Print all internal nodes of a Binary tree

Given a Binary tree, the task is to print all the internal nodes in a tree.

An internal node is a node which carries at least one child or in other words, an internal node is not a leaf node. Here we intend to print all such internal nodes in level order. Consider the following Binary Tree:

Input:



Output: 15 10 20

The way to solve this involves a BFS of the tree. The algorithm is as follows:

  • Do a level order traversal by pushing nodes in the queue one by one.
  • Pop the elements from the queue one by one and keep a track of following cases:
    1. The node has a left child only.
    2. The node has a right child only.
    3. The node has both left and right child.
    4. The node has no children at all.
  • Except for case 4, print the data in the node for all the other 3 cases.

Below is the implementation of the above approach:

C++

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// C++ program to print all internal
// nodes in tree
#include <bits/stdc++.h>
using namespace std;
  
// A node in the Binary tree
struct Node {
    int data;
    Node *left, *right;
    Node(int data)
    {
       left = right = NULL;
       this->data = data;
    }
};
  
// Function to print all internal nodes 
// in level order from left to right
void printInternalNodes(Node* root)
{
    // Using a queue for a level order traversal
    queue<Node*> q;
    q.push(root);
    while (!q.empty()) {
  
        // Check and pop the element in 
        // the front of the queue
        Node* curr = q.front();
        q.pop();
  
        // The variable flag keeps track of 
        // whether a node is an internal node
        bool isInternal = 0;
  
        // The node has a left child
        if (curr->left) {
            isInternal = 1;
            q.push(curr->left);
        }
  
        // The node has a right child
        if (curr->right) {
            isInternal = 1;
            q.push(curr->right);
        }
  
        // In case the node has either a left 
        // or right child or both print the data
        if (isInternal)
            cout << curr->data << " ";
    }
}
  
// Driver program to build a sample tree
int main()
{
    Node* root = new Node(1);
    root->left = new Node(2);
    root->right = new Node(3);
    root->left->left = new Node(4);
    root->right->left = new Node(5);
    root->right->right = new Node(6);
    root->right->right->right = new Node(10);
    root->right->right->left = new Node(7);
    root->right->left->left = new Node(8);
    root->right->left->right = new Node(9);
  
    // A call to the function
    printInternalNodes(root);
  
    return 0;
}

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Java

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// Java program to print all internal 
// nodes in tree
import java.util.*;
class GfG 
{
  
// A node in the Binary tree 
static class Node 
    int data; 
    Node left, right; 
    Node(int data) 
    
        left = right = null
        this.data = data; 
    
}
  
// Function to print all internal nodes 
// in level order from left to right 
static void printInternalNodes(Node root) 
    // Using a queue for a level order traversal 
    Queue<Node> q = new LinkedList<Node>(); 
    q.add(root); 
    while (!q.isEmpty()) 
    
  
        // Check and pop the element in 
        // the front of the queue 
        Node curr = q.peek(); 
        q.remove(); 
  
        // The variable flag keeps track of 
        // whether a node is an internal node 
        boolean isInternal = false
  
        // The node has a left child 
        if (curr.left != null
        
            isInternal = true
            q.add(curr.left); 
        
  
        // The node has a right child 
        if (curr.right != null
        
            isInternal = true
            q.add(curr.right); 
        
  
        // In case the node has either a left 
        // or right child or both print the data 
        if (isInternal == true
            System.out.print(curr.data + " "); 
    
  
// Driver code
public static void main(String[] args) 
    Node root = new Node(1); 
    root.left = new Node(2); 
    root.right = new Node(3); 
    root.left.left = new Node(4); 
    root.right.left = new Node(5); 
    root.right.right = new Node(6); 
    root.right.right.right = new Node(10); 
    root.right.right.left = new Node(7); 
    root.right.left.left = new Node(8); 
    root.right.left.right = new Node(9); 
  
    // A call to the function 
    printInternalNodes(root); 
}
  
// This code is contributed by 
// Prerna Saini.

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Python3

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# Python3 program to print all internal
# nodes in tree
  
# A node in the Binary tree
class new_Node: 
      
    # Constructor to create a new_Node 
    def __init__(self, data): 
        self.data = data 
        self.left = None
        self.right = None
      
# Function to prall internal nodes 
# in level order from left to right
def printInternalNodes(root):
      
    # Using a queue for a level order traversal
    q = []
    q.append(root)
    while (len(q)):
          
        # Check and pop the element in 
        # the front of the queue
        curr = q[0]
        q.pop(0)
          
        # The variable flag keeps track of 
        # whether a node is an internal node
        isInternal = 0
          
        # The node has a left child
        if (curr.left):
            isInternal = 1
            q.append(curr.left)
          
        # The node has a right child
        if (curr.right):
            isInternal = 1
            q.append(curr.right)
          
        # In case the node has either a left 
        # or right child or both prthe data
        if (isInternal):
            print(curr.data, end = " ")
      
# Driver Code
root = new_Node(1)
root.left = new_Node(2)
root.right = new_Node(3)
root.left.left = new_Node(4)
root.right.left = new_Node(5)
root.right.right = new_Node(6)
root.right.right.right = new_Node(10)
root.right.right.left = new_Node(7)
root.right.left.left = new_Node(8)
root.right.left.right = new_Node(9)
  
# A call to the function
printInternalNodes(root)
  
# This code is contributed by SHUBHAMSINGH10

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C#

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// C# program to print all internal 
// nodes in tree 
using System;
using System.Collections.Generic;
  
class GFG 
  
// A node in the Binary tree 
public class Node 
    public int data; 
    public Node left, right; 
    public Node(int data) 
    
        left = right = null
        this.data = data; 
    
  
// Function to print all internal nodes 
// in level order from left to right 
static void printInternalNodes(Node root) 
    // Using a queue for a level order traversal 
    Queue<Node> q = new Queue<Node>(); 
    q.Enqueue(root); 
    while (q.Count != 0) 
    
  
        // Check and pop the element in 
        // the front of the queue 
        Node curr = q.Peek(); 
        q.Dequeue(); 
  
        // The variable flag keeps track of 
        // whether a node is an internal node 
        Boolean isInternal = false
  
        // The node has a left child 
        if (curr.left != null
        
            isInternal = true
            q.Enqueue(curr.left); 
        
  
        // The node has a right child 
        if (curr.right != null
        
            isInternal = true
            q.Enqueue(curr.right); 
        
  
        // In case the node has either a left 
        // or right child or both print the data 
        if (isInternal == true
            Console.Write(curr.data + " "); 
    
  
// Driver code 
public static void Main(String[] args) 
    Node root = new Node(1); 
    root.left = new Node(2); 
    root.right = new Node(3); 
    root.left.left = new Node(4); 
    root.right.left = new Node(5); 
    root.right.right = new Node(6); 
    root.right.right.right = new Node(10); 
    root.right.right.left = new Node(7); 
    root.right.left.left = new Node(8); 
    root.right.left.right = new Node(9); 
  
    // A call to the function 
    printInternalNodes(root); 
  
// This code contributed by Rajput-Ji

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Output:

1 2 3 5 6


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