Skip to content
Related Articles

Related Articles

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

Improve Article
Save Article
Like Article
  • Difficulty Level : Easy
  • Last Updated : 07 Feb, 2022

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++




// 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;
}

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 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

Python3




# Python implementation of the approach
from collections import dequeue
# 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 leftmost in every level
# of the given binary tree
# with level order traversal
def excludeLeftMost(root: Node):
 
    # Base Case
    if root is None:
        return
 
    # Create an empty queue for level
    # order traversal
    q = dequeue()
 
    # Enqueue root
    q.append(root)
 
    while 1:
 
        # nodeCount (queue size) indicates
        # number of nodes at current level
        nodeCount = len(q)
        if nodeCount == 0:
            break
 
        # Initialize leftmost as true
        # just before the beginning
        # of each level
        leftmost = True
 
        # Dequeue all nodes of current level
        # and Enqueue all nodes of next level
        while nodeCount > 0:
            node = q[0]
 
            # Switch leftmost flag after processing
            # the leftmost node
            if leftmost:
                leftmost = not leftmost
 
            # Print all the nodes except leftmost
            else:
                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)
    excludeLeftMost(root)
 
# This code is contributed by
# sanjeev2552

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 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

Javascript




<script>
 
// Javascript implementation of the approach
 
// Structure of the tree node
class Node
{
    constructor(data)
    {
        this.left = null;
        this.right = null;
        this.data = data;
    }
}
 
// Utility method to create a node
function newNode(data)
{
    let node = new Node(data);
    return (node);
}
 
// Function to print all the nodes
// except the leftmost in every level
// of the given binary tree
// with level order traversal
function excludeLeftmost(root)
{
     
    // Base Case
    if (root == null)
        return;
 
    // Create an empty queue for level
    // order traversal
    let q = [];
 
    // Enqueue root
    q.push(root);
 
    while (true)
    {
         
        // nodeCount (queue size) indicates
        // number of nodes at current level.
        let nodeCount = q.length;
        if (nodeCount == 0)
            break;
 
        // Initialize leftmost as true
        // just before the beginning
        // of each level
        let leftmost = true;
 
        // Dequeue all nodes of current level
        // and Enqueue all nodes of next level
        while (nodeCount > 0)
        {
            let node = q[0];
 
            // Switch leftmost flag after processing
            // the leftmost node
            if (leftmost)
                leftmost = !leftmost;
 
            // Print all the nodes except leftmost
            else
                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
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.right.right = newNode(10);
 
excludeLeftmost(root);
 
// This code is contributed by divyeshrabadiya07
 
</script>
Output: 
3 
5 6 7 
9

 


My Personal Notes arrow_drop_up
Recommended Articles
Page :

Start Your Coding Journey Now!