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

Last Updated : 23 Feb, 2023
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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.

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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
        // they 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
Python3 
# 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
        // they 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: O(n) where n is number of nodes in given binary tree.
Auxiliary Space: O(n) for call stack since using recursion

Iterative Approach:

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

  1. Create a queue to store nodes
  2. Store the root of the tree in the queue
  3. Traverse until the queue is not empty
    1. If the current node is not a leaf insert root->left and root->right in the queue.
    2. If the current node is NULL return false.
  4. If the queue is empty return true.

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

C++ 
// c++ program to check whether a given BT is full or not
#include <bits/stdc++.h>
using namespace std;

// Tree node structure
struct Node {
    int val;
    Node *left, *right;
};

// fun that creates and returns a new node
Node* newNode(int data)
{
    Node* node = new Node();
    node->val = data;
    node->left = node->right = NULL;
    return node;
}

// helper fun to check leafnode
bool isleafnode(Node* root)
{
    return !root->left && !root->right;
}

// fun checks whether the given BT is a full BT or not
bool isFullTree(Node* root)
{
    // if tree is empty
    if (!root)
        return true;

    queue<Node*> q;
    q.push(root);

    while (!q.empty()) {

        root = q.front();
        q.pop();

        // null indicates - not a full BT
        if (root == NULL)
            return false;

        // if its not a leafnode then the current node
        // should contain both left and right pointers.
        if (!isleafnode(root)) {
            q.push(root->left);
            q.push(root->right);
        }
    }

    return true;
}

int main()
{
    Node* root = newNode(1);
    root->left = newNode(2);
    root->right = newNode(3);
    root->left->left = newNode(4);
    root->left->right = newNode(5);

    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 Modem Upendra.
Java 
// Java program to check whether a given BT is full or not
import java.util.ArrayDeque;
import java.util.Queue;

public class GFG 
{

  /*  Tree node structure */
  static class Node {
    int data;
    Node left, right;

    Node(int item)
    {
      data = item;
      left = right = null;
    }
  }

  // helper fun to check leafnode
  static boolean isleafnode(Node root)
  {
    return root.left == null && root.right == null;
  }

  // fun checks whether the given BT is a full BT or not
  static boolean isFullTree(Node root)
  {

    // if tree is empty
    if (root == null)
      return true;

    Queue<Node> q = new ArrayDeque<>();
    q.add(root);

    while (!q.isEmpty()) {

      root = q.peek();
      q.remove();

      // null indicates - not a full BT
      if (root == null)
        return false;

      // if its not a leafnode then the current node
      // should contain both left and right pointers.
      if (!isleafnode(root)) {
        q.add(root.left);
        q.add(root.right);
      }
    }

    return true;
  }

  // Driver Code
  public static void main(String[] args)
  {
    Node root = new Node(1);
    root.left = new Node(2);
    root.right = new Node(3);
    root.left.left = new Node(4);
    root.left.right = new Node(5);

    if (isFullTree(root))
      System.out.println("The Binary Tree is full");
    else
      System.out.println(
      "The Binary Tree is not full");
  }
}

// This code is contributed by karandeep1234
Python3 
# Python program to check whether a given BT is full or not
# Tree Structure
class Node:
    def __init__(self, key):
        self.data = key
        self.left = None
        self.right = None

# function that creates and returns a new node
def newNode(data):
    node = Node(data)
    return node

# helper function to check leafnode
def isleafnode(root):
    return root.left is not None and root.right is not None

# function checks whether the given BT is a full BT or not
def isFullTree(root):
    # if tree is empty
    if root is None:
        return True

    q = []
    q.append(root)

    while(len(q) > 0):
        root = q.pop(0)

        # null indicates - not a full BT
        if root is None:
            return False

        # if its not a leafnode then the current node
        # should contain both left and right pointers
        if isleafnode(root) is False:
            q.append(root.left)
            q.append(root.right)

    return True

# Driver program to test above function
root = newNode(1)
root.left = newNode(2)
root.right = newNode(3)
root.left.left = newNode(4)
root.left.right = newNode(5)

if isFullTree(root) is True:
    print("The Binary Tree is full")
else:
    print("The Binary Tree is not full")

# This code is contributed by Yash Agarwal(yashagarwal2852002)
C# 
// C# program to check whether a given BT is full or not

using System;
using System.Collections.Generic;

public class GFG {

    /*  Tree node structure */
    public class Node {
        public int data;
        public Node left, right;

        public Node(int item)
        {
            data = item;
            left = right = null;
        }
    }

    // helper fun to check leafnode
    static bool isleafnode(Node root)
    {
        return root.left == null && root.right == null;
    }

    // fun checks whether the given BT is a full BT or not
    static bool isFullTree(Node root)
    {

        // if tree is empty
        if (root == null)
            return true;

        Queue<Node> q = new Queue<Node>();
        q.Enqueue(root);

        while (q.Count != 0) {

            root = q.Dequeue();

            // null indicates - not a full BT
            if (root == null)
                return false;

            // if its not a leafnode then the current node
            // should contain both left and right pointers.
            if (!isleafnode(root)) {
                q.Enqueue(root.left);
                q.Enqueue(root.right);
            }
        }

        return true;
    }

    // Driver Code
    public static void Main(string[] args)
    {
        Node root = new Node(1);
        root.left = new Node(2);
        root.right = new Node(3);
        root.left.left = new Node(4);
        root.left.right = new Node(5);

        if (isFullTree(root))
            Console.WriteLine("The Binary Tree is full");
        else
            Console.WriteLine(
                "The Binary Tree is not full");
    }
}
// This code is contributed by karandeep1234.
JavaScript 
// JAVASCRIPT program to check whether a given BT is full or not
class Queue {
    constructor() {
        this.items = [];
    }
    
    // add element to the queue
    enqueue(element) {
        return this.items.push(element);
    }
    
    // remove element from the queue
    dequeue() {
        if(this.items.length > 0) {
            return this.items.shift();
        }
    }
    
    // view the last element
    peek() {
        return this.items[0];
    }
    
    // check if the queue is empty
    isEmpty(){
       return this.items.length == 0;
    }
   
    // the size of the queue
    size(){
        return this.items.length;
    }
 
    // empty the queue
    clear(){
        this.items = [];
    }
}
// Tree node structure
class Node {
    constructor(item) {
        this.data = item;
        this.left = this.right = null;
    }
}

// helper fun to check leafnode
function isleafnode(root)
{
    if(root.left==null && root.right==null)
        return true;
    return false;
    
}

// fun checks whether the given BT is a full BT or not
function isFullTree( root)
{
    // if tree is empty
    if (root==null)
        return true;
  let q = new Queue();
    q.enqueue(root)
    while (q.size()!=0) {

        root = q.peek();
        q.dequeue();
        // null indicates - not a full BT
        if (root == null)
            return false;
        
        // if its not a leafnode then the current node
        // should contain both left and right pointers.
        if (isleafnode(root)==false) {
            q.enqueue(root.left);
            q.enqueue(root.right);
        }
    }

    return true;
}

    let root = new Node(1);
    root.left = new Node(2);
    root.right = new Node(3);
    root.left.left = new Node(4);
    root.left.right = new Node(5);

    if (isFullTree(root)== true)
        console.log("The Binary Tree is full");
    else
        console.log("The Binary Tree is not full");

   // This code is contributed by garg28harsh.

Output
The Binary Tree is full

Time Complexity: O(N), Where N is the total nodes in a given binary tree.
Auxiliary Space: O(N), in most cases the last level contains nodes as half of the total nodes. O(N/2) ~ O(N)


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