Print the nodes of Binary Tree having a grandchild

Given a Binary Tree, the task is to print the nodes that have grandchildren.

Examples: 

Input: 
 

Output: 20 8 
Explanation: 
20 and 8 are the grandparents of 4, 12 and 10, 14.



Input: 
 

Output:
Explanation: 
1 is the grandparent of 4, 5. 

Approach: The idea uses Recursion. Below are the steps: 

  1. Traverse the given tree at every node.
  2. Check if each node has grandchildren node or not.
  3. For any tree node(say temp) if one of the below node exists then current node is the grandparent node: 
    • temp->left->left.
    • temp->left->right.
    • temp->right->left.
    • temp->right->right.
  4. If any of the above exist for any node temp then the node temp is the grandparent node.

Below is the implementation of the above approach:

C++

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// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
 
// A Binary Tree Node
struct node {
    struct node *left, *right;
    int key;
};
 
// Function to create new tree node
node* newNode(int key)
{
    node* temp = new node;
    temp->key = key;
    temp->left = temp->right = NULL;
    return temp;
}
 
// Function to print the nodes of
// the Binary Tree having a grandchild
void cal(struct node* root)
{
    // Base case to check
    // if the tree exists
 
    if (root == NULL)
        return;
 
    else {
 
        // Check if there is a left and
        // right child of the curr node
        if (root->left != NULL
            && root->right != NULL) {
 
            // Check for grandchildren
            if (root->left->left != NULL
                || root->left->right != NULL
                || root->right->left != NULL
                || root->right->right != NULL) {
 
                // Print the node's key
                cout << root->key << " ";
            }
        }
 
        // Check if the left child
        // of node is not null
        else if (root->left != NULL) {
 
            // Check for grandchildren
            if (root->left->left != NULL
                || root->left->right != NULL) {
                cout << root->key << " ";
            }
        }
 
        // Check if the right child
        // of node is not null
        else if (root->right != NULL) {
 
            // Check for grandchildren
            if (root->right->left != NULL
                || root->right->right != NULL) {
                cout << root->key << " ";
            }
        }
 
        // Recursive call on left and
        // right subtree
        cal(root->left);
        cal(root->right);
    }
}
 
// Driver Code
int main()
{
    // Given Tree
    struct node* root = newNode(1);
    root->left = newNode(2);
    root->right = newNode(3);
    root->left->left = newNode(4);
    root->left->right = newNode(5);
 
    // Function Call
    cal(root);
    return 0;
}

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Java

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// Java program for the above approach
import java.util.*;
 
class GFG{
 
// A Binary Tree Node
static class node
{
    node left, right;
    int key;
};
 
// Function to create new tree node
static node newNode(int key)
{
    node temp = new node();
    temp.key = key;
    temp.left = temp.right = null;
    return temp;
}
 
// Function to print the nodes of
// the Binary Tree having a grandchild
static void cal(node root)
{
     
    // Base case to check
    // if the tree exists
    if (root == null)
        return;
 
    else
    {
         
        // Check if there is a left and
        // right child of the curr node
        if (root.left != null &&
           root.right != null)
        {
             
            // Check for grandchildren
            if (root.left.left != null ||
               root.left.right != null ||
               root.right.left != null ||
              root.right.right != null)
            {
                 
                // Print the node's key
                System.out.print(root.key + " ");
            }
        }
 
        // Check if the left child
        // of node is not null
        else if (root.left != null)
        {
             
            // Check for grandchildren
            if (root.left.left != null ||
               root.left.right != null)
            {
                System.out.print(root.key + " ");
            }
        }
 
        // Check if the right child
        // of node is not null
        else if (root.right != null)
        {
             
            // Check for grandchildren
            if (root.right.left != null ||
               root.right.right != null)
            {
                System.out.print(root.key + " ");
            }
        }
 
        // Recursive call on left and
        // right subtree
        cal(root.left);
        cal(root.right);
    }
}
 
// Driver Code
public static void main(String[] args)
{
     
    // Given Tree
    node root = newNode(1);
    root.left = newNode(2);
    root.right = newNode(3);
    root.left.left = newNode(4);
    root.left.right = newNode(5);
 
    // Function call
    cal(root);
}
}
 
// This code is contributed by Amit Katiyar

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Python3

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# Python3 program for the
# above approach
 
# A Binary Tree Node
class newNode:
   
    def __init__(self, key):
       
        self.key = key
        self.left = None
        self.right = None
 
# Function to print the nodes
# of the Binary Tree having a
# grandchild
def cal(root):
   
    # Base case to check
    # if the tree exists
    if (root == None):
        return
 
    else:
        # Check if there is a left
        # and right child of the
        # curr node
        if (root.left != None and
            root.right != None):
           
            # Check for grandchildren
            if (root.left.left != None or
                root.left.right != None or
                root.right.left != None or
                root.right.right != None):
               
                # Print the node's key
                print(root.key, end = " ")
 
        # Check if the left child
        # of node is not None
        elif (root.left != None):
           
            # Check for grandchildren
            if (root.left.left != None or
                root.left.right != None):
                print(root.key, end = " ")
 
        # Check if the right child
        # of node is not None
        elif(root.right != None):
           
            # Check for grandchildren
            if (root.right.left != None or
                root.right.right != None):
                print(root.key, end = " ")
 
        # Recursive call on left and
        # right subtree
        cal(root.left)
        cal(root.right)
 
# Driver Code
if __name__ == '__main__':
   
    # Given Tree
    root = newNode(1)
    root.left = newNode(2)
    root.right = newNode(3)
    root.left.left = newNode(4)
    root.left.right = newNode(5)
 
    # Function Call
    cal(root)
 
# This code is contributed by SURENDRA_GANGWAR

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C#

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// C# program for the
// above approach
using System;
class GFG{
 
// A Binary Tree Node
public class node
{
  public node left, right;
  public int key;
};
 
// Function to create new
// tree node
static node newNode(int key)
{
  node temp = new node();
  temp.key = key;
  temp.left = temp.right = null;
  return temp;
}
 
// Function to print the
// nodes of the Binary Tree
// having a grandchild
static void cal(node root)
{
  // Base case to check
  // if the tree exists
  if (root == null)
    return;
   
  else
  {
    // Check if there is a left and
    // right child of the curr node
    if (root.left != null &&
        root.right != null)
    {
      // Check for grandchildren
      if (root.left.left != null ||
          root.left.right != null ||
          root.right.left != null ||
          root.right.right != null)
      {
        // Print the node's key
        Console.Write(root.key + " ");
      }
    }
 
    // Check if the left child
    // of node is not null
    else if (root.left != null)
    {
      // Check for grandchildren
      if (root.left.left != null ||
          root.left.right != null)
      {
        Console.Write(root.key + " ");
      }
    }
 
    // Check if the right child
    // of node is not null
    else if (root.right != null)
    {
      // Check for grandchildren
      if (root.right.left != null ||
          root.right.right != null)
      {
        Console.Write(root.key + " ");
      }
    }
 
    // Recursive call on left and
    // right subtree
    cal(root.left);
    cal(root.right);
  }
}
 
// Driver Code
public static void Main(String[] args)
{
  // Given Tree
  node root = newNode(1);
  root.left = newNode(2);
  root.right = newNode(3);
  root.left.left = newNode(4);
  root.left.right = newNode(5);
 
  // Function call
  cal(root);
}
}
 
// This code is contributed by Princi Singh

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Output: 

1









 

Time Complexity: O(N), where N is the number of nodes. 
Auxiliary Space: O(1)

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