Product of all leaf nodes of binary tree
Given a binary tree, find the product of all the leaf nodes.
Examples:
Input :
1
/ \
2 3
/ \ / \
4 5 6 7
\
8
Output :
product = 4 * 5 * 8 * 7 = 1120
The idea is to traverse the tree in any fashion and check if the node is the leaf node or not. If the node is the leaf node, multiply node data to a variable prod used to store the products of leaf nodes.
Following is the implementation of above approach.
C++
// CPP program to find product of // all leaf nodes of binary tree #include <bits/stdc++.h> using namespace std; // struct binary tree node struct Node { int data; Node *left, *right; }; // return new node Node* newNode(int data) { Node* temp = new Node(); temp->data = data; temp->left = temp->right = NULL; return temp; } // utility function which calculates // product of all leaf nodes void leafProduct(Node* root, int& prod) { if (!root) return; // product root data to prod if // root is a leaf node if (!root->left && !root->right) prod *= root->data; // propagate recursively in left // and right subtree leafProduct(root->left, prod); leafProduct(root->right, prod); } // Driver program int main() { // contruct binary tree Node* root = newNode(1); root->left = newNode(2); root->left->left = newNode(4); root->left->right = newNode(5); root->right = newNode(3); root->right->right = newNode(7); root->right->left = newNode(6); root->right->left->right = newNode(8); // variable to store product of leaf nodes int prod = 1; leafProduct(root, prod); cout << prod << endl; return 0; } |
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Java
// Java program to find product of // all leaf nodes of binary tree class GFG { // struct binary tree node static class Node { int data; Node left, right; }; // return new node static Node newNode(int data) { Node temp = new Node(); temp.data = data; temp.left = temp.right = null; return temp; } // product static int prod = 1; // utility function which calculates // product of all leaf nodes static void leafProduct(Node root ) { if (root == null) return; // product root data to prod if // root is a leaf node if (root.left == null && root.right == null) prod *= root.data; // propagate recursively in left // and right subtree leafProduct(root.left); leafProduct(root.right); } // Driver program public static void main(String args[]) { // contruct binary tree Node root = newNode(1); root.left = newNode(2); root.left.left = newNode(4); root.left.right = newNode(5); root.right = newNode(3); root.right.right = newNode(7); root.right.left = newNode(6); root.right.left.right = newNode(8); // variable to store product of leaf nodes prod = 1; leafProduct(root); System.out.println(prod ); } } // This code is contributed by Arnab Kundu |
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C#
// C# program to find product of // all leaf nodes of binary tree using System; class GFG { // struct binary tree node public class Node { public int data; public Node left, right; }; // return new node static Node newNode(int data) { Node temp = new Node(); temp.data = data; temp.left = temp.right = null; return temp; } // product static int prod = 1; // utility function which calculates // product of all leaf nodes static void leafProduct(Node root ) { if (root == null) return; // product root data to prod if // root is a leaf node if (root.left == null && root.right == null) prod *= root.data; // propagate recursively in left // and right subtree leafProduct(root.left); leafProduct(root.right); } // Driver code public static void Main(String []args) { // contruct binary tree Node root = newNode(1); root.left = newNode(2); root.left.left = newNode(4); root.left.right = newNode(5); root.right = newNode(3); root.right.right = newNode(7); root.right.left = newNode(6); root.right.left.right = newNode(8); // variable to store product of leaf nodes prod = 1; leafProduct(root); Console.WriteLine(prod ); } } // This code has been contributed by 29AjayKumar |
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Output:
1120
Time Complexity : O(n), where n is the total number of nodes in the tree.
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