Count Non-Leaf nodes in a Binary Tree

Given a Binary tree, count total number of non-leaf nodes in the tree

Examples:

Input :


Output :2
Explanation
In the above tree only two nodes 1 and 2 are non-leaf nodes

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

We recursively traverse the given tree. While traversing, we count non-leaf nodes in left and right subtrees and add 1 to the result.

C++

// CPP program to count total number of 
// non-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; 
    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 = new Node; 
    node->data = data; 
    node->left = node->right = NULL; 
    return (node); 
} 
  
/* Computes the number of non-leaf nodes in a tree. */
int countNonleaf(struct Node* root) 
{ 
    // Base cases. 
    if (root == NULL || (root->left == NULL &&  
                         root->right == NULL)) 
        return 0; 
  
    // If root is Not NULL and its one of its 
    // child is also not NULL 
    return 1 + countNonleaf(root->left) +  
               countNonleaf(root->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); 
    cout << countNonleaf(root); 
    return 0; 
} 

Java

// Java program to count total number of  
// non-leaf nodes in a binary tree  
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;  
    Node right;  
} 
  
/* 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 = null; 
    node.right = null;  
    return (node);  
}  
  
/* Computes the number of non-leaf nodes in a tree. */
static int countNonleaf(Node root)  
{  
    // Base cases.  
    if (root == null || (root.left == null &&  
                        root.right == null))  
        return 0;  
  
    // If root is Not NULL and its one of its  
    // child is also not NULL  
    return 1 + countNonleaf(root.left) +  
                countNonleaf(root.right);  
}  
  
/* Driver program to test size function*/
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);  
    System.out.println(countNonleaf(root));  
} 
}  

Python3

# Python3 program to count total number  
# of non-leaf nodes in a binary tree 
  
# class 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
  
# Computes the number of non-leaf  
# nodes in a tree.  
def countNonleaf(root): 
      
    # Base cases.  
    if (root == None or (root.left == None and 
                         root.right == None)):  
        return 0
  
    # If root is Not None and its one of   
    # its child is also not None  
    return (1 + countNonleaf(root.left) + 
                countNonleaf(root.right)) 
  
# Driver Code 
if __name__ == '__main__': 
  
    root = newNode(1)  
    root.left = newNode(2)  
    root.right = newNode(3)  
    root.left.left = newNode(4)  
    root.left.right = newNode(5)  
    print(countNonleaf(root)) 
  
# This code is contributed by PranchalK 

C#

// C# program to count total number of  
// non-leaf nodes in a binary tree 
using System; 
  
class GfG  
{  
  
    /* A binary tree node has data, pointer to  
    left child and a pointer to right child */
    class Node {  
        public int data;  
        public Node left;  
        public Node right;  
    } 
  
    /* 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 = null; 
        node.right = null;  
        return (node);  
    }  
  
    /* Computes the number of non-leaf nodes in a tree. */
    static int countNonleaf(Node root)  
    {  
        // Base cases.  
        if (root == null || (root.left == null &&  
                            root.right == null))  
            return 0;  
  
        // If root is Not NULL and its one of its  
        // child is also not NULL  
        return 1 + countNonleaf(root.left) +  
                    countNonleaf(root.right);  
    }  
  
    /* 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);  
        Console.WriteLine(countNonleaf(root));  
    } 
} 
  
// This code is contributed by PrinciRaj1992 

Output:

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My Personal Notes

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

C++

Java

Python3

C#

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