Iterative program to count leaf nodes in a Binary Tree

• Difficulty Level : Easy
• Last Updated : 14 May, 2021

Given a binary tree, count leaves in the tree without using recursion. A node is a leaf node if both left and right children of it are NULL.

Example Tree Leaves count for the above tree is 3.

The idea is to use level order traversal. During traversal, if we find a node whose left and right children are NULL, we increment count.

C++

 // C++ program to count leaf nodes in a Binary Tree#include using namespace std; /* A binary tree Node has data, pointer to left   child  and a pointer to right child */struct Node{    int data;    struct Node* left, *right;}; /* Function to get the count of leaf Nodes in   a binary tree*/unsigned int getLeafCount(struct Node* node){    // If tree is empty    if (!node)        return 0;     // Initialize empty queue.    queue q;     // Do level order traversal starting from root    int count = 0; // Initialize count of leaves    q.push(node);    while (!q.empty())    {        struct Node *temp = q.front();        q.pop();         if (temp->left != NULL)            q.push(temp->left);        if (temp->right != NULL)            q.push(temp->right);        if (temp->left == NULL && temp->right == NULL)            count++;    }    return count;} /* 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 = new Node;    node->data = data;    node->left = node->right = NULL;    return (node);} /* Driver program to test above functions*/int main(){    /*   1        /  \      2     3     / \    4   5    Let us create Binary Tree shown in    above example */     struct Node *root = newNode(1);    root->left        = newNode(2);    root->right       = newNode(3);    root->left->left  = newNode(4);    root->left->right = newNode(5);     /* get leaf count of the above created tree */    cout << getLeafCount(root);     return 0;}

Java

 // Java program to count leaf nodes// in a Binary Treeimport java.util.*; class GFG{    /* A binary tree Node has data,    pointer to left child and    a pointer to right child */    static class Node    {        int data;        Node left, right;    }     /* Function to get the count of    leaf Nodes in a binary tree*/    static int getLeafCount(Node node)    {        // If tree is empty        if (node == null)        {            return 0;        }         // Initialize empty queue.        Queue q = new LinkedList<>();         // Do level order traversal starting from root        int count = 0; // Initialize count of leaves        q.add(node);        while (!q.isEmpty())        {            Node temp = q.peek();            q.poll();             if (temp.left != null)            {                q.add(temp.left);            }            if (temp.right != null)            {                q.add(temp.right);            }            if (temp.left == null &&                temp.right == null)            {                count++;            }        }        return count;    }     /* Helper function that allocates    a new Node with the given data    and null left and right pointers. */    static Node newNode(int data)    {        Node node = new Node();        node.data = data;        node.left = node.right = null;        return (node);    }     // Driver Code    public static void main(String[] args)    {        {            /*   1                / \               2   3              / \             4   5            Let us create Binary Tree shown in            above example */            Node root = newNode(1);            root.left = newNode(2);            root.right = newNode(3);            root.left.left = newNode(4);            root.left.right = newNode(5);             /* get leaf count of the above created tree */            System.out.println(getLeafCount(root));        }    }} // This code is contributed by Rajput-Ji

Python3

 # Python3 program to count leaf nodes# in a Binary Treefrom queue import Queue # Helper function that allocates a new# Node with the given data and None# left and right pointers.class newNode:    def __init__(self, data):        self.data = data        self.left = self.right = None         # Function to get the count of leaf# Nodes in a binary treedef getLeafCount(node):         # If tree is empty    if (not node):        return 0     # Initialize empty queue.    q = Queue()     # Do level order traversal    # starting from root    count = 0 # Initialize count of leaves    q.put(node)    while (not q.empty()):        temp = q.queue        q.get()         if (temp.left != None):            q.put(temp.left)        if (temp.right != None):            q.put(temp.right)        if (temp.left == None and            temp.right == None):            count += 1    return count # Driver Codeif __name__ == '__main__':         # 1    # / \    # 2 3    # / \    # 4 5    # Let us create Binary Tree shown    # in above example    root = newNode(1)    root.left = newNode(2)    root.right = newNode(3)    root.left.left = newNode(4)    root.left.right = newNode(5)     # get leaf count of the above    # created tree    print(getLeafCount(root)) # This code is contributed by PranchalK

