Given root of a binary tree, The task is to print level order traversal in a way that nodes of all levels are printed in separate lines.
Examples:
Input:
Output:
20
8 22
4 12
10 14Input:
1
/ \
2 3
/ \
4 5Output:
1
2 3
4 5
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 on different lines.
Approach: Below is the idea to solve the problem:
A simple solution is to print using the recursive function discussed in the level order traversal post and print a new line after every call to printGivenLevel().
Find height of tree and run depth first search and maintain current height, print nodes for every height from root and for 1 to height.
Below is the implementation of the above approach:
C++
/* Function to line by line print level order traversal a * tree*/#include <bits/stdc++.h>using namespace std;/* A binary tree node has data,pointer to left childand a pointer to right child */class node {public: int data; node *left, *right;};/* Function prototypes */void printCurrentLevel(node* root, int level);int height(node* node);node* newNode(int data);/* 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); }}void printLevelOrder(struct node* root){ int h = height(root); int i; for (i = 1; i <= h; i++) { printGivenLevel(root, i); printf("\n"); }}/* Compute the "height" of a tree -- the number of nodes along the longest path from the root node down to the farthest leaf node.*/int height(node* node){ if (node == NULL) return 0; else { /* compute the height of each subtree */ int lheight = height(node->left); int rheight = height(node->right); /* use the larger one */ if (lheight > rheight) { return (lheight + 1); } else { return (rheight + 1); } }}/* Helper function that allocatesa new node with the given data andNULL left and right pointers. */node* newNode(int data){ node* Node = new node(); Node->data = data; Node->left = NULL; Node->right = NULL; return (Node);}/* Driver code*/int main(){ node* root = newNode(1); root->left = newNode(2); root->right = newNode(3); root->left->left = newNode(4); root->left->right = newNode(5); cout << "Level Order traversal of binary tree is \n"; printLevelOrder(root); return 0;} |
Java
/* Function to line by line print level order traversal a * tree*//* Class containing left and right child of currentnode and key value*//* Class containing left and right child of currentnode and key value*/class Node { int data; Node left, right; public Node(int item) { data = item; left = right = null; }}class BinaryTree { // Root of the Binary Tree Node root; public BinaryTree() { root = null; } /* Compute the "height" of a tree -- the number of nodes along the longest path from the root node down to the farthest leaf node.*/ int height(Node root) { if (root == null) return 0; else { /* compute height of each subtree */ int lheight = height(root.left); int rheight = height(root.right); /* use the larger one */ if (lheight > rheight) return (lheight + 1); else return (rheight + 1); } } /* Print nodes at a given level */ void printGivenLevel(Node root, int level) { if (root == null) return; if (level == 1) { System.out.print(root.data + " "); } else if (level > 1) { printGivenLevel(root.left, level - 1); printGivenLevel(root.right, level - 1); } } /* function to print level order traversal of tree*/ void printLevelOrder() { int h = height(root); int i; for (i = 1; i <= h; i++) { printGivenLevel(root, i); System.out.print(System.lineSeparator()); } } /* Driver program to test above functions */ public static void main(String args[]) { BinaryTree tree = new BinaryTree(); tree.root = new Node(1); tree.root.left = new Node(2); tree.root.right = new Node(3); tree.root.left.left = new Node(4); tree.root.left.right = new Node(5); System.out.println( "Level order traversal of binary tree is "); tree.printLevelOrder(); }} |
Python3
# Python3 program for above approachclass Node: # A utility function to create a new node def __init__(self, key): self.data = key self.left = None self.right = None# Function to print level order traversal of treedef printlevelorder(root): h = height(root) for i in range(1, h + 1): printGivenLevel(root, i) print()def printGivenLevel(root, level): if root is None: return root if level == 1: print(root.data, end=' ') elif level > 1: printGivenLevel(root.left, level - 1) printGivenLevel(root.right, level - 1)""" Compute the height of a tree--the number of nodes along the longest path from the root node down to the farthest leaf node"""def height(node): if node is None: return 0 else: # Compute the height of each subtree lheight = height(node.left) rheight = height(node.right) # Use the larger one if lheight > rheight: return lheight+1 else: return rheight+1# Driver program to test above functionroot = Node(1)root.left = Node(2)root.right = Node(3)root.left.left = Node(4)root.left.right = Node(5)print("Level order traversal of binary tree is -")printlevelorder(root)# Contributed by Praveen kumar & Vivek Radhakrishna |
C#
