Check whether a given Binary Tree is Complete or not (Iterative Solution)
Last Updated :
23 Jul, 2025
Given a Binary Tree, the task is to check whether the given Binary Tree is a Complete Binary Tree or not.
A complete binary tree is a binary tree in which every level, except possibly the last, is completely filled, and all nodes are as far left as possible.
Examples:
The following trees are examples of Complete Binary Trees:
The following trees are examples of Non-Complete Binary Trees:
[Expected Approach - 1] Using Level-Order Traversal - O(n) Time and O(n) Space
The idea is to do a level order traversal starting from the root. In the traversal, once a node is found which is Not a Full Node, all the following nodes must be leaf nodes. A node is ‘Full Node’ if both left and right children are not empty (or not NULL).
Also, one more thing needs to be checked to handle the below case: If a node has an empty left child, then the right child must be empty.
Below is the implementation of the above approach:
C++
// C++ implementation to check if a
// Binary Tree is complete
#include <bits/stdc++.h>
using namespace std;
class Node {
public:
int data;
Node* left;
Node* right;
Node(int x) {
data = x;
left = nullptr;
right = nullptr;
}
};
// Function to check if the binary tree is complete
bool isCompleteBinaryTree(Node* root) {
if (root == nullptr)
return true;
queue<Node*> q;
q.push(root);
bool end = false;
while (!q.empty()) {
Node* current = q.front();
q.pop();
// Check left child
if (current->left) {
if (end)
return false;
q.push(current->left);
}
else {
// If left child is missing,
// mark the end
end = true;
}
// Check right child
if (current->right) {
if (end)
return false;
q.push(current->right);
}
else {
// If right child is missing,
// mark the end
end = true;
}
}
return true;
}
int main() {
// Representation of Input tree
// 1
// / \
// 2 3
// / \ /
// 4 5 6
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);
root->right->left = new Node(6);
if (isCompleteBinaryTree(root))
cout << "True" << endl;
else
cout << "False" << endl;
return 0;
}
// Java implementation to check if a
// Binary Tree is complete
import java.util.LinkedList;
import java.util.Queue;
class Node {
int data;
Node left;
Node right;
Node(int x) {
data = x;
left = null;
right = null;
}
}
class GfG {
// Function to check if the binary
// tree is complete
static boolean isCompleteBinaryTree(Node root) {
if (root == null)
return true;
Queue<Node> q = new LinkedList<>();
q.add(root);
boolean end = false;
while (!q.isEmpty()) {
Node current = q.poll();
// Check left child
if (current.left != null) {
if (end)
return false;
q.add(current.left);
}
else {
// If left child is missing,
// mark the end
end = true;
}
// Check right child
if (current.right != null) {
if (end)
return false;
q.add(current.right);
}
else {
// If right child is missing,
// mark the end
end = true;
}
}
return true;
}
public static void main(String[] args) {
// Representation of Input tree
// 1
// / \
// 2 3
// / \ /
// 4 5 6
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);
root.right.left = new Node(6);
if (isCompleteBinaryTree(root))
System.out.println("True");
else
System.out.println("False");
}
}
# Python implementation to check if a
# Binary Tree is complete
from collections import deque
class Node:
def __init__(self, x):
self.data = x
self.left = None
self.right = None
# Function to check if the binary
# tree is complete
def is_complete_binary_tree(root):
if root is None:
return True
q = deque([root])
end = False
while q:
current = q.popleft()
# Check left child
if current.left:
if end:
return False
q.append(current.left)
else:
# If left child is missing,
# mark the end
end = True
# Check right child
if current.right:
if end:
return False
q.append(current.right)
else:
# If right child is missing,
# mark the end
end = True
return True
if __name__ == "__main__":
# Representation of Input tree
# 1
# / \
# 2 3
# / \ /
# 4 5 6
root = Node(1)
root.left = Node(2)
root.right = Node(3)
root.left.left = Node(4)
