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:-
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 |
chevron_right
filter_none
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); } |
chevron_right
filter_none
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 |
chevron_right
filter_none
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) |
chevron_right
filter_none
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 |
chevron_right
filter_none
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
Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.
