Check whether a binary tree is a full binary tree or not

A full binary tree is defined as a binary tree in which all nodes have either zero or two child nodes. Conversely, there is no node in a full binary tree, which has one child node. More information about full binary trees can be found here.

For Example :
Full

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

To check whether a binary tree is a full binary tree we need to test the following cases:-

1) If a binary tree node is NULL then it is a full binary tree.
2) If a binary tree node does have empty left and right sub-trees, then it is a full binary tree by definition.
3) If a binary tree node has left and right sub-trees, then it is a part of a full binary tree by definition. In this case recursively check if the left and right sub-trees are also binary trees themselves.
4) In all other combinations of right and left sub-trees, the binary tree is not a full binary tree.

Following is the implementation for checking if a binary tree is a full binary tree.

C

// C program to check whether a given Binary Tree is full or not 
#include<stdio.h> 
#include<stdlib.h> 
#include<stdbool.h> 
  
/*  Tree node structure */
struct Node 
{ 
    int key; 
    struct Node *left, *right; 
}; 
  
/* Helper function that allocates a new node with the 
   given key and NULL left and right pointer. */
struct Node *newNode(char k) 
{ 
    struct Node *node = (struct Node*)malloc(sizeof(struct Node)); 
    node->key = k; 
    node->right = node->left = NULL; 
    return node; 
} 
  
/* This function tests if a binary tree is a full binary tree. */
bool isFullTree (struct Node* root) 
{ 
    // If empty tree 
    if (root == NULL) 
        return true; 
  
    // If leaf node 
    if (root->left == NULL && root->right == NULL) 
        return true; 
  
    // If both left and right are not NULL, and left & right subtrees 
    // are full 
    if ((root->left) && (root->right)) 
        return (isFullTree(root->left) && isFullTree(root->right)); 
  
    // We reach here when none of the above if conditions work 
    return false; 
} 
  
// Driver Program 
int main() 
{ 
    struct Node* root = NULL; 
    root = newNode(10); 
    root->left = newNode(20); 
    root->right = newNode(30); 
  
    root->left->right = newNode(40); 
    root->left->left = newNode(50); 
    root->right->left = newNode(60); 
    root->right->right = newNode(70); 
  
    root->left->left->left = newNode(80); 
    root->left->left->right = newNode(90); 
    root->left->right->left = newNode(80); 
    root->left->right->right = newNode(90); 
    root->right->left->left = newNode(80); 
    root->right->left->right = newNode(90); 
    root->right->right->left = newNode(80); 
    root->right->right->right = newNode(90); 
  
    if (isFullTree(root)) 
        printf("The Binary Tree is full\n"); 
    else
        printf("The Binary Tree is not full\n"); 
  
    return(0); 
} 

Java

// Java program to check if binay tree is full or not 
  
/*  Tree node structure */
class Node  
{ 
    int data; 
    Node left, right; 
   
    Node(int item)  
    { 
        data = item; 
        left = right = null; 
    } 
} 
   
class BinaryTree  
{ 
    Node root; 
       
    /* this function checks if a binary tree is full or not */
    boolean isFullTree(Node node) 
    { 
        // if empty tree 
        if(node == null) 
        return true; 
           
        // if leaf node 
        if(node.left == null && node.right == null ) 
            return true; 
           
        // if both left and right subtrees are not null 
        // the are full 
        if((node.left!=null) && (node.right!=null)) 
            return (isFullTree(node.left) && isFullTree(node.right)); 
           
        // if none work 
        return false; 
    } 
   
       
    // Driver program 
    public static void main(String args[])  
    { 
        BinaryTree tree = new BinaryTree(); 
        tree.root = new Node(10); 
        tree.root.left = new Node(20); 
        tree.root.right = new Node(30); 
        tree.root.left.right = new Node(40); 
        tree.root.left.left = new Node(50); 
        tree.root.right.left = new Node(60); 
        tree.root.left.left.left = new Node(80); 
        tree.root.right.right = new Node(70); 
        tree.root.left.left.right = new Node(90); 
        tree.root.left.right.left = new Node(80); 
        tree.root.left.right.right = new Node(90); 
        tree.root.right.left.left = new Node(80); 
        tree.root.right.left.right = new Node(90); 
        tree.root.right.right.left = new Node(80); 
        tree.root.right.right.right = new Node(90); 
           
        if(tree.isFullTree(tree.root)) 
            System.out.print("The binary tree is full"); 
        else
            System.out.print("The binary tree is not full");  
    } 
} 
   