root.left.right = Node(5)
root.right.left = Node(6)
if is_complete_binary_tree(root):
print("True")
else:
print("False")
// C# implementation to check if a
// Binary Tree is complete
using System;
using System.Collections.Generic;
class Node {
public int data;
public Node left;
public Node right;
public Node(int x) {
data = x;
left = null;
right = null;
}
}
// Function to check if the binary
// tree is complete
class GfG {
static bool isCompleteBinaryTree(Node root) {
if (root == null)
return true;
Queue<Node> q = new Queue<Node>();
q.Enqueue(root);
bool end = false;
while (q.Count > 0) {
Node current = q.Dequeue();
// Check left child
if (current.left != null) {
if (end)
return false;
q.Enqueue(current.left);
}
else {
// If left child is missing,
// mark the end
end = true;
}
// Check right child
if (current.right != null) {
if (end)
return false;
q.Enqueue(current.right);
}
else {
// If right child is missing,
// mark the end
end = true;
}
}
return true;
}
static void Main(string[] args) {
// Representation of Input tree
// 1
// / \
// 2 3
// / \ /
// 4 5 6
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);
root.right.left = new Node(6);
if (isCompleteBinaryTree(root))
Console.WriteLine("True");
else
Console.WriteLine("False");
}
}
// JavaScript implementation to check if a
// Binary Tree is complete
class Node {
constructor(x) {
this.data = x;
this.left = null;
this.right = null;
}
}
// Function to check if the binary
// tree is complete
function isCompleteBinaryTree(root) {
if (root === null)
return true;
const q = [];
q.push(root);
let end = false;
while (q.length > 0) {
const current = q.shift();
// Check left child
if (current.left) {
if (end)
return false;
q.push(current.left);
}
else {
// If left child is missing,
// mark the end
end = true;
}
// Check right child
if (current.right) {
if (end)
return false;
q.push(current.right);
}
else {
// If right child is missing,
// mark the end
end = true;
}
}
return true;
}
// Representation of Input tree
// 1
// / \
// 2 3
// / \ /
// 4 5 6
const 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);
root.right.left = new Node(6);
if (isCompleteBinaryTree(root))
console.log("True");
else
console.log("False");
[Expected Approach - 2] Checking position of NULL - O(n) Time and O(n) Space
A simple idea would be to check whether the NULL Node encountered is the last node of the Binary Tree. If the null node encountered in the binary tree is the last node then it is a complete binary tree and if there exists a valid node even after encountering a null node then the tree is not a complete binary tree.
Below is the implementation of the above approach:
C++
// C++ implementation to check if a binary
// tree is complete by position of null
#include <bits/stdc++.h>
using namespace std;
class Node {
public:
int data;
Node* left;
Node* right;
Node(int x) {
data = x;
left = nullptr;
right = nullptr;
}
};
// Function to check if the binary
// tree is complete
bool isCompleteBinaryTree(Node* root) {
if (root == nullptr) {
return true;
}
queue<Node*> q;
q.push(root);
bool nullEncountered = false;
while (!q.empty()) {
Node* curr = q.front();
q.pop();
if (curr == NULL) {
// If we have seen a NULL node, we
// set the flag to true
nullEncountered= true;
}
else {
// If that NULL node is not the last node then
// return false
if (nullEncountered == true) {
return false;
}
// Push both nodes even if
// there are null
q.push(curr->left);
q.push(curr->right);
}
}
return true;
}
int main() {
// Representation of Input tree
// 1
// / \
// 2 3
// / \ /
// 4 5 6
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);
root->right->left = new Node(6);
if (isCompleteBinaryTree(root)) {
cout << "True" << endl;
}
else {
cout << "False" << endl;
}
return 0;
}
// Java implementation to check if a binary
// tree is complete by position of null
import java.util.*;
class Node {
int data;
Node left;
Node right;
Node(int x) {
data = x;
left = null;
right = null;
}
}
class GfG {
// Function to check if the binary tree
