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Factor Tree of a given Number

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  • Difficulty Level : Medium
  • Last Updated : 20 Aug, 2022
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Factor Tree is an intuitive method to understand the factors of a number. It shows how all the factors are been derived from the number. It is a special diagram where you find the factors of a number, then the factors of those numbers, etc until you can’t factor anymore. The ends are all the prime factors of the original number.

Example: 

Input : v = 48
Output : Root of below tree
   48
   /\
  2  24
     /\
    2  12
       /\
      2  6
         /\
        2  3

The factor tree is created recursively. A binary tree is used. 

  1. We start with a number and find the minimum divisor possible.
  2. Then, we divide the parent number by the minimum divisor.
  3. We store both the divisor and quotient as two children of the parent number.
  4. Both the children are sent into function recursively.
  5. If a divisor less than half the number is not found, two children are stored as NULL.

Implementation:

C++




// C++ program to construct Factor Tree for
// a given number
#include<bits/stdc++.h>
using namespace std;
 
// Tree node
struct Node
{
    struct Node *left, *right;
    int key;
};
 
// Utility function to create a new tree Node
Node* newNode(int key)
{
    Node* temp = new Node;
    temp->key = key;
    temp->left = temp->right = NULL;
    return temp;
}
 
// Constructs factor tree for given value and stores
// root of tree at given reference.
void createFactorTree(struct Node **node_ref, int v)
{
    (*node_ref) = newNode(v);
 
    // the number is factorized
    for (int i = 2 ; i < v/2 ; i++)
    {
        if (v % i != 0)
          continue;
 
        // If we found a factor, we construct left
        // and right subtrees and return. Since we
        // traverse factors starting from smaller
        // to greater, left child will always have
        // smaller factor
        createFactorTree(&((*node_ref)->left), i);
        createFactorTree(&((*node_ref)->right), v/i);
        return;
    }
}
 
// Iterative method to find the height of Binary Tree
void printLevelOrder(Node *root)
{
    // Base Case
    if (root == NULL)  return;
 
    queue<Node *> q;
    q.push(root);
 
    while (q.empty() == false)
    {
        // Print front of queue and remove
        // it from queue
        Node *node = q.front();
        cout << node->key << " ";
        q.pop();
        if (node->left != NULL)
            q.push(node->left);
        if (node->right != NULL)
            q.push(node->right);
    }
}
 
// driver program
int main()
{
    int val = 48;// sample value
    struct Node *root = NULL;
    createFactorTree(&root, val);
    cout << "Level order traversal of "
            "constructed factor tree";
    printLevelOrder(root);
    return 0;
}

Java




// Java program to construct Factor Tree for
// a given number
import java.util.*;
class GFG
{
 
  // Tree node
  static class Node
  {
    Node left, right;
    int key;
  };
  static Node root;
 
  // Utility function to create a new tree Node
  static Node newNode(int key)
  {
    Node temp = new Node();
    temp.key = key;
    temp.left = temp.right = null;
    return temp;
  }
 
  // Constructs factor tree for given value and stores
  // root of tree at given reference.
  static Node createFactorTree(Node node_ref, int v)
  {
    (node_ref) = newNode(v);
 
    // the number is factorized
    for (int i = 2 ; i < v/2 ; i++)
    {
      if (v % i != 0)
        continue;
 
      // If we found a factor, we construct left
      // and right subtrees and return. Since we
      // traverse factors starting from smaller
      // to greater, left child will always have
      // smaller factor
      node_ref.left = createFactorTree(((node_ref).left), i);
      node_ref.right =  createFactorTree(((node_ref).right), v/i);
      return node_ref;
    }
    return node_ref;
  }
 
  // Iterative method to find the height of Binary Tree
  static void printLevelOrder(Node root)
  {
 
    // Base Case
    if (root == nullreturn;
    Queue<Node > q = new LinkedList<>();
    q.add(root);
    while (q.isEmpty() == false)
    {
 
      // Print front of queue and remove
      // it from queue
      Node node = q.peek();
      System.out.print(node.key + " ");
      q.remove();
      if (node.left != null)
        q.add(node.left);
      if (node.right != null)
        q.add(node.right);
    }
  }
 
  // Driver program
  public static void main(String[] args)
  {
    int val = 48;// sample value
    root = null;
    root = createFactorTree(root, val);
    System.out.println("Level order traversal of "+
                       "constructed factor tree");
    printLevelOrder(root);
  }
}
 
// This code is contributed by Rajput-Ji

Python3




# Python program to construct Factor Tree for
# a given number
class Node:
    def __init__(self, key):
        self.left = None
        self.right = None
        self.key = key
 
# Utility function to create a new tree Node
def newNode(key):
    temp = Node(key)
    return temp
 
# Constructs factor tree for given value and stores
# root of tree at given reference.
def createFactorTree(node_ref, v):
    node_ref = newNode(v)
 
