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

  • Difficulty Level : Easy
  • Last Updated : 04 Oct, 2021
Geek Week

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

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 <bits/stdc++.h>
 
using namespace std;
  
/*  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 = new  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))
        cout << "The Binary Tree is full\n";
    else
        cout << "The Binary Tree is not full\n";
  
    return(0);
}
 
// This code is contributed by shubhamsingh10

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 binary 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 conditions 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

Javascript




<script>
// javascript program to check if binary tree is full or not
 
/*  Tree node structure */
class Node {
    constructor(item) {
        this.data = item;
        this.left = this.right = null;
    }
}
 
 
    var root;
 
    /* this function checks if a binary tree is full or not */
    function isFullTree( 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
     
     
        root = new Node(10);
        root.left = new Node(20);
        root.right = new Node(30);
        root.left.right = new Node(40);
        root.left.left = new Node(50);
        root.right.left = new Node(60);
        root.left.left.left = new Node(80);
        root.right.right = new Node(70);
        root.left.left.right = new Node(90);
        root.left.right.left = new Node(80);
        root.left.right.right = new Node(90);
        root.right.left.left = new Node(80);
        root.right.left.right = new Node(90);
        root.right.right.left = new Node(80);
        root.right.right.right = new Node(90);
         if(isFullTree(root))
            document.write("The binary tree is full");
        else
            document.write("The binary tree is not full");
             
// This code contributed by gauravrajput1
</script>
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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