Perfect Binary Tree Specific Level Order Traversal
Last Updated :
23 Jul, 2025
Given a Perfect Binary Tree like below:
Print the level order of nodes in following specific manner:
1 2 3 4 7 5 6 8 15 9 14 10 13 11 12 16 31 17 30 18 29 19 28 20 27 21 26 22 25 23 24
i.e. print nodes in level order but nodes should be from left and
right side alternatively. Here 1st and 2nd levels are trivial.
While 3rd level: 4(left), 7(right), 5(left), 6(right) are printed.
While 4th level: 8(left), 15(right), 9(left), 14(right), .. are printed.
While 5th level: 16(left), 31(right), 17(left), 30(right), .. are printed.
We strongly recommend to minimize your browser and try this yourself first.
In standard Level Order Traversal, we enqueue root into a queue 1st, then we dequeue ONE node from queue, process (print) it, enqueue its children into queue. We keep doing this until queue is empty.
Approach 1: We can do standard level order traversal here too but instead of printing nodes directly, we have to store nodes in current level in a temporary array or list 1st and then take nodes from alternate ends (left and right) and print nodes. Keep repeating this for all levels.
This approach takes more memory than standard traversal.
C++
// C++ program for special order traversal
#include <iostream>
#include <queue>
using namespace std;
// A binary tree node has data, pointer to left child
// and a pointer to right child
struct Node{
int data;
Node *left;
Node *right;
};
// Helper function that allocates a new node with the
// given data and NULL left and right pointers.
Node *newNode(int data){
Node *node = new Node;
node->data = data;
node->right = node->left = NULL;
return node;
}
// Given a perfect binary tree, print its nodes in specific
// level order
void printSpecificLevelOrder(Node *root){
if(root == NULL) return;
// declare queue for level order traversal
queue<Node*> q;
q.push(root);
vector<Node*> vec;
cout<<root->data<<" ";
while(!q.empty()){
int n = q.size();
for(int i = 0; i<n; i++){
Node* temp = q.front();
q.pop();
if(temp->left != NULL){
q.push(temp->left);
vec.push_back(temp->left);
}
if(temp->right != NULL){
q.push(temp->right);
vec.push_back(temp->right);
}
}
auto it1 = vec.begin();
auto it2 = vec.end()-1;
while(it1 < it2){
cout<<(*it1)->data<<" ";
cout<<(*it2)->data<<" ";
it1++;
it2--;
}
vec.clear();
}
}
// Driver program to test above functions
int main()
{
//Perfect Binary Tree of Height 4
Node *root = newNode(1);
root->left = newNode(2);
root->right = newNode(3);
root->left->left = newNode(4);
root->left->right = newNode(5);
root->right->left = newNode(6);
root->right->right = newNode(7);
root->left->left->left = newNode(8);
root->left->left->right = newNode(9);
root->left->right->left = newNode(10);
root->left->right->right = newNode(11);
root->right->left->left = newNode(12);
root->right->left->right = newNode(13);
root->right->right->left = newNode(14);
root->right->right->right = newNode(15);
root->left->left->left->left = newNode(16);
root->left->left->left->right = newNode(17);
root->left->left->right->left = newNode(18);
root->left->left->right->right = newNode(19);
root->left->right->left->left = newNode(20);
root->left->right->left->right = newNode(21);
root->left->right->right->left = newNode(22);
root->left->right->right->right = newNode(23);
root->right->left->left->left = newNode(24);
root->right->left->left->right = newNode(25);
root->right->left->right->left = newNode(26);
root->right->left->right->right = newNode(27);
root->right->right->left->left = newNode(28);
root->right->right->left->right = newNode(29);
root->right->right->right->left = newNode(30);
root->right->right->right->right = newNode(31);
cout << "Specific Level Order traversal of binary tree is \n";
printSpecificLevelOrder(root);
return 0;
}
// this code is contributed by Kirti Agarwal(kirtiagarwal23121999)
// Java program for the above approach
import java.util.*;
// A binary tree node has data, pointer to left child
// and a pointer to right child
class Node {
int data;
Node left, right;
// Helper function that allocates a new node with the
// given data and NULL left and right pointers.
