Print all the nodes except the leftmost node in every level of the given binary tree

Given a binary tree, the task is to print all the nodes except the leftmost in every level of the tree. The root is considered at level 0, and left most 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:
3
5  6
8
10

Input:
          1
        /   \
       2     3
        \     \
         4     5
Output:
3
5

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 mark leftmost flag true just before the processing of each level and mark it false just after processing of the first node at each level.



Below is the implementation of the above approach:

C++

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// 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 leftmost in every level
// of the given binary tree
// with level order traversal
void excludeLeftmost(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;
  
        // Initialize leftmost as true
        // just before the beginning
        // of each level
        bool leftmost = true;
  
        // Dequeue all nodes of current level
        // and Enqueue all nodes of next level
        while (nodeCount > 0) {
            Node* node = q.front();
  
            // Switch leftmost flag after processing
            // the leftmost node
            if (leftmost)
                leftmost = !leftmost;
  
            // Print all the nodes except leftmost
            else
                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);
  
    excludeLeftmost(root);
  
    return 0;
}

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Java

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// 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 leftmost in every level 
// of the given binary tree 
// with level order traversal 
static void excludeLeftmost(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
  
        // Initialize leftmost as true 
        // just before the beginning 
        // of each level 
        boolean leftmost = true
  
        // Dequeue all nodes of current level 
        // and Enqueue all nodes of next level 
        while (nodeCount > 0)
        
            Node node = q.peek(); 
  
            // Switch leftmost flag after processing 
            // the leftmost node 
            if (leftmost) 
                leftmost = !leftmost; 
  
            // Print all the nodes except leftmost 
            else
                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); 
  
    excludeLeftmost(root); 
}
  
// This code is contributed by Arnab Kundu

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C#

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// 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 leftmost in every level 
// of the given binary tree 
// with level order traversal 
static void excludeLeftmost(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
  
        // Initialize leftmost as true 
        // just before the beginning 
        // of each level 
        Boolean leftmost = true
  
        // Dequeue all nodes of current level 
        // and Enqueue all nodes of next level 
        while (nodeCount > 0)
        
            Node node = q.Peek(); 
  
            // Switch leftmost flag after processing 
            // the leftmost node 
            if (leftmost) 
                leftmost = !leftmost; 
  
            // Print all the nodes except leftmost 
            else
                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); 
  
    excludeLeftmost(root); 
}
}
  
// This code is contributed by PrinciRaj1992 

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Output:

3 
5 6 7 
9


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