Size of a tree is the number of elements present in the tree. Size of the below tree is 5.
Example Tree
Size() function recursively calculates the size of a tree. It works as follows:
Size of a tree = Size of left subtree + 1 + Size of right subtree.
Algorithm:
size(tree)
1. If tree is empty then return 0
2. Else
(a) Get the size of left subtree recursively i.e., call
size( tree->left-subtree)
(a) Get the size of right subtree recursively i.e., call
size( tree->right-subtree)
(c) Calculate size of the tree as following:
tree_size = size(left-subtree) + size(right-
subtree) + 1
(d) Return tree_size
C++
// A recursive C++ program to // calculate the size of the tree #include <bits/stdc++.h> using namespace std; /* A binary tree node has data, pointer to left child and a pointer to right child */class node { public: int data; node* left; node* right; }; /* Helper function that allocates a new node with the given data and NULL left and right pointers. */node* newNode(int data) { node* Node = new node(); Node->data = data; Node->left = NULL; Node->right = NULL; return(Node); } /* Computes the number of nodes in a tree. */int size(node* node) { if (node == NULL) return 0; else return(size(node->left) + 1 + size(node->right)); } /* Driver code*/int main() { node *root = newNode(1); root->left = newNode(2); root->right = newNode(3); root->left->left = newNode(4); root->left->right = newNode(5); cout << "Size of the tree is " << size(root); return 0; } // This code is contributed by rathbhupendra |
C
#include <stdio.h> #include <stdlib.h> /* A binary tree node has data, pointer to left child and a pointer to right child */struct node { int data; struct node* left; struct node* right; }; /* Helper function that allocates a new node with the given data and NULL left and right pointers. */struct node* newNode(int data) { struct node* node = (struct node*) malloc(sizeof(struct node)); node->data = data; node->left = NULL; node->right = NULL; return(node); } /* Computes the number of nodes in a tree. */int size(struct node* node) { if (node==NULL) return 0; else return(size(node->left) + 1 + size(node->right)); } /* Driver program to test size function*/ int main() { struct node *root = newNode(1); root->left = newNode(2); root->right = newNode(3); root->left->left = newNode(4); root->left->right = newNode(5); printf("Size of the tree is %d", size(root)); getchar(); return 0; } |
Java
// A recursive Java program to calculate the size of the tree /* Class containing left and right child of current node and key value*/class Node { int data; Node left, right; public Node(int item) { data = item; left = right = null; } } /* Class to find size of Binary Tree */class BinaryTree { Node root; /* Given a binary tree. Print its nodes in level order using array for implementing queue */ int size() { return size(root); } /* computes number of nodes in tree */ int size(Node node) { if (node == null) return 0; else return(size(node.left) + 1 + size(node.right)); } 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()); } } |
Python
# Python Program to find the size of binary tree # A binary tree node class Node: # Constructor to create a new node def __init__(self, data): self.data = data self.left = None self.right = None # Computes the number of nodes in tree def size(node): if node is None: return 0 else: return (size(node.left)+ 1 + size(node.right)) # Driver program to test above function root = Node(1) root.left = Node(2) root.right = Node(3) root.left.left = Node(4) root.left.right = Node(5) print "Size of the tree is %d" %(size(root)) # This code is contributed by Nikhil Kumar Singh(nickzuck_007) |
C#
using System; // A recursive C# program to calculate the size of the tree /* Class containing left and right child of current node and key value*/public class Node { public int data; public Node left, right; public Node(int item) { data = item; left = right = null; } } /* Class to find size of Binary Tree */public class BinaryTree { public Node root; /* Given a binary tree. Print its nodes in level order using array for implementing queue */ public virtual int size() { return size(root); } /* computes number of nodes in tree */ public virtual int size(Node node) { if (node == null) { return 0; } else { return (size(node.left) + 1 + size(node.right)); } } 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 is contributed by Shrikant13 |
Output:
Size of the tree is 5
Time & Space Complexities: Since this program is similar to traversal of tree, time and space complexities will be same as Tree traversal (Please see our Tree Traversal post for details)
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