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 solution in below articles.
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 the top of the queue. Continue this process till the queues become empty.
C++
/* CPP program to print levels line by line */#include <bits/stdc++.h> 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 tarversal 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 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. |
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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<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 // occurence 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 |
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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 // occurence 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 |
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Output :
1 2 3 4 5 6
Time Complexity : O(n)
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