// is complete
static boolean isCompleteBinaryTree(Node root) {
if (root == null) {
return true;
}
Queue<Node> q = new LinkedList<>();
q.add(root);
boolean nullEncountered = false;
while (!q.isEmpty()) {
Node curr = q.poll();
if (curr == null) {
// If we have seen a null node,
// we set the flag
nullEncountered = true;
}
else {
// If that null node is not the
// last node, return false
if (nullEncountered) {
return false;
}
// Push both nodes even if
// there are null
q.add(curr.left);
q.add(curr.right);
}
}
return true;
}
public static void main(String[] args) {
// Representation of Input tree
// 1
// / \
// 2 3
// / \ /
// 4 5 6
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);
root.right.left = new Node(6);
if (isCompleteBinaryTree(root)) {
System.out.println("True");
}
else {
System.out.println("False");
}
}
}
# Python implementation to check if a binary
# tree is complete by position of null
class Node:
def __init__(self, x):
self.data = x
self.left = None
self.right = None
# Function to check if the binary
# tree is complete
def isCompleteBinaryTree(root):
if root is None:
return True
queue = []
queue.append(root)
nullEncountered = False
while queue:
curr = queue.pop(0)
if curr is None:
# If we have seen a null node,
# set the flag
nullEncountered = True
else:
# If that null node is not the last
# node, return false
if nullEncountered:
return False
# Push both nodes even if
# there are null
queue.append(curr.left)
queue.append(curr.right)
return True
if __name__ == '__main__':
# Representation of Input tree
# 1
# / \
# 2 3
# / \ /
# 4 5 6
root = Node(1)
root.left = Node(2)
root.right = Node(3)
root.left.left = Node(4)
root.left.right = Node(5)
root.right.left = Node(6)
if isCompleteBinaryTree(root):
print("True")
else:
print("False")
// C# implementation to check if a binary
// tree is complete by position of null
using System;
using System.Collections.Generic;
class Node {
public int data;
public Node left;
public Node right;
public Node(int x) {
data = x;
left = null;
right = null;
}
}
class GfG {
// Function to check if the binary tree
// is complete
static bool isCompleteBinaryTree(Node root) {
if (root == null) {
return true;
}
Queue<Node> q = new Queue<Node>();
q.Enqueue(root);
bool nullEncountered = false;
while (q.Count > 0) {
Node curr = q.Dequeue();
if (curr == null) {
// If we have seen a null node,
// set the flag
nullEncountered = true;
} else {
// If that null node is not the
// last node, return false
if (nullEncountered) {
return false;
}
// Push both nodes even if there
// are null
q.Enqueue(curr.left);
q.Enqueue(curr.right);
}
}
return true;
}
static void Main() {
// Representation of Input tree
// 1
// / \
// 2 3
// / \ /
// 4 5 6
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);
root.right.left = new Node(6);
if (isCompleteBinaryTree(root)) {
Console.WriteLine("True");
}
else {
Console.WriteLine("False");
}
}
}
// JavaScript implementation to check if a binary
// tree is complete by position of null
class Node {
constructor(x) {
this.data = x;
this.left = null;
this.right = null;
}
}
// Function to check if the binary
// tree is complete
function isCompleteBinaryTree(root) {
if (root === null) {
return true;
}
const queue = [];
queue.push(root);
let nullEncountered = false;
while (queue.length > 0) {
const curr = queue.shift();
if (curr === null) {
// If we have seen a null node,
// set the flag
nullEncountered = true;
}
else {
// If that null node is not the last
// node, return false
if (nullEncountered) {
return false;
}
// Push both nodes even if
// there are null
queue.push(curr.left);
queue.push(curr.right);
}
}
return true;
}
// Representation of Input tree
// 1
// / \
// 2 3
// / \ /
// 4 5 6
const 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);
root.right.left = new Node(6);
if (isCompleteBinaryTree(root)) {
console.log("True");
}
else {
console.log("False");
}
Check whether a given Binary Tree is Complete or not | Set 1 (Iterative Solution)
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