Count full nodes in a Binary tree (Iterative and Recursive)

Given A binary Tree, how do you count all the full nodes (Nodes which have both children as not NULL) without using recursion and with recursion? Note leaves should not be touched as they have both children as NULL.


Nodes 2 and 6 are full nodes has both child’s. So count of full nodes in the above tree is 2

Method: Iterative
The idea is to use level-order traversal to solve this problem efficiently.

1) Create an empty Queue Node and push root node to Queue.
2) Do following while nodeQeue is not empty.
   a) Pop an item from Queue and process it.
      a.1) If it is full node then increment count++.
   b) Push left child of popped item to Queue, if available.
   c) Push right child of popped item to Queue, if available.

Below is the implementation of this idea.

C++

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// C++ program to count 
// full 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 full Nodes in
// a binary tree
unsigned int getfullCount(struct Node* node)
{
    // If tree is empty
    if (!node)
        return 0;
    queue<Node *> q;
      
    // Do level order traversal starting from root
    int count = 0; // Initialize count of full nodes
    q.push(node);
    while (!q.empty())
    {
        struct Node *temp = q.front();
        q.pop();
  
        if (temp->left && temp->right)
            count++;
              
        if (temp->left != NULL)
            q.push(temp->left);
        if (temp->right != NULL)
            q.push(temp->right);
    }
    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 
int main(void)
{
    /* 2
     / \
    7     5
    \     \
     6     9
    / \ /
    1 11 4
    Let us create Binary Tree as shown 
    */
      
    struct Node *root = newNode(2);
    root->left     = newNode(7);
    root->right     = newNode(5);
    root->left->right = newNode(6);
    root->left->right->left = newNode(1);
    root->left->right->right = newNode(11);
    root->right->right = newNode(9);
    root->right->right->left = newNode(4);
      
    cout << getfullCount(root);
      
    return 0;
}

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Java

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// Java program to count 
// full nodes in a Binary Tree 
// using Iterative approach
import java.util.Queue;
import java.util.LinkedList;
  
// Class to represent Tree node 
class Node 
{
    int data;
    Node left, right;
  
    public Node(int item) 
    {
        data = item;
        left = null;
        right = null;
    }
}
  
// Class to count full nodes of Tree 
class BinaryTree 
{
  
    Node root;
  
    /* Function to get the count of full Nodes in
    a binary tree*/
    int getfullCount() 
    {
        // If tree is empty
        if (root==null)
        return 0
          
        // Initialize empty queue.
        Queue<Node> queue = new LinkedList<Node>();
          
        // Do level order traversal starting from root
        queue.add(root);
          
        int count=0; // Initialize count of full nodes
        while (!queue.isEmpty()) 
        {
  
            Node temp = queue.poll();
            if (temp.left!=null && temp.right!=null)
            count++;
  
            // Enqueue left child 
            if (temp.left != null
            {
                queue.add(temp.left);
            }
  
            // Enqueue right child 
            if (temp.right != null
            {
                queue.add(temp.right);
            }
        }
        return count;
    }
  
    public static void main(String args[]) 
    {
        /* 2
          / \
        7     5
        \     \
        6     9
        / \ /
        1 11 4
        Let us create Binary Tree as shown
        */
        BinaryTree tree_level = new BinaryTree();
        tree_level.root = new Node(2);
        tree_level.root.left = new Node(7);
        tree_level.root.right = new Node(5);
        tree_level.root.left.right = new Node(6);
        tree_level.root.left.right.left = new Node(1);
        tree_level.root.left.right.right = new Node(11);
        tree_level.root.right.right = new Node(9);
        tree_level.root.right.right.left = new Node(4);
  
        System.out.println(tree_level.getfullCount());
          
    }
}

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Python

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# Python program to count 
# full nodes in a Binary Tree
# using iterative approach
  
# A node structure
class Node:
    # A utility function to create a new node
    def __init__(self ,key):
        self.data = key
        self.left = None
        self.right = None
  
# Iterative Method to count full nodes of binary tree
def getfullCount(root):
    # Base Case
    if root is None:
        return 0
      
    # Create an empty queue for level order traversal
    queue = []
  
    # Enqueue Root and initialize count
    queue.append(root)
          
    count = 0 #initialize count for full nodes
    while(len(queue) > 0):
        node = queue.pop(0)
  
        # if it is full node then increment count
        if node.left is not None and node.right is not None:
            count = count+1
  
        # Enqueue left child
        if node.left is not None:
            queue.append(node.left)
  
        # Enqueue right child
        if node.right is not None:
            queue.append(node.right)
              
    return count
  
# Driver Program to test above function
root = Node(2)
root.left = Node(7)
root.right = Node(5)
root.left.right = Node(6)
root.left.right.left = Node(1)
root.left.right.right = Node(11)
root.right.right = Node(9)
root.right.right.left = Node(4)
  
  
print "%d" %(getfullCount(root))

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C#

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// C# program to count 
// full nodes in a Binary Tree 
// using Iterative approach 
using System; 
using System.Collections.Generic;
  
// Class to represent Tree node 
public class Node 
    public int data; 
    public Node left, right; 
  
    public Node(int item) 
    
        data = item; 
        left = null
        right = null
    
  
// Class to count full nodes of Tree 
public class BinaryTree 
  
    Node root; 
  
    /* Function to get the count of full Nodes in 
    a binary tree*/
    int getfullCount() 
    
        // If tree is empty 
        if (root == null
        return 0; 
          
        // Initialize empty queue. 
        Queue<Node> queue = new Queue<Node>(); 
          