public Node(int data) {
this.data = data;
this.left = null;
this.right = null;
}
}
class SpecialOrderTraversal {
// Given a perfect binary tree, print its nodes in specific
// level order
public static void printSpecificLevelOrder(Node root) {
if (root == null) return;
// declare queue for level order traversal
Queue<Node> q = new LinkedList<>();
Vector<Node> vec = new Vector<>();
System.out.print(root.data + " ");
q.add(root);
while (!q.isEmpty()) {
int n = q.size();
for (int i = 0; i < n; i++) {
Node temp = q.poll();
if (temp.left != null) {
q.add(temp.left);
vec.add(temp.left);
}
if (temp.right != null) {
q.add(temp.right);
vec.add(temp.right);
}
}
int size = vec.size();
for (int i = 0; i < size / 2; i++) {
System.out.print(vec.get(i).data + " ");
System.out.print(vec.get(size - 1 - i).data + " ");
}
if (size % 2 != 0) {
System.out.print(vec.get(size / 2).data + " ");
}
vec.clear();
}
}
// Driver program to test above functions
public static void main(String[] args) {
// Perfect Binary Tree of Height 4
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);
root.right.right = new Node(7);
root.left.left.left = new Node(8);
root.left.left.right = new Node(9);
root.left.right.left = new Node(10);
root.left.right.right = new Node(11);
root.right.left.left = new Node(12);
root.right.left.right = new Node(13);
root.right.right.left = new Node(14);
root.right.right.right = new Node(15);
root.left.left.left.left = new Node(16);
root.left.left.left.right = new Node(17);
root.left.left.right.left = new Node(18);
root.left.left.right.right = new Node(19);
root.left.right.left.left = new Node(20);
root.left.right.left.right = new Node(21);
root.left.right.right.left = new Node(22);
root.left.right.right.right = new Node(23);
root.right.left.left.left = new Node(24);
root.right.left.left.right = new Node(25);
root.right.left.right.left = new Node(26);
root.right.left.right.right = new Node(27);
root.right.right.left.left = new Node(28);
root.right.right.left.right = new Node(29);
root.right.right.right.left = new Node(30);
root.right.right.right.right = new Node(31);
System.out.println("Specific Level Order traversal of binary tree is:");
printSpecificLevelOrder(root);
}
}
// This code is contributed by codebraxnzt
from queue import Queue
# A binary tree node has data, pointer to left child
# and a pointer to right child
class Node:
def __init__(self, data):
self.data = data
self.left = None
self.right = None
# Given a perfect binary tree, print its nodes in specific
# level order
def printSpecificLevelOrder(root):
if root is None:
return
# declare queue for level order traversal
q = Queue()
vec = []
print(root.data, end=" ")
q.put(root)
while not q.empty():
n = q.qsize()
for i in range(n):
temp = q.get()
if temp.left is not None:
q.put(temp.left)
vec.append(temp.left)
if temp.right is not None:
q.put(temp.right)
vec.append(temp.right)
size = len(vec)
for i in range(size // 2):
print(vec[i].data, end=" ")
print(vec[size - 1 - i].data, end=" ")
if size % 2 != 0:
print(vec[size // 2].data, end=" ")
vec.clear()
# Driver program to test above functions
if __name__ == '__main__':
# Perfect Binary Tree of Height 4
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)
root.right.right = Node(7)
root.left.left.left = Node(8)
root.left.left.right = Node(9)
root.left.right.left = Node(10)
root.left.right.right = Node(11)
root.right.left.left = Node(12)
root.right.left.right = Node(13)
root.right.right.left = Node(14)
root.right.right.right = Node(15)
root.left.left.left.left = Node(16)
root.left.left.left.right = Node(17)
root.left.left.right.left = Node(18)
root.left.left.right.right = Node(19)
root.left.right.left.left = Node(20)
root.left.right.left.right = Node(21)
root.left.right.right.left = Node(22)
root.left.right.right.right = Node(23)
root.right.left.left.left = Node(24)
root.right.left.left.right = Node(25)
root.right.left.right.left = Node(26)
root.right.left.right.right = Node(27)
root.right.right.left.left = Node(28)
root.right.right.left.right = Node(29)
root.right.right.right.left = Node(30)
root.right.right.right.right = Node(31)
print("Specific Level Order traversal of binary tree is:")
printSpecificLevelOrder(root)
using System;
using System.Collections.Generic;
public class Node {
public int data;
public Node left;
public Node right;
public Node(int data) {
this.data = data;
this.left = null;
this.right = null;
}
}
public class BinaryTree {
public Node root;
public BinaryTree(Node node) {
root = node;
}
// Given a perfect binary tree, print its nodes in specific level order
public void PrintSpecificLevelOrder() {
if (root == null) return;
// declare queue for level order traversal
Queue<Node> q = new Queue<Node>();
q.Enqueue(root);