    # the number is factorized
    for i in range(2, int(v/2)):
 
        if (v % i != 0):
            continue
 
        # If we found a factor, we construct left
        # and right subtrees and return. Since we
        # traverse factors starting from smaller
        # to greater, left child will always have
        # smaller factor
        node_ref.left = createFactorTree(((node_ref).left), i)
        node_ref.right = createFactorTree(((node_ref).right), int(v/i))
        return node_ref
 
    return node_ref
 
# Iterative method to find the height of Binary Tree
def printLevelOrder(root):
 
    # Base Case
    if (root == None):
        return
    q = [];
    q.append(root);
    while (len(q) > 0):
 
        # Print front of queue and remove
        # it from queue
        node = q[0]
        print(node.key, end = " ")
        q = q[1:]
        if (node.left != None):
                q.append(node.left)
        if (node.right != None):
                q.append(node.right)
 
val = 48# sample value
root = None
root = createFactorTree(root, val)
print("Level order traversal of constructed factor tree")
printLevelOrder(root)
 
# This code is contributed by shinjanpatra

C#




// C# program to construct Factor Tree for
// a given number
using System;
using System.Collections.Generic;
 
public class GFG
{
 
  // Tree node
  public
    class Node
    {
      public
        Node left, right;
      public
        int key;
    };
  static Node root;
 
  // Utility function to create a new tree Node
  static Node newNode(int key)
  {
    Node temp = new Node();
    temp.key = key;
    temp.left = temp.right = null;
    return temp;
  }
 
  // Constructs factor tree for given value and stores
  // root of tree at given reference.
  static Node createFactorTree(Node node_ref, int v)
  {
    (node_ref) = newNode(v);
 
    // the number is factorized
    for (int i = 2 ; i < v/2 ; i++)
    {
      if (v % i != 0)
        continue;
 
      // If we found a factor, we construct left
      // and right subtrees and return. Since we
      // traverse factors starting from smaller
      // to greater, left child will always have
      // smaller factor
      node_ref.left = createFactorTree(((node_ref).left), i);
      node_ref.right =  createFactorTree(((node_ref).right), v/i);
      return node_ref;
    }
    return node_ref;
  }
 
  // Iterative method to find the height of Binary Tree
  static void printLevelOrder(Node root)
  {
 
    // Base Case
    if (root == nullreturn;
    Queue<Node > q = new Queue<Node>();
    q.Enqueue(root);
    while (q.Count != 0)
    {
 
      // Print front of queue and remove
      // it from queue
      Node node = q.Peek();
      Console.Write(node.key + " ");
      q.Dequeue();
      if (node.left != null)
        q.Enqueue(node.left);
      if (node.right != null)
        q.Enqueue(node.right);
    }
  }
 
  // Driver program
  public static void Main(String[] args)
  {
    int val = 48;// sample value
    root = null;
    root = createFactorTree(root, val);
    Console.WriteLine("Level order traversal of "+
                      "constructed factor tree");
    printLevelOrder(root);
  }
}
 
// This code is contributed by gauravrajput1

Javascript




<script>
    // Javascript program to construct Factor Tree for
    // a given number
     
    class Node
    {
        constructor(key) {
           this.left = null;
           this.right = null;
           this.key = key;
        }
    }
     
    let root;
  
    // Utility function to create a new tree Node
    function newNode(key)
    {
      let temp = new Node(key);
      return temp;
    }
 
    // Constructs factor tree for given value and stores
    // root of tree at given reference.
    function createFactorTree(node_ref, v)
    {
      (node_ref) = newNode(v);
 
      // the number is factorized
      for (let i = 2 ; i < parseInt(v/2, 10) ; i++)
      {
        if (v % i != 0)
          continue;
 
        // If we found a factor, we construct left
        // and right subtrees and return. Since we
        // traverse factors starting from smaller
        // to greater, left child will always have
        // smaller factor
        node_ref.left = createFactorTree(((node_ref).left), i);
        node_ref.right =  createFactorTree(((node_ref).right), parseInt(v/i, 10));
        return node_ref;
      }
      return node_ref;
    }
 
    // Iterative method to find the height of Binary Tree
    function printLevelOrder(root)
    {
 
      // Base Case
      if (root == nullreturn;
      let q = [];
      q.push(root);
      while (q.length > 0)
      {
 
        // Print front of queue and remove
        // it from queue
        let node = q[0];
        document.write(node.key + " ");
        q.shift();
        if (node.left != null)
          q.push(node.left);
        if (node.right != null)
          q.push(node.right);
      }
    }
     
    let val = 48;// sample value
    root = null;
    root = createFactorTree(root, val);
    document.write("Level order traversal of "+
                       "constructed factor tree" + "</br>");
    printLevelOrder(root);
    
   // This code is contributed by suresh07.
</script>

Output

Level order traversal of constructed factor tree48 2 24 2 12 2 6 2 3 

Time Complexity: O(n), where n is the given number.

Space Complexity: O(k), where k is the factor of the number.

This article is contributed by Suprotik Dey. If you like GeeksforGeeks and would like to contribute, you can also write an article using write.geeksforgeeks.org or mail your article to review-team@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks. 


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