Print all nodes except rightmost node of every level of the Binary Tree
Given a binary tree, the task is to print all the nodes except the rightmost in every level of the tree. The root is considered at level 0, and rightmost node of any level is considered as a node at position 0.
Examples:
Input: 1 / \ 2 3 / \ \ 4 5 6 / \ 7 8 / \ 9 10 Output: 2 4 5 7 9 Input: 1 / \ 2 3 \ \ 4 5 Output: 2 4
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 if the node in the queue of level order is the last node then that node will be the rightmost node and don’t print that node.
Below is the implementation of the above approach:
C++
// C++ implementation of the approach #include <bits/stdc++.h> using namespace std; // Structure of the tree node struct Node { int data; Node *left, *right; }; // Utility method to create a node struct Node* newNode( int data) { struct Node* node = new Node; node->data = data; node->left = node->right = NULL; return (node); } // Function to print all the nodes // except the rightmost in every level // of the given binary tree // with level order traversal void excluderightmost(Node* root) { // Base Case if (root == NULL) return ; // Create an empty queue for level // order traversal queue<Node*> q; // Enqueue root q.push(root); while (1) { // 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.front(); // If node is not rightmost print if (nodeCount != 1) cout << node->data << " " ; q.pop(); if (node->left != NULL) q.push(node->left); if (node->right != NULL) q.push(node->right); nodeCount--; } cout << "\n" ; } } // Driver code int 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->right->right = newNode(10); excluderightmost(root); return 0; } |
Java
// Java implementation of the approach import java.util.*; class Sol { // Structure of the tree node static class Node { int data; Node left, right; }; // Utility method to create a node static Node newNode( int data) { Node node = new Node(); node.data = data; node.left = node.right = null ; return (node); } // Function to print all the nodes // except the rightmost in every level // of the given binary tree // with level order traversal static void excluderightmost(Node root) { // Base Case if (root == null ) return ; // Create an empty queue for level // order traversal Queue<Node> q = new LinkedList<Node>(); // Enqueue root 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(); // If node is not rightmost print if (nodeCount != 1 ) 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 code public 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.right.right = newNode( 10 ); excluderightmost(root); } } |
Python
# Python implementation of the approach from collections import deque # Structure of the tree node class Node: def __init__( self ): self .data = 0 self .left = None self .right = None # Utility method to create a node def newNode(data: int ) - > Node: node = Node() node.data = data node.left = None node.right = None return node # Function to print all the nodes # except the rightmost in every level # of the given binary tree # with level order traversal def excluderightmost(root: Node): # Base Case if root is None : return # Create an empty queue for level # order traversal q = deque() # Enqueue root q.append(root) while 1 : # nodeCount (queue size) indicates # number of nodes at current level nodeCount = len (q) if nodeCount = = 0 : break # Dequeue all nodes of current level # and Enqueue all nodes of next level while nodeCount > 0 : node = q[ 0 ] # If Node is not right most print if nodeCount ! = 1 : print (node.data, end = " " ) q.popleft() if node.left is not None : q.append(node.left) if node.right is not None : q.append(node.right) nodeCount - = 1 print () # Driver Code if __name__ = = "__main__" : root = 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.right.right = newNode( 10 ) excluderightmost(root) |
C#
// C# implementation of the above approach using System; using System.Collections.Generic; class GFG { // Structure of the tree node public class Node { public int data; public Node left, right; }; // Utility method to create a node static Node newNode( int data) { Node node = new Node(); node.data = data; node.left = node.right = null ; return (node); } // Function to print all the nodes // except the rightmost in every level // of the given binary tree // with level order traversal static void excluderightmost(Node root) { // Base Case if (root == null ) return ; // Create an empty queue for level // order traversal Queue<Node> q = new Queue<Node>(); // Enqueue root 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(); // if Node is not right most print if (nodeCount != 1) 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) { 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.right.right = newNode(10); excluderightmost(root); } } |
Javascript
<script> // JavaScript implementation of the above approach // Structure of the tree node class Node { constructor() { this .data = 0; this .left = null ; this .right = null ; } }; // Utility method to create a node function newNode(data) { var node = new Node(); node.data = data; node.left = node.right = null ; return (node); } // Function to print all the nodes // except the rightmost in every level // of the given binary tree // with level order traversal function excluderightmost(root) { // Base Case if (root == null ) return ; // Create an empty queue for level // order traversal var q = []; // push root q.push(root); while ( true ) { // nodeCount (queue size) indicates // number of nodes at current level. var nodeCount = q.length; if (nodeCount == 0) break ; // Dequeue all nodes of current level // and push all nodes of next level while (nodeCount > 0) { var node = q[0]; // if Node is not right most print if (nodeCount != 1) 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>" ); } } // Driver code var 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.right.right = newNode(10); excluderightmost(root); </script> |
Output:
2 4 5 6 8