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Print all nodes except rightmost node of every level of the Binary Tree

  • Last Updated : 22 Jun, 2021

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

 

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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

 




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