C#

 // C# program to count leaf nodes// in a Binary Treeusing System;using System.Collections.Generic;     class GFG{    /* A binary tree Node has data,    pointer to left child and    a pointer to right child */    public class Node    {        public int data;        public Node left, right;    }     /* Function to get the count of    leaf Nodes in a binary tree*/    static int getLeafCount(Node node)    {        // If tree is empty        if (node == null)        {            return 0;        }         // Initialize empty queue.        Queue q = new Queue();         // Do level order traversal starting from root        int count = 0; // Initialize count of leaves        q.Enqueue(node);        while (q.Count!=0)        {            Node temp = q.Peek();            q.Dequeue();             if (temp.left != null)            {                q.Enqueue(temp.left);            }            if (temp.right != null)            {                q.Enqueue(temp.right);            }            if (temp.left == null &&                temp.right == null)            {                count++;            }        }        return count;    }     /* Helper function that allocates    a new Node with the given data    and null left and right pointers. */    static Node newNode(int data)    {        Node node = new Node();        node.data = data;        node.left = node.right = null;        return (node);    }     // Driver Code    public static void Main(String[] args)    {        {            /* 1                / \            2 3            / \            4 5            Let us create Binary Tree shown in            above example */            Node root = newNode(1);            root.left = newNode(2);            root.right = newNode(3);            root.left.left = newNode(4);            root.left.right = newNode(5);             /* get leaf count of the above created tree */            Console.WriteLine(getLeafCount(root));        }    }} // This code is contributed by 29AjayKumar

Javascript


Output
3

Time Complexity: O(n)

Here is a recursive solution for the same problem:

The idea is simple we break the larger tree into smaller sub-trees and solve for them to get the final answer.

Below is the implementation of the above approach.

C++

 // C++ Program for above approach#include using namespace std; // Node classstruct node{  int data;  struct node* left;  struct node* right;}; // Program to count leavesint countLeaves(struct node *node){   // If the node itself is "null"  // return 0, as there  // are no leaves  if(node == NULL)  {    return 0;  }   // It the node is a leaf then  // both right and left  // children will be "null"  if(node->left == NULL && node->right == NULL)  {    return 1;  }   // Now we count the leaves in  // the left and right  // subtrees and return the sum  return countLeaves(node->left) + countLeaves(node->right);} // Class newNode of Node typestruct node* newNode(int data){  struct node* node =    (struct node*)malloc(sizeof(struct node));  node->data = data;  node->left = NULL;  node->right = NULL;  return(node);} // Driver Codeint main(){  struct node *root  = newNode(1);  root->left = newNode(2);  root->right = newNode(3);  root->left->left = newNode(4);  root->left->right = newNode(5);   /* get leaf count of the above      created tree */  cout<

Java

 // Java Program for above approachimport java.util.LinkedList;import java.util.Queue;import java.io.*;import java.util.*; class GfG{     // Node class  static class Node  {    int data;    Node left, right;  }   // Program to count leaves  static int countLeaves(Node node)  {     // If the node itself is "null"    // return 0, as there    // are no leaves    if (node == null)      return 0;     // It the node is a leaf then    // both right and left    // children will be "null"    if (node.left == null &&                node.right == null)      return 1;     // Now we count the leaves in    // the left and right    // subtrees and return the sum    return countLeaves(node.left)             + countLeaves(node.right);  }    // Class newNode of Node type  static Node newNode(int data)  {    Node node = new Node();    node.data = data;    node.left = node.right = null;    return (node);  }   // Driver Code  public static void main(String[] args)  {          Node root = newNode(1);      root.left = newNode(2);      root.right = newNode(3);      root.left.left = newNode(4);      root.left.right = newNode(5);       /* get leaf count of the above      created tree */      System.out.println(countLeaves(root));  }}//Code by Rounik Prashar

Python3

 # Python3 Program for above approach # Node classclass Node:    def __init__(self,x):        self.data = x        self.left = None        self.right = None # Program to count leavesdef countLeaves(node):     # If the node itself is "None"    # return 0, as there    # are no leaves    if (node == None):        return 0     # It the node is a leaf then    # both right and left    # children will be "None"    if (node.left == None and node.right == None):        return 1     # Now we count the leaves in    # the left and right    # subtrees and return the sum    return countLeaves(node.left) + countLeaves(node.right) # Driver Codeif __name__ == '__main__':     root = Node(1)    root.left = Node(2)    root.right = Node(3)    root.left.left = Node(4)    root.left.right = Node(5)       # /* get leaf count of the above      # created tree */    print(countLeaves(root))     # This code is contributed by mohit kumar 29

C#

 // C# Program for above approachusing System; // Node classpublic class Node{    public int data;    public Node left, right;       public Node(int d)    {        data = d;        left = right = null;    }} public class BinaryTree{     public static Node root; // Program to count leavesstatic int countLeaves(Node node){         // If the node itself is "null"    // return 0, as there    // are no leaves    if (node == null)    {        return 0;    }         // It the node is a leaf then    // both right and left    // children will be "null"    if (node.left == null &&        node.right == null)    {        return 1;    }         // Now we count the leaves in    // the left and right    // subtrees and return the sum    return countLeaves(node.left) +           countLeaves(node.right);} // Driver Codestatic public void Main(){    BinaryTree.root = new Node(1);    BinaryTree.root.left = new Node(2);    BinaryTree.root.right = new Node(3);    BinaryTree.root.left.left = new Node(4);    BinaryTree.root.left.right = new Node(5);         // Get leaf count of the above created tree    Console.WriteLine(countLeaves(root));}} // This code is contributed by avanitrachhadiya2155

Javascript


Output
3

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