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 approach class newNode: def __init__( self , data): self .val = data self .left = None self .right = None # Iterative method to do level order traversal # line by line def 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 Code 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); # 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)
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