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

Approach: The idea use 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 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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Output:

1

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

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