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: 1
Explanation:
1 is the grandparent of 4, 5.
Approach: The idea uses Recursion. Below are the steps:
- Traverse the given tree at every node.
- Check if each node has grandchildren node or not.
- 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.
- 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++
// 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; } |
Java
// 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 |
Python3
# 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 |
C#
// 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 |
Javascript
<script> // Javascript program for the above approach // A Binary Tree Node class node { constructor(key) { this.left = null; this.right = null; this.key = key; } } // Function to create new tree node function newNode(key) { let temp = new node(key); return temp; } // Function to print the nodes of // the Binary Tree having a grandchild function cal(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 document.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) { document.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) { document.write(root.key + " "); } } // Recursive call on left and // right subtree cal(root.left); cal(root.right); } } // Given Tree let 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 mukesh07. </script> |
Output:
1
Time Complexity: O(N), where N is the number of nodes.
Auxiliary Space: O(N)
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