Check whether a binary tree is a complete tree or not | Set 2 (Recursive Solution)
Last Updated :
31 May, 2023
A complete binary tree is a binary tree whose all levels except the last level are completely filled and all the leaves in the last level are all to the left side. More information about complete binary trees can be found here.
Example:
Below tree is a Complete Binary Tree (All nodes till the second last nodes are filled and all leaves are to the left side)
An iterative solution for this problem is discussed in below post.
Check whether a given Binary Tree is Complete or not | Set 1 (Using Level Order Traversal)
In this post a recursive solution is discussed.
In the array representation of a binary tree, if the parent node is assigned an index of ‘i’ and left child gets assigned an index of ‘2*i + 1’ while the right child is assigned an index of ‘2*i + 2’. If we represent the above binary tree as an array with the respective indices assigned to the different nodes of the tree above from top to down and left to right.
Hence we proceed in the following manner in order to check if the binary tree is complete binary tree.
- Calculate the number of nodes (count) in the binary tree.
- Start recursion of the binary tree from the root node of the binary tree with index (i) being set as 0 and the number of nodes in the binary (count).
- If the current node under examination is NULL, then the tree is a complete binary tree. Return true.
- If index (i) of the current node is greater than or equal to the number of nodes in the binary tree (count) i.e. (i>= count), then the tree is not a complete binary. Return false.
- Recursively check the left and right sub-trees of the binary tree for same condition. For the left sub-tree use the index as (2*i + 1) while for the right sub-tree use the index as (2*i + 2).
The time complexity of the above algorithm is O(n). Following is the code for checking if a binary tree is a complete binary tree.
Implementation:
C++
/* C++ program to checks if a binary tree complete or not */
#include<bits/stdc++.h>
#include<stdbool.h>
using namespace std;
/* Tree node structure */
class Node
{
public:
int key;
Node *left, *right;
Node *newNode(char k)
{
Node *node = ( Node*)malloc(sizeof( Node));
node->key = k;
node->right = node->left = NULL;
return node;
}
};
/* Helper function that allocates a new node with the
given key and NULL left and right pointer. */
/* This function counts the number of nodes
in a binary tree */
unsigned int countNodes(Node* root)
{
if (root == NULL)
return (0);
return (1 + countNodes(root->left) +
countNodes(root->right));
}
/* This function checks if the binary tree
is complete or not */
bool isComplete ( Node* root, unsigned int index,
unsigned int number_nodes)
{
// An empty tree is complete
if (root == NULL)
return (true);
// If index assigned to current node is more than
// number of nodes in tree, then tree is not complete
if (index >= number_nodes)
return (false);
// Recur for left and right subtrees
return (isComplete(root->left, 2*index + 1, number_nodes) &&
isComplete(root->right, 2*index + 2, number_nodes));
}
// Driver code
int main()
{
Node n1;
// Let us create tree in the last diagram above
Node* root = NULL;
root = n1.newNode(1);
root->left = n1.newNode(2);
root->right = n1.newNode(3);
root->left->left = n1.newNode(4);
root->left->right = n1.newNode(5);
root->right->right = n1.newNode(6);
unsigned int node_count = countNodes(root);
unsigned int index = 0;
if (isComplete(root, index, node_count))
cout << "The Binary Tree is complete\n";
else
cout << "The Binary Tree is not complete\n";
return (0);
}
// This code is contributed by SoumikMondal
/* C program to checks if a binary tree complete 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 counts the number of nodes in a binary tree */
unsigned int countNodes(struct Node* root)
{
if (root == NULL)
return (0);
return (1 + countNodes(root->left) + countNodes(root->right));
}
/* This function checks if the binary tree is complete or not */
bool isComplete (struct Node* root, unsigned int index,
unsigned int number_nodes)
{
// An empty tree is complete
if (root == NULL)
return (true);
// If index assigned to current node is more than
// number of nodes in tree, then tree is not complete
if (index >= number_nodes)
return (false);
// Recur for left and right subtrees
return (isComplete(root->left, 2*index + 1, number_nodes) &&
isComplete(root->right, 2*index + 2, number_nodes));
}
// Driver program
int main()
{
// Le us create tree in the last diagram above
struct Node* root = NULL;
root = newNode(1);
root->left = newNode(2);
root->right = newNode(3);
root->left->left = newNode(4);
root->left->right = newNode(5);
root->right->right = newNode(6);
unsigned int node_count = countNodes(root);
unsigned int index = 0;
if (isComplete(root, index, node_count))
printf("The Binary Tree is complete\n");
else
printf("The Binary Tree is not complete\n");
return (0);
}
// Java program to check if binary tree is complete 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 counts the number of nodes in a binary tree */
int countNodes(Node root)
{
if (root == null)
return (0);
return (1 + countNodes(root.left) + countNodes(root.right));
}
/* This function checks if the binary tree is complete or not */