/* Print nodes at a given level */using System;/* Class containing left and right child of current node and key value*/public class Node { public int data; public Node left, right; public Node(int item) { data = item; left = right = null; }}class GFG { // Root of the Binary Tree public Node root; public void BinaryTree() { root = null; } /* function to print level order traversal of tree*/ public virtual void printLevelOrder() { int h = height(root); int i; for (i = 1; i <= h; i++) { printGivenLevel(root, i); Console.WriteLine(); } } /* Compute the "height" of a tree -- the number of nodes along the longest path from the root node down to the farthest leaf node.*/ public virtual int height(Node root) { if (root == null) { return 0; } else { /* compute height of each subtree */ int lheight = height(root.left); int rheight = height(root.right); /* use the larger one */ if (lheight > rheight) { return (lheight + 1); } else { return (rheight + 1); } } } /* Print nodes at the current level */ static void printGivenLevel(Node root, int level) { if (root == null) return; if (level == 1) { Console.Write(root.data + " "); } else if (level > 1) { printGivenLevel(root.left, level - 1); printGivenLevel(root.right, level - 1); } } // Driver Code public static void Main(string[] args) { GFG tree = new GFG(); tree.root = new Node(1); tree.root.left = new Node(2); tree.root.right = new Node(3); tree.root.left.left = new Node(4); tree.root.left.right = new Node(5); Console.WriteLine("Level order traversal " + "of binary tree is "); tree.printLevelOrder(); }} |
Javascript
/* Print nodes at a given level */function printGivenLevel(root, level){ if (root == null) return; if (level == 1) document.write(root.data); else if (level > 1) { printGivenLevel(root.left, level-1); printGivenLevel(root.right, level-1); }} |
Output
Level Order traversal of binary tree is 1 2 3 4 5
Time complexity: O(N2)
Auxiliary Space: O(N)
Print level order traversal line by line using iterative level order traversal
The idea is to keep a queue that stores nodes of the current level. Starting from root, calculate the size of queue sz and for each one of sz nodes enqueue its children to queue and print the node. After printing sz nodes of every iteration print a line break.
Follow the below steps to Implement the idea:
- Initialize a queue q. Push root in q.
- while q is not empty
- Create a variable nodeCount = q.size().
- while (nodeCount > 0)
- Create temporary node node *node = q.front() and print node->data.
- Pop front element from q.
- If node->left != NULL push node->left in q.
- If node->right != NULL push node->right in q.
- Print endline.
Below is the implementation of the above approach:
C++
/* Iterative program to print levels line by line */#include <iostream>#include <queue>using namespace std;// A Binary Tree Nodestruct node { struct node* left; int data; struct node* right;};// Iterative method to do level order traversal// line by linevoid printLevelOrder(node* root){ // Base Case if (root == NULL) return; // Create an empty queue for level order traversal queue<node*> q; // Enqueue Root and initialize height q.push(root); while (q.empty() == false) { // nodeCount (queue size) indicates number // of nodes at current level. 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 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); 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 traversal 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); 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 level. 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)printLevelOrder(root)# This code is contributed by Praveen kumar |
C#
/* An Iterative C# program to printlevels 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 traversal 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); printLevelOrder(root); }}// This code is contributed 29AjayKumar |
Javascript
<script> /* An Iterative Javascript program to print levels line by line */ class Node { constructor(data) { this.left = null; this.right = null; this.data = data; } } // Iterative method to do level order traversal line by line function printLevelOrder(root) { // Base Case if(root == null) return; // Create an empty queue for level order traversal let q = []; // Enqueue Root and initialize height q.push(root); while(true) { // nodeCount (queue size) indicates number of nodes // at current level. let nodeCount = q.length; if(nodeCount == 0) break; // Dequeue all nodes of current level and Enqueue all // nodes of next level while(nodeCount > 0) { let node = q[0]; document.write(node.data + " "); q.shift(); if(node.left != null) q.push(node.left); if(node.right != null) q.push(node.right); nodeCount--; } document.write("</br>"); } } // Let us create binary tree shown in above diagram /* 1 / \ 2 3 / \ \ 4 5 6 */ let 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); printLevelOrder(root); </script> |
Output
1 2 3 4 5
Time complexity: O(N) where n is no of nodes of binary tree
Auxiliary Space: O(N) for queue
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)
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