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
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# implementation of the above approach
// 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;
// Function to calculate density of Binary Tree
static float density(Node root)
// add root to queue first
// add null as a seperator
int height = 1, size = 0;
while (q.Count > 0)
Node t = q.Peek();
if (t != null)
// If after removeping null queue is
// empty then get out of loop i.e
// stop the level order traversal.
if (q.Count == 0)
// if t has left child
// then add it to queue
if (t.left !=null)
// if t has right child
// then add it to queue
if (t.right != null)
return ((float)size ) / height;
// Driver code
public static void Main(String args)
Node root = newNode(1);
root.left = newNode(2);
root.right = newNode(3);
// This code is contributed by PrinciRaj1992
Time Complexity : O(N)
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