Iterative program to count leaf nodes in a Binary Tree

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 <bits/stdc++.h> 
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<Node *> 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; 
} 

Python3

# Python3 program to count leaf nodes  
# in a Binary Tree  
from 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 tree 
def 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[0]  
        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 Code 
if __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 

Output:

Time Complexity: O(n)

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Iterative program to count leaf nodes in a Binary Tree

C++

Python3

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