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++

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// 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;
}

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Python3














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# 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 



















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Output:


 3



Time Complexity: O(n)







This article is contributed by Mr. Somesh Awasthi. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.


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