List<Node> vec = new List<Node>();
Console.Write(root.data + " ");
while (q.Count != 0) {
int n = q.Count;
for (int i = 0; i < n; i++) {
Node temp = q.Peek();
q.Dequeue();
if (temp.left != null) {
q.Enqueue(temp.left);
vec.Add(temp.left);
}
if (temp.right != null) {
q.Enqueue(temp.right);
vec.Add(temp.right);
}
}
int it1 = 0;
int it2 = vec.Count-1;
while (it1 < it2) {
Console.Write(vec[it1].data + " ");
Console.Write(vec[it2].data + " ");
it1++;
it2--;
}
vec.Clear();
}
}
static void Main(string[] args) {
// Perfect Binary Tree of Height 4
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);
root.right.right = new Node(7);
root.left.left.left = new Node(8);
root.left.left.right = new Node(9);
root.left.right.left = new Node(10);
root.left.right.right = new Node(11);
root.right.left.left = new Node(12);
root.right.left.right = new Node(13);
root.right.right.left = new Node(14);
root.right.right.right = new Node(15);
root.left.left.left.left = new Node(16);
root.left.left.left.right = new Node(17);
root.left.left.right.left = new Node(18);
root.left.left.right.right = new Node(19);
root.left.right.left.left = new Node(20);
root.left.right.left.right = new Node(21);
root.left.right.right.left = new Node(22);
root.left.right.right.right = new Node(23);
root.right.left.left.left = new Node(24);
root.right.left.left.right = new Node(25);
root.right.left.right.left = new Node(26);
root.right.left.right.right = new Node(27);
root.right.right.left.left = new Node(28);
root.right.right.left.right = new Node(29);
root.right.right.right.left = new Node(30);
root.right.right.right.right = new Node(31);
BinaryTree tree = new BinaryTree(root);
Console.WriteLine("Specific Level Order traversal of binary tree is:");
tree.PrintSpecificLevelOrder();
Console.ReadKey();
}
}
// THIS CODE IS CONTRIBUTED BY KIRTI AGARWAL(KIRTIAGARWAL23121999)
// JavaScript program for special order traversal
// A binary tree node has data, pointer to left child
// and a pointer to right child
class Node{
constructor(data){
this.data = data;
this.left = null;
this.right = null;
}
}
// Helper function that allocates a new node with the
// given data and NULL left and right pointers.
function newNode(data){
return new Node(data);
}
// Given a perfect binary tree, print its nodes in specific
// level order
function printSpecificLevelOrder(root){
if(root == null) return;
// declare queue for level order traversal
let q = [];
q.push(root);
let vec = [];
console.log(root.data + " ");
while(q.length > 0){
let n = q.length;
for(let i = 0; i<n; i++){
let temp = q.shift();
if(temp.left != null){
q.push(temp.left);
vec.push(temp.left);
}
if(temp.right != null){
q.push(temp.right);
vec.push(temp.right);
}
}
let it1 = 0;
let it2 = vec.length-1;
while(it1 < it2){
console.log(vec[it1].data + " ") ;
console.log(vec[it2].data + " ") ;
it1++;
it2--;
}
vec = [];
}
}
// driver program to test above functions
let root = newNode(1);
root.left = newNode(2);
root.right = newNode(3);
root.left.left = newNode(4);
root.left.right = newNode(5);
root.right.left = newNode(6);
root.right.right = newNode(7);
root.left.left.left = newNode(8);
root.left.left.right = newNode(9);
root.left.right.left = newNode(10);
root.left.right.right = newNode(11);
root.right.left.left = newNode(12);
root.right.left.right = newNode(13);
root.right.right.left = newNode(14);
root.right.right.right = newNode(15);
root.left.left.left.left = newNode(16);
root.left.left.left.right = newNode(17);
root.left.left.right.left = newNode(18);
root.left.left.right.right = newNode(19);
root.left.right.left.left = newNode(20);
root.left.right.left.right = newNode(21);
root.left.right.right.left = newNode(22);
root.left.right.right.right = newNode(23);
root.right.left.left.left = newNode(24);
root.right.left.left.right = newNode(25);
root.right.left.right.left = newNode(26);
root.right.left.right.right = newNode(27);
root.right.right.left.left = newNode(28);
root.right.right.left.right = newNode(29);
root.right.right.right.left = newNode(30);
root.right.right.right.right = newNode(31);
console.log("Specific Level Order Traversal of binary tree is : ");
printSpecificLevelOrder(root);
OutputSpecific Level Order traversal of binary tree is
1 2 3 4 7 5 6 8 15 9 14 10 13 11 12 16 31 17 30 18 29 19 28 20 27 21 26 22 25 23 24
Time Complexity: O(N) where N is the number of nodes in given binary tree.
Auxiliary Space: O(N) due to queue and vector data structure.
Approach 2: The standard level order traversal idea will slightly change here. Instead of processing ONE node at a time, we will process TWO nodes at a time. And while pushing children into queue, the enqueue order will be: 1st node's left child, 2nd node's right child, 1st node's right child and 2nd node's left child.
