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주어진 숫자의 요인 트리

Factor Tree는 숫자의 요인을 이해하는 직관적인 방법입니다. 모든 요소가 숫자에서 어떻게 파생되었는지 보여줍니다. 이는 더 이상 인수분해할 수 없을 때까지 숫자의 인수를 찾은 다음 해당 숫자의 인수 등을 찾는 특수 다이어그램입니다. 끝은 모두 원래 숫자의 소인수입니다.

예: 

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

요인 트리는 재귀적으로 생성됩니다. 이진 트리가 사용됩니다. 



  1. 우리는 숫자로 시작하여 가능한 최소 제수를 찾습니다.
  2. 그런 다음 상위 숫자를 최소 제수로 나눕니다.
  3. 우리는 제수와 몫을 모두 상위 숫자의 두 하위 항목으로 저장합니다.
  4. 두 하위 항목 모두 재귀적으로 기능하도록 전송됩니다.
  5. 숫자의 절반보다 작은 제수가 발견되지 않으면 두 하위 항목은 NULL로 저장됩니다.

구현:

C++
// C++ program to construct Factor Tree for // a given number #include   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 == null) return;  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 == null) return;  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 == null) return;  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' + '
'
); printLevelOrder(root); // This code is contributed by suresh07. </script>

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

시간 복잡도: O(n) 여기서 n은 주어진 숫자입니다.

공간 복잡도: O(k) 여기서 k는 숫자의 인수입니다.

 

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