Given a binary tree, the task is to print the odd positioned nodes of even levels in the level order traversal of the tree. The root is considered at level 0, and the leftmost node of any level is considered as a node at position 0.
Example:
Input:
1
/ \
2 3
/ \ / \
4 5 6 7
/ \
8 9
/ \
10 11
Output: 5 7 11
Input:
2
/ \
4 15
/ /
45 17
Output: 17
Prerequisite – Even positioned elements at even level
Approach: To print nodes level by level, use level order traversal. The idea is based on Print level order traversal line by line. For that, traverse nodes level by level and switch odd level flag after every level. Similarly, mark 2nd node in every level as odd position and switch it after each time the next node is processed.
Below is the implementation of the above approach:
C++
// C++ implementation of the approach#include <bits/stdc++.h>using namespace std;struct Node { int data; Node *left, *right;};// Iterative method to do level order// traversal line by linevoid printEvenLevelOddNodes(Node* root){ // Base Case if (root == NULL) return; // Create an empty queue for level // order traversal queue<Node*> q; // Enqueue root and initialize level as even q.push(root); bool evenLevel = true; while (1) { // nodeCount (queue size) indicates // number of nodes in the current level int nodeCount = q.size(); if (nodeCount == 0) break; // Mark 1st node as even positioned bool evenNodePosition = true; // Dequeue all the nodes of current level // and Enqueue all the nodes of next level while (nodeCount > 0) { Node* node = q.front(); // Print only even positioned // nodes of even levels if (evenLevel && !evenNodePosition) cout << node->data << " "; q.pop(); if (node->left != NULL) q.push(node->left); if (node->right != NULL) q.push(node->right); nodeCount--; // Switch the even position flag evenNodePosition = !evenNodePosition; } // Switch the even level flag evenLevel = !evenLevel; }}// Utility method to create a nodestruct Node* newNode(int data){ struct Node* node = new Node; node->data = data; node->left = node->right = NULL; return (node);}// Driver codeint main(){ struct Node* root = newNode(1); root->left = newNode(2); root->right = newNode(3); root->left->left = newNode(4); root->left->right = newNode(5); root->right->left = newNode(6); root->right->right = newNode(7); root->left->right->left = newNode(8); root->left->right->right = newNode(9); root->left->right->left->left = newNode(10); root->left->right->right->right = newNode(11); printEvenLevelOddNodes(root); return 0;} |
Java
// Java implementation of the above approachimport java.util.*;class GFG{static class Node { int data; Node left, right;};// Iterative method to do level order// traversal line by linestatic void printEvenLevelOddNodes(Node root){ // Base Case if (root == null) return; // Create an empty queue for level // order traversal Queue<Node> q = new LinkedList<>(); // Enqueue root and initialize level as even q.add(root); boolean evenLevel = true; while (true) { // nodeCount (queue size) indicates // number of nodes in the current level int nodeCount = q.size(); if (nodeCount == 0) break; // Mark 1st node as even positioned boolean evenNodePosition = true; // Dequeue all the nodes of current level // and Enqueue all the nodes of next level while (nodeCount > 0) { Node node = q.peek(); // Print only even positioned // nodes of even levels if (evenLevel && !evenNodePosition) System.out.print(node.data + " "); q.remove(); if (node.left != null) q.add(node.left); if (node.right != null) q.add(node.right); nodeCount--; // Switch the even position flag evenNodePosition = !evenNodePosition; } // Switch the even level flag evenLevel = !evenLevel; }}// Utility method to create a nodestatic Node newNode(int data){ Node node = new Node(); node.data = data; node.left = node.right = null; return (node);}// Driver codepublic static void main(String[] args){ Node root = newNode(1); root.left = newNode(2); root.right = newNode(3); root.left.left = newNode(4); root.left.right = newNode(5); root.right.left = newNode(6); root.right.right = newNode(7); root.left.right.left = newNode(8); root.left.right.right = newNode(9); root.left.right.left.left = newNode(10); root.left.right.right.right = newNode(11); printEvenLevelOddNodes(root);}} // This code is contributed by Princi Singh |
Python3