        // Do level order traversal starting from root 
        queue.Enqueue(root); 
          
        int count = 0; // Initialize count of full nodes 
        while (queue.Count != 0) 
        
  
            Node temp = queue.Dequeue(); 
            if (temp.left != null && temp.right != null
            count++; 
  
            // Enqueue left child 
            if (temp.left != null
            
                queue.Enqueue(temp.left); 
            
  
            // Enqueue right child 
            if (temp.right != null
            
                queue.Enqueue(temp.right); 
            
        
        return count; 
    
  
    // Driver code
    public static void Main(String []args) 
    
        /* 2 
        / \ 
        7 5 
        \ \ 
        6 9 
        / \ / 
        1 11 4 
        Let us create Binary Tree as shown 
        */
        BinaryTree tree_level = new BinaryTree(); 
        tree_level.root = new Node(2); 
        tree_level.root.left = new Node(7); 
        tree_level.root.right = new Node(5); 
        tree_level.root.left.right = new Node(6); 
        tree_level.root.left.right.left = new Node(1); 
        tree_level.root.left.right.right = new Node(11); 
        tree_level.root.right.right = new Node(9); 
        tree_level.root.right.right.left = new Node(4); 
  
        Console.WriteLine(tree_level.getfullCount()); 
    
  
// This code has been contributed by 29AjayKumar

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Output:

 2

Time Complexity: O(n)
Auxiliary Space : O(n) where, n is number of nodes in given binary tree

Method: Recursive

The idea is to traverse the tree in postorder. If the current node is full, we increment result by 1 and add returned values of left and right subtrees.

C++

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// C++ program to count full 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 full Nodes in
// a binary tree
unsigned int getfullCount(struct Node* root)
{
    if (root == NULL)
       return 0;
   
    int res = 0;
    if  (root->left && root->right) 
       res++;
   
    res += (getfullCount(root->left) + 
            getfullCount(root->right));
    return res;
}
   
/* 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
int main(void)
{
    /* 2
     / \
    7    5
    \    \
     6   9
    / \ /
    1 11 4
    Let us create Binary Tree as shown
    */
   
    struct Node *root = newNode(2);
    root->left    = newNode(7);
    root->right   = newNode(5);
    root->left->right = newNode(6);
    root->left->right->left = newNode(1);
    root->left->right->right = newNode(11);
    root->right->right = newNode(9);
    root->right->right->left = newNode(4);
   
    cout << getfullCount(root);
   
    return 0;
}

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Java

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// Java program to count full nodes in a Binary Tree 
import 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 full Nodes in 
// a binary tree 
static int getfullCount(Node root) 
    if (root == null
    return 0
  
    int res = 0
    if (root.left != null && root.right != null
    res++; 
  
    res += (getfullCount(root.left) + getfullCount(root.right)); 
    return res; 
  
/* 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); 
  
// Driver program 
public static void main(String[] args) 
    /* 2 
    / \ 
    7 5 
    \ \ 
    6 9 
    / \ / 
    1 11 4 
    Let us create Binary Tree as shown 
    */
  
    Node root = newNode(2); 
    root.left = newNode(7); 
    root.right = newNode(5); 
    root.left.right = newNode(6); 
    root.left.right.left = newNode(1); 
    root.left.right.right = newNode(11); 
    root.right.right = newNode(9); 
    root.right.right.left = newNode(4); 
  
    System.out.println(getfullCount(root)); 
  
}

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Python3

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# Python program to count full 
# nodes in a Binary Tree
class newNode(): 
  
    def __init__(self, data): 
        self.data = data
        self.left = None
        self.right = None
          
          
# Function to get the count of  
# full Nodes in a binary tree 
def getfullCount(root):
  
    if (root == None):
        return 0
      
    res = 0
    if (root.left and root.right):
        res += 1
      
    res += (getfullCount(root.left) + 
            getfullCount(root.right)) 
    return res 
  
          
# Driver code 
if __name__ == '__main__':
    """ 2 
    / \ 
    7 5 
    \ \ 
    6 9 
    / \ / 
    1 11 4 
    Let us create Binary Tree as shown 
    """
      
    root = newNode(2
    root.left = newNode(7
    root.right = newNode(5
    root.left.right = newNode(6
    root.left.right.left = newNode(1
    root.left.right.right = newNode(11
    root.right.right = newNode(9
    root.right.right.left = newNode(4
      
    print(getfullCount(root))
  
# This code is contributed by SHUBHAMSINGH10

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C#

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// C# program to count full 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 
public class Node 
    public int data; 
    public Node left, right; 
}
  
// Function to get the count of full Nodes in 
// a binary tree 
static int getfullCount(Node root) 
    if (root == null
    return 0; 
  
    int res = 0; 
    if (root.left != null && root.right != null
    res++; 
  
    res += (getfullCount(root.left) + getfullCount(root.right)); 
    return res; 
  
/* 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); 
  
// Driver program 
public static void Main() 
    /* 2 
    / \ 
    7 5 
    \ \ 
    6 9 
    / \ / 
    1 11 4 
    Let us create Binary Tree as shown 
    */
  
    Node root = newNode(2); 
    root.left = newNode(7); 
    root.right = newNode(5); 
    root.left.right = newNode(6); 
    root.left.right.left = newNode(1); 
    root.left.right.right = newNode(11); 
    root.right.right = newNode(9); 
    root.right.right.left = newNode(4); 
  
    Console.WriteLine(getfullCount(root)); 
}
  
/* This code contributed by PrinciRaj1992 */

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Output:

 2

Time Complexity: O(n)
Auxiliary Space : O(n)
where, n is number of nodes in given binary tree

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