# Density of Binary Tree using Level Order Traversal

Given a Binary Tree, find the density of it by doing one traversal of it.

The density of binary tree is defined as:

Density of Binary Tree = Size / Height

Examples:

Input :
Root of following tree
10
/   \
20   30

Output :  1.5
Height of given tree = 2
Size of given tree = 3

Input :
Root of the following tree
10
/
20
/
30
Output : 1
Height of given tree = 3
Size of given tree = 3

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

The size and height of the tree can be found in single traversal using level order traversal.

To calculate the height of the binary tree the idea is to use a “NULL” pointer as a separator between two levels. Whenever “NULL” occurs during the traversal, height is incremented.

To calculate the size of the binary tree, increment the counter for every new node encountered during the level order traversal.

Finally, use the above formula to calculate the density of the Binary Tree.

Below is the implementation of the above approach:

 // C++ implementation of the above approach #include using namespace std;    // A binary tree node struct Node {     int data;     Node *left, *right; };    // Helper function to allocates a new node Node* newNode(int data) {     Node* node = new Node;     node->data = data;     node->left = node->right = NULL;     return node; }    // Function to calculate density of Binary Tree float density(Node* root) {     queue q;        // push root to queue first     q.push(root);            // push NULL as a seperator     q.push(NULL);     int height = 1, size = 0;     while (!q.empty()) {         Node* t = q.front();         q.pop();         if (t)             size++;         else {                // If after popping NULL queue is             // empty then get out of loop i.e             // stop the level order traversal.             if (q.empty())                 break;             q.push(NULL);             height++;             continue;         }            // if t has left child         // then push it to queue         if (t->left) {             q.push(t->left);         }            // if t has right child         // then push it to queue         if (t->right) {             q.push(t->right);         }     }     return (float)size / height; }    // Driver code int main() {     Node* root = newNode(1);     root->left = newNode(2);     root->right = newNode(3);        cout << density(root) << endl;     return 0; }

 // Java implementation of the above approach  import java.util.*;    class Solution {    // A binary tree node  static class Node {      int data;      Node left, right;  }    // Helper function to allocates a new node  static Node newNode(int data)  {      Node node = new Node();      node.data = data;      node.left = node.right = null;      return node;  }     // Function to calculate density of Binary Tree  static float density(Node root)  {      Queue q = new LinkedList();         // add root to queue first      q.add(root);             // add null as a seperator      q.add(null);      int height = 1, size = 0;      while (q.size() > 0)     {          Node t = q.peek();          q.remove();          if (t != null)              size++;          else         {                 // If after removeping null queue is              // empty then get out of loop i.e              // stop the level order traversal.              if (q.size() == 0)                  break;              q.add(null);              height++;              continue;          }             // if t has left child          // then add it to queue          if (t.left !=null)          {              q.add(t.left);          }             // if t has right child          // then add it to queue          if (t.right != null)         {              q.add(t.right);          }      }      return ((float)size )/ height;  }     // Driver code  public static void main(String args[]) {      Node root = newNode(1);      root.left = newNode(2);      root.right = newNode(3);         System.out.println(density(root));  } }     // This code is contributed by Arnab Kundu

 # Python implementation of the above approach     # Linked List node  class Node:      def __init__(self, data):          self.data = data          self.left = None         self.right = None    # Helper function to allocates a new node  def newNode( data) :        node = Node(0)      node.data = data      node.left = node.right = None     return node     # Function to calculate density of Binary Tree  def density(root) :        q = []         # append root to queue first      q.append(root)             # append None as a seperator      q.append(None)      height = 1     size = 0     while (len(q) > 0):         t = q[0]          q.pop(0)          if (t != None):              size = size + 1         else:                # If after removeping None queue is              # empty then get out of loop i.e              # stop the level order traversal.              if (len(q) == 0):                  break             q.append(None)              height = height + 1             continue            # if t has left child          # then append it to queue          if (t.left != None) :             q.append(t.left)             # if t has right child          # then append it to queue          if (t.right != None):                        q.append(t.right)         return (size ) / height     # Driver code    root = newNode(1)  root.left = newNode(2)  root.right = newNode(3)     print(density(root))     # This code is contributed by Arnab Kundu

 // C# implementation of the above approach  using System;  using System.Collections.Generic;     class GFG {    // A binary tree node  public class Node {      public int data;      public Node left, right;  }    // Helper function to allocates a new node  static Node newNode(int data)  {      Node node = new Node();      node.data = data;      node.left = node.right = null;      return node;  }     // Function to calculate density of Binary Tree  static float density(Node root)  {      Queue q = new Queue();         // add root to queue first      q.Enqueue(root);             // add null as a seperator      q.Enqueue(null);      int height = 1, size = 0;      while (q.Count > 0)     {          Node t = q.Peek();          q.Dequeue();          if (t != null)              size++;          else         {                 // If after removeping null queue is              // empty then get out of loop i.e              // stop the level order traversal.              if (q.Count == 0)                  break;              q.Enqueue(null);              height++;              continue;          }             // if t has left child          // then add it to queue          if (t.left !=null)          {              q.Enqueue(t.left);          }             // if t has right child          // then add it to queue          if (t.right != null)         {              q.Enqueue(t.right);          }      }      return ((float)size ) / height;  }     // Driver code  public static void Main(String []args) {      Node root = newNode(1);      root.left = newNode(2);      root.right = newNode(3);         Console.WriteLine(density(root));  } }    // This code is contributed by PrinciRaj1992

Output:
1.5

Time Complexity : O(N)

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