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Density of Binary Tree using Level Order Traversal
  • Last Updated : 29 Jan, 2020

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 

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++




// C++ implementation of the above approach
#include <bits/stdc++.h>
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<Node*> 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




// 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<Node> q = new LinkedList<Node>(); 
  
    // 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




# 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#




// 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<Node> q = new Queue<Node>(); 
  
    // 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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