// This code is contributed by Mayank Jaiswal  

Python

# Python program to check whether given Binary tree is full or not 
  
# Tree node structure 
class Node: 
  
    # Constructor of the node class for creating the node 
    def __init__(self , key): 
        self.key = key 
        self.left = None
        self.right = None
  
# Checks if the binary tree is full or not 
def isFullTree(root): 
  
    # If empty tree 
    if root is None:     
        return True
      
    # If leaf node 
    if root.left is None and root.right is None: 
        return True
  
    # If both left and right subtress are not None and 
    # left and right subtress are full 
    if root.left is not None and root.right is not None: 
        return (isFullTree(root.left) and isFullTree(root.right)) 
      
    # We reach here when none of the above if condiitions work 
    return False
  
# Driver Program 
root = Node(10); 
root.left = Node(20); 
root.right = Node(30); 
  
root.left.right = Node(40); 
root.left.left = Node(50); 
root.right.left = Node(60); 
root.right.right = Node(70); 
  
root.left.left.left = Node(80); 
root.left.left.right = Node(90); 
root.left.right.left = Node(80); 
root.left.right.right = Node(90); 
root.right.left.left = Node(80); 
root.right.left.right = Node(90); 
root.right.right.left = Node(80); 
root.right.right.right = Node(90); 
  
if isFullTree(root): 
    print "The Binary tree is full"
else: 
    print "Binary tree is not full"
  
# This code is contributed by Nikhil Kumar Singh(nickzuck_007) 

C#

// C# program to check if binary tree  
// is full or not  
using System; 
  
/* Tree node structure */
public class Node 
{ 
    public int data; 
    public Node left, right; 
  
    public Node(int item) 
    { 
        data = item; 
        left = right = null; 
    } 
} 
  
class GFG 
{ 
public Node root; 
  
/* This function checks if a binary 
tree is full or not */
public virtual bool isFullTree(Node node) 
{ 
    // if empty tree  
    if (node == null) 
    { 
    return true; 
    } 
  
    // if leaf node  
    if (node.left == null && node.right == null) 
    { 
        return true; 
    } 
  
    // if both left and right subtrees  
    // are not null they are full  
    if ((node.left != null) && (node.right != null)) 
    { 
        return (isFullTree(node.left) &&  
                isFullTree(node.right)); 
    } 
  
    // if none work  
    return false; 
} 
  
// Driver Code  
public static void Main(string[] args) 
{ 
    GFG tree = new GFG(); 
    tree.root = new Node(10); 
    tree.root.left = new Node(20); 
    tree.root.right = new Node(30); 
    tree.root.left.right = new Node(40); 
    tree.root.left.left = new Node(50); 
    tree.root.right.left = new Node(60); 
    tree.root.left.left.left = new Node(80); 
    tree.root.right.right = new Node(70); 
    tree.root.left.left.right = new Node(90); 
    tree.root.left.right.left = new Node(80); 
    tree.root.left.right.right = new Node(90); 
    tree.root.right.left.left = new Node(80); 
    tree.root.right.left.right = new Node(90); 
    tree.root.right.right.left = new Node(80); 
    tree.root.right.right.right = new Node(90); 
  
    if (tree.isFullTree(tree.root)) 
    { 
        Console.Write("The binary tree is full"); 
    } 
    else
    { 
        Console.Write("The binary tree is not full"); 
    } 
} 
} 
  
// This code is contributed by Shrikant13 

Output:

The Binary Tree is full

Time complexity of the above code is O(n) where n is number of nodes in given binary tree.

This article is contributed by Gaurav Gupta. Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above

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

C

Java

Python

C#

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