Iterative program to Calculate Size of a tree

Size of a tree is the number of elements present in the tree. Size of the below tree is 5. Example Tree

Recommended: Please solve it on “PRACTICE” first, before moving on to the solution.

Approach
The idea is to use Level Order Traversing

1) Create an empty queue q
2) temp_node = root /*start from root*/
3) Loop while temp_node is not NULL
a) Enqueue temp_node’s children (first left then right children) to q
b) Increase count with every enqueuing.
c) Dequeue a node from q and assign it’s value to temp_node

Java

 // Java programn to calculate // Size of a tree import java.util.LinkedList; import java.util.Queue;    class Node {     int data;     Node left, right;        public Node(int item)     {         data = item;         left = right = null;     } }    class BinaryTree {     Node root;                public int size()     {         if (root == null)             return 0;                    // Using level order Traversal .         Queue q = new LinkedList();         q.offer(root);                    int count = 1;          while (!q.isEmpty())         {             Node tmp = q.poll();                    // when the queue is empty:             // the poll() method returns null.             if (tmp!=null)             {                 if (tmp.left!=null)                 {                     // Increment count                     count++;                                            // Enqueue left child                      q.offer(tmp.left);                 }                 if (tmp.right!=null)                 {                     // Increment count                     count++;                                            // Enqueue left child                      q.offer(tmp.right);                 }             }         }                    return count;     }        public static void main(String args[])     {         /* creating a binary tree and entering            the nodes */         BinaryTree tree = new BinaryTree();         tree.root = new Node(1);         tree.root.left = new Node(2);         tree.root.right = new Node(3);         tree.root.left.left = new Node(4);         tree.root.left.right = new Node(5);            System.out.println("The size of binary tree" +                           " is : " + tree.size());     } }

C#

 // C# programn to calculate // Size of a tree using System; using System.Collections.Generic;    public class Node {     public int data;     public Node left, right;        public Node(int item)     {         data = item;         left = right = null;     } }    public class BinaryTree {     Node root;                public int size()     {         if (root == null)             return 0;                    // Using level order Traversal .         Queue q = new Queue();         q.Enqueue(root);                    int count = 1;          while (q.Count != 0)         {             Node tmp = q.Dequeue();                    // when the queue is empty:             // the poll() method returns null.             if (tmp != null)             {                 if (tmp.left != null)                 {                     // Increment count                     count++;                                            // Enqueue left child                      q.Enqueue(tmp.left);                 }                 if (tmp.right != null)                 {                     // Increment count                     count++;                                            // Enqueue left child                      q.Enqueue(tmp.right);                 }             }         }                    return count;     }        // Driver code     public static void Main(String []args)     {         /* creating a binary tree and entering          the nodes */         BinaryTree tree = new BinaryTree();         tree.root = new Node(1);         tree.root.left = new Node(2);         tree.root.right = new Node(3);         tree.root.left.left = new Node(4);         tree.root.left.right = new Node(5);            Console.WriteLine("The size of binary tree" +                          " is : " + tree.size());     } }    // This code has been contributed by 29AjayKumar

Output:

Size of the tree is 5

Time Complexity: O(n)
Auxiliary Space : O(level_max) where level max is maximum number of node in any level.

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