# Python implementation of the approach # Utility method to create a node class newNode: # Construct to create a new node def __init__(self, key): self.data = key self.left = None self.right = None# Iterative method to do level order # traversal line by line def printEvenLevelOddNodes(root): # Base Case if (root == None): return # Create an empty queue for level # order traversal q =[] # Enqueue root and initialize level as even q.append(root) evenLevel = True while (1): # nodeCount (queue size) indicates # number of nodes in the current level nodeCount = len(q) if (nodeCount == 0): break # Mark 1st node as even positioned evenNodePosition = True # Dequeue all the nodes of current level # and Enqueue all the nodes of next level while (nodeCount > 0): node = q[0] # Pronly even positioned # nodes of even levels if evenLevel and not evenNodePosition: print(node.data, end =" ") q.pop(0) if (node.left != None): q.append(node.left) if (node.right != None): q.append(node.right) nodeCount-= 1 # Switch the even position flag evenNodePosition = not evenNodePosition # Switch the even level flag evenLevel = not evenLevel # Driver code if __name__ == '__main__': root = newNode(1) root.left = newNode(2) root.right = newNode(3) root.left.left = newNode(4) root.left.right = newNode(5) root.right.left = newNode(6) root.right.right = newNode(7) root.left.right.left = newNode(8) root.left.right.right = newNode(9) root.left.right.left.left = newNode(10) root.left.right.right.right = newNode(11) printEvenLevelOddNodes(root) |
C#
// C# implementation of the above approachusing System;using System.Collections.Generic; class GFG{public class Node { public int data; public Node left, right;};// Iterative method to do level order// traversal line by linestatic void printEvenLevelOddNodes(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 level as even q.Enqueue(root); bool evenLevel = true; while (true) { // nodeCount (queue size) indicates // number of nodes in the current level int nodeCount = q.Count; if (nodeCount == 0) break; // Mark 1st node as even positioned bool evenNodePosition = true; // Dequeue all the nodes of current level // and Enqueue all the nodes of next level while (nodeCount > 0) { Node node = q.Peek(); // Print only even positioned // nodes of even levels if (evenLevel && !evenNodePosition) Console.Write(node.data + " "); q.Dequeue(); if (node.left != null) q.Enqueue(node.left); if (node.right != null) q.Enqueue(node.right); nodeCount--; // Switch the even position flag evenNodePosition = !evenNodePosition; } // Switch the even level flag evenLevel = !evenLevel; }}// Utility method to create a nodestatic Node newNode(int data){ Node node = new Node(); node.data = data; node.left = node.right = null; return (node);}// Driver codepublic static void Main(String[] args){ Node root = newNode(1); root.left = newNode(2); root.right = newNode(3); root.left.left = newNode(4); root.left.right = newNode(5); root.right.left = newNode(6); root.right.right = newNode(7); root.left.right.left = newNode(8); root.left.right.right = newNode(9); root.left.right.left.left = newNode(10); root.left.right.right.right = newNode(11); printEvenLevelOddNodes(root);}}// This code is contributed by 29AjayKumar |
Javascript
<script> // JavaScript implementation of the above approach class Node { constructor(data) { this.left = null; this.right = null; this.data = data; } } // Iterative method to do level order // traversal line by line function printEvenLevelOddNodes(root) { // Base Case if (root == null) return; // Create an empty queue for level // order traversal let q = []; // Enqueue root and initialize level as even q.push(root); let evenLevel = true; while (true) { // nodeCount (queue size) indicates // number of nodes in the current level let nodeCount = q.length; if (nodeCount == 0) break; // Mark 1st node as even positioned let evenNodePosition = true; // Dequeue all the nodes of current level // and Enqueue all the nodes of next level while (nodeCount > 0) { let node = q[0]; // Print only even positioned // nodes of even levels if (evenLevel && !evenNodePosition) document.write(node.data + " "); q.shift(); if (node.left != null) q.push(node.left); if (node.right != null) q.push(node.right); nodeCount--; // Switch the even position flag evenNodePosition = !evenNodePosition; } // Switch the even level flag evenLevel = !evenLevel; } } // Utility method to create a node function newNode(data) { let node = new Node(data); return (node); } let root = newNode(1); root.left = newNode(2); root.right = newNode(3); root.left.left = newNode(4); root.left.right = newNode(5); root.right.left = newNode(6); root.right.right = newNode(7); root.left.right.left = newNode(8); root.left.right.right = newNode(9); root.left.right.left.left = newNode(10); root.left.right.right.right = newNode(11); printEvenLevelOddNodes(root);</script> |
Output:
5 7 11
Time Complexity: O(n) where n is the number of nodes in the binary tree.
Auxiliary Space: O(n) where n is the number of nodes in the binary tree.
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