Print level order traversal line by line | Set 1
Given a binary tree, print level order traversal in a way that nodes of all levels are printed in separate lines.
For example consider the following tree
Example 1:Output for above tree should be 20 8 22 4 12 10 14 Example 2: 1 / \ 2 3 / \ \ 4 5 6 / \ / 7 8 9 Output for above tree should be 1 2 3 4 5 6 7 8 9<
Note that this is different from simple level order traversal where we need to print all nodes together. Here we need to print nodes of different levels in different lines.
A simple solution is to print use the recursive function discussed in the level order traversal post and print a new line after every call to printGivenLevel().
C++
/* Function to line by line print level order traversal a tree*/void printLevelOrder(struct node* root){ int h = height(root); int i; for (i=1; i<=h; i++) { printGivenLevel(root, i); printf("\n"); }} /* Print nodes at a given level */void printGivenLevel(struct node* root, int level){ if (root == NULL) return; if (level == 1) printf("%d ", root->data); else if (level > 1) { printGivenLevel(root->left, level-1); printGivenLevel(root->right, level-1); }} |
Java
/* Function to line by line print level order traversal a tree*/static void printLevelOrder(Node root){ int h = height(root); int i; for (i=1; i<=h; i++) { printGivenLevel(root, i); System.out.println(); }}/* Print nodes at a given level */void printGivenLevel(Node root, int level){ if (root == null) return; if (level == 1) System.out.println(root.data); else if (level > 1) { printGivenLevel(root.left, level-1); printGivenLevel(root.right, level-1); }} |
Python3
# Python3 program for above approach def printlevelorder(root): h = height(root) for i in range(1, h + 1): givenspirallevel(root, i) def printGivenLevel(root, level): if root is None: return root if level == 1: print(root.val, end = ' ') elif level > 1: printGivenLevel(root.left, level - 1) printGivenLevel(root.right, level - 1) # This code is contributed by Praveen kumar |
C#
/* Print nodes at a given level */static void printGivenLevel(Node root, int level){ if (root == null) return; if (level == 1) Console.WriteLine(root.data); else if (level > 1) { printGivenLevel(root.left, level-1); printGivenLevel(root.right, level-1); }} |
The time complexity of the above solution is O(n2)
How to modify the iterative level order traversal (Method 2 of this) to levels line by line?
The idea is similar to this post. We count the nodes at current level. And for every node, we enqueue its children to queue.
C++
/* Iterative program to print levels line by line */#include <iostream> #include <queue> using namespace std; // A Binary Tree Node struct node { struct node *left; int data; struct node *right; }; // Iterative method to do level order traversal// line by line void printLevelOrder(node *root) { // Base Case if (root == NULL) return; // Create an empty queue for level order tarversal queue<node *> q; // Enqueue Root and initialize height q.push(root); while (q.empty() == false) { // nodeCount (queue size) indicates number // of nodes at current lelvel. int nodeCount = q.size(); // Dequeue all nodes of current level and // Enqueue all nodes of next level while (nodeCount > 0) { node *node = q.front(); cout << node->data << " "; q.pop(); if (node->left != NULL) q.push(node->left); if (node->right != NULL) q.push(node->right); nodeCount--; } cout << endl; } } // 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); printLevelOrder(root); return 0; } |
Java
/* An Iterative Java program to print levels line by line */ import java.util.LinkedList;import java.util.Queue; public class LevelOrder { // A Binary Tree Node static class Node { int data; Node left; Node right; // constructor Node(int data){ this.data = data; left = null; right =null; } } // Iterative method to do level order traversal line by line static void printLevelOrder(Node root) { // Base Case if(root == null) return; // Create an empty queue for level order tarversal Queue<Node> q =new LinkedList<Node>(); // Enqueue Root and initialize height q.add(root); while(true) { // nodeCount (queue size) indicates number of nodes // at current level. int nodeCount = q.size(); if(nodeCount == 0) break; // Dequeue all nodes of current level and Enqueue all // nodes of next level while(nodeCount > 0) { Node node = q.peek(); System.out.print(node.data + " "); q.remove(); if(node.left != null) q.add(node.left); if(node.right != null) q.add(node.right); nodeCount--; } System.out.println(); } } // Driver program to test above functions public static void main(String[] args) { // Let us create binary tree shown in above diagram /* 1 / \ 2 3 / \ \ 4 5 6 */ 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); printLevelOrder(root); } }//This code is contributed by Sumit Ghosh |
Python3
# Python3 program for above approachclass newNode: def __init__(self, data): self.val = data self.left = None self.right = None # Iterative method to do level order traversal# line by linedef printLevelOrder(root): # Base case if root is None: return # Create an empty queue for level order traversal q = [] # Enqueue root and initialize height q.append(root) while q: # nodeCount (queue size) indicates number # of nodes at current lelvel. count = len(q) # Dequeue all nodes of current level and # Enqueue all nodes of next level while count > 0: temp = q.pop(0) print(temp.val, end = ' ') if temp.left: q.append(temp.left) if temp.right: q.append(temp.right) count -= 1 print(' ') # Driver Coderoot = 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); printLevelOrder(root); # This code is contributed by Praveen kumar |
C#
/* An Iterative C# program to print levels line by line */using System;using System.Collections.Generic; public class LevelOrder { // A Binary Tree Node class Node { public int data; public Node left; public Node right; // constructor public Node(int data) { this.data = data; left = null; right =null; } } // Iterative method to do level order // traversal line by line static void printLevelOrder(Node root) { // Base Case if(root == null) return; // Create an empty queue for level // order tarversal Queue<Node> q =new Queue<Node>(); // Enqueue Root and initialize height q.Enqueue(root); while(true) { // nodeCount (queue size) indicates // number of nodes at current level. int nodeCount = q.Count; if(nodeCount == 0) break; // Dequeue all nodes of current level // and Enqueue all nodes of next level while(nodeCount > 0) { Node node = q.Peek(); Console.Write(node.data + " "); q.Dequeue(); if(node.left != null) q.Enqueue(node.left); if(node.right != null) q.Enqueue(node.right); nodeCount--; } Console.WriteLine(); } } // Driver Code public static void Main(String[] args) { // Let us create binary tree shown // in above diagram /* 1 / \ 2 3 / \ \ 4 5 6 */ 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); printLevelOrder(root); } } // This code is contributed 29AjayKumar |
Output:
1 2 3 4 5 6
Time complexity of this method is O(n) where n is number of nodes in given binary tree.
Level order traversal line by line | Set 2 (Using Two Queues)
Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above
Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.