boolean isComplete(Node root, int index, int number_nodes)
{
// An empty tree is complete
if (root == null)
return true;
// If index assigned to current node is more than
// number of nodes in tree, then tree is not complete
if (index >= number_nodes)
return false;
// Recur for left and right subtrees
return (isComplete(root.left, 2 * index + 1, number_nodes)
&& isComplete(root.right, 2 * index + 2, number_nodes));
}
// Driver program
public static void main(String args[])
{
BinaryTree tree = new BinaryTree();
// Le us create tree in the last diagram above
Node NewRoot = null;
tree.root = new Node(1);
tree.root.left = new Node(2);
tree.root.right = new Node(3);
tree.root.left.right = new Node(5);
tree.root.left.left = new Node(4);
tree.root.right.right = new Node(6);
int node_count = tree.countNodes(tree.root);
int index = 0;
if (tree.isComplete(tree.root, index, node_count))
System.out.print("The binary tree is complete");
else
System.out.print("The binary tree is not complete");
}
}
// This code is contributed by Mayank Jaiswal
# Python program to check if a binary tree complete or not
# Tree node structure
class Node:
# Constructor to create a new node
def __init__(self, key):
self.key = key
self.left = None
self.right = None
# This function counts the number of nodes in a binary tree
def countNodes(root):
if root is None:
return 0
return (1+ countNodes(root.left) + countNodes(root.right))
# This function checks if binary tree is complete or not
def isComplete(root, index, number_nodes):
# An empty is complete
if root is None:
return True
# If index assigned to current nodes is more than
# number of nodes in tree, then tree is not complete
if index >= number_nodes :
return False
# Recur for left and right subtrees
return (isComplete(root.left , 2*index+1 , number_nodes)
and isComplete(root.right, 2*index+2, number_nodes)
)
# Driver Program
root = Node(1)
root.left = Node(2)
root.right = Node(3)
root.left.left = Node(4)
root.left.right = Node(5)
root.right.right = Node(6)
node_count = countNodes(root)
index = 0
if isComplete(root, index, node_count):
print ("The Binary Tree is complete")
else:
print ("The Binary Tree is not complete")
# This code is contributed by Nikhil Kumar Singh(nickzuck_007)
// C# program to check if binary
// tree is complete or not
using System;
/* Tree node structure */
class Node
{
public int data;
public Node left, right;
public Node(int item)
{
data = item;
left = right = null;
}
}
public class BinaryTree
{
Node root;
/* This function counts the number
of nodes in a binary tree */
int countNodes(Node root)
{
if (root == null)
return (0);
return (1 + countNodes(root.left) +
countNodes(root.right));
}
/* This function checks if the
binary tree is complete or not */
bool isComplete(Node root, int index,
int number_nodes)
{
// An empty tree is complete
if (root == null)
return true;
// If index assigned to current node is more than
// number of nodes in tree, then tree is not complete
if (index >= number_nodes)
return false;
// Recur for left and right subtrees
return (isComplete(root.left, 2 * index + 1, number_nodes)
&& isComplete(root.right, 2 * index + 2, number_nodes));
}
// Driver code
public static void Main()
{
BinaryTree tree = new BinaryTree();
// Let us create tree in the last diagram above
tree.root = new Node(1);
tree.root.left = new Node(2);
tree.root.right = new Node(3);
tree.root.left.right = new Node(5);
tree.root.left.left = new Node(4);
tree.root.right.right = new Node(6);
int node_count = tree.countNodes(tree.root);
int index = 0;
if (tree.isComplete(tree.root, index, node_count))
Console.WriteLine("The binary tree is complete");
else
Console.WriteLine("The binary tree is not complete");
}
}
/* This code is contributed by Rajput-Ji*/
<script>
// JavaScript program to check if
// binary tree is complete or not
/* Tree node structure */
class Node {
constructor(val) {
this.data = val;
this.left = null;
this.right = null;
}
}
var root;
/* This function counts the number of
nodes in a binary tree */
function countNodes(root) {
if (root == null)
return (0);
return (1 + countNodes(root.left) + countNodes(root.right));
}
/* This function checks if the binary tree is complete or not */
function isComplete(root , index , number_nodes) {
// An empty tree is complete
if (root == null)
return true;
// If index assigned to current node is more than
// number of nodes in tree, then tree is not complete
if (index >= number_nodes)
return false;
// Recur for left and right subtrees
return (isComplete(root.left, 2 * index + 1, number_nodes)
&& isComplete(root.right, 2 * index + 2, number_nodes));
}
// Driver program
// Le us create tree in the last diagram above
var NewRoot = null;
root = new Node(1);
root.left = new Node(2);
root.right = new Node(3);
root.left.right = new Node(5);
root.left.left = new Node(4);
root.right.right = new Node(6);
var node_count = countNodes(root);
var index = 0;
if (isComplete(root, index, node_count))
document.write("The binary tree is complete");
else
document.write("The binary tree is not complete");
// This code contributed by umadevi9616
</script>
OutputThe Binary Tree is not complete
Time Complexity: O(N) where N is the number of nodes in the tree.
Space Complexity: O(h) where h is the height of given tree due to recursion call.
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