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 :
To check whether a binary tree is a full binary tree we need to test the following cases:-
- If a binary tree node is NULL then it is a full binary tree.
- If a binary tree node does have empty left and right sub-trees, then it is a full binary tree by definition.
- 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.
- 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++ 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 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 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 |
# 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# 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 |
<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> |
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:-
- Create a queue to store nodes
- Store the root of the tree in the queue
- Traverse until the queue is not empty
- If the current node is not a leaf insert root->left and root->right in the queue.
- If the current node is NULL return false.
- If the queue is empty return true.
Following is the implementation for checking if a binary tree is a full binary tree.
// 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 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 |
# 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# 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 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.
|
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)