NULL; temp->right = NULL; return temp;}// Driver program to test above// functionsint 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 lineimport 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<Node> 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 linefrom collections import deque as queue# A Binary Tree Nodeclass Node: def __init__(self, key): self.data = key self.left = None self.right = None# Function to do level order# traversal line by linedef 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 codeif __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 lineusing 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 codestatic 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
<script>// Javascript program to do level order// traversal line by lineclass Node{ constructor(data) { this.data = data; this.left = null; this.right = null; }}// Prints level order traversal line// by line using two queues.function levelOrder(root){ if (root == null) return; let q= []; // Pushing root node into the queue. q.push(root); // Pushing delimiter into the queue. q.push(null); // Executing loop till queue becomes // empty while (q.length!=0) { let curr = q.shift(); // condition to check the // occurrence of next level if (curr == null) { if (q.length!=0) { q.push(null); document.write("<br>"); } } else { // Pushing left child current node if (curr.left != null) q.push(curr.left); // Pushing right child current node if (curr.right != null) q.push(curr.right); document.write(curr.data + " "); } }}// Driver functionlet 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 patel2127</script> |
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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