Implementation:
C++
/* C++ program for special order traversal */
#include <iostream>
#include <queue>
using namespace std;
/* A binary tree node has data, pointer to left child
and a pointer to right child */
struct Node
{
int data;
Node *left;
Node *right;
};
/* Helper function that allocates a new node with the
given data and NULL left and right pointers. */
Node *newNode(int data)
{
Node *node = new Node;
node->data = data;
node->right = node->left = NULL;
return node;
}
/* Given a perfect binary tree, print its nodes in specific
level order */
void printSpecificLevelOrder(Node *root)
{
if (root == NULL)
return;
// Let us print root and next level first
cout << root->data;
// / Since it is perfect Binary Tree, right is not checked
if (root->left != NULL)
cout << " " << root->left->data << " " << root->right->data;
// Do anything more if there are nodes at next level in
// given perfect Binary Tree
if (root->left->left == NULL)
return;
// Create a queue and enqueue left and right children of root
queue <Node *> q;
q.push(root->left);
q.push(root->right);
// We process two nodes at a time, so we need two variables
// to store two front items of queue
Node *first = NULL, *second = NULL;
// traversal loop
while (!q.empty())
{
// Pop two items from queue
first = q.front();
q.pop();
second = q.front();
q.pop();
// Print children of first and second in reverse order
cout << " " << first->left->data << " " << second->right->data;
cout << " " << first->right->data << " " << second->left->data;
// If first and second have grandchildren, enqueue them
// in reverse order
if (first->left->left != NULL)
{
q.push(first->left);
q.push(second->right);
q.push(first->right);
q.push(second->left);
}
}
}
/* Driver program to test above functions*/
int main()
{
//Perfect Binary Tree of Height 4
Node *root = newNode(1);
root->left = newNode(2);
root->right = newNode(3);
root->left->left = newNode(4);
root->left->right = newNode(5);
root->right->left = newNode(6);
root->right->right = newNode(7);
root->left->left->left = newNode(8);
root->left->left->right = newNode(9);
root->left->right->left = newNode(10);
root->left->right->right = newNode(11);
root->right->left->left = newNode(12);
root->right->left->right = newNode(13);
root->right->right->left = newNode(14);
root->right->right->right = newNode(15);
root->left->left->left->left = newNode(16);
root->left->left->left->right = newNode(17);
root->left->left->right->left = newNode(18);
root->left->left->right->right = newNode(19);
root->left->right->left->left = newNode(20);
root->left->right->left->right = newNode(21);
root->left->right->right->left = newNode(22);
root->left->right->right->right = newNode(23);
root->right->left->left->left = newNode(24);
root->right->left->left->right = newNode(25);
root->right->left->right->left = newNode(26);
root->right->left->right->right = newNode(27);
root->right->right->left->left = newNode(28);
root->right->right->left->right = newNode(29);
root->right->right->right->left = newNode(30);
root->right->right->right->right = newNode(31);
cout << "Specific Level Order traversal of binary tree is \n";
printSpecificLevelOrder(root);
return 0;
}
// Java program for special level order traversal
import java.util.LinkedList;
import java.util.Queue;
/* Class containing left and right child of current
node and key value*/
class Node
{
int data;
Node left, right;
public Node(int item)
{
data = item;
left = right = null;
}
}
class BinaryTree
{
Node root;
/* Given a perfect binary tree, print its nodes in specific
level order */
void printSpecificLevelOrder(Node node)
{
if (node == null)
return;
// Let us print root and next level first
System.out.print(node.data);
// Since it is perfect Binary Tree, right is not checked
if (node.left != null)
System.out.print(" " + node.left.data + " " + node.right.data);
// Do anything more if there are nodes at next level in
// given perfect Binary Tree
if (node.left.left == null)
return;
// Create a queue and enqueue left and right children of root
Queue<Node> q = new LinkedList<Node>();
q.add(node.left);
q.add(node.right);
// We process two nodes at a time, so we need two variables
// to store two front items of queue
Node first = null, second = null;
// traversal loop
while (!q.isEmpty())
{
// Pop two items from queue
first = q.peek();
q.remove();
second = q.peek();
q.remove();
// Print children of first and second in reverse order
System.out.print(" " + first.left.data + " " +second.right.data);
System.out.print(" " + first.right.data + " " +second.left.data);
// If first and second have grandchildren, enqueue them
// in reverse order
if (first.left.left != null)
{
q.add(first.left);
q.add(second.right);
q.add(first.right);
q.add(second.left);
}
}
}
// Driver program to test for above functions
public static void main(String args[])
{
BinaryTree tree = new BinaryTree();
tree.root = new Node(1);
tree.root.left = new Node(2);
tree.root.right = new Node(3);
tree.root.left.left = new Node(4);
tree.root.left.right = new Node(5);
tree.root.right.left = new Node(6);
tree.root.right.right = new Node(7);
tree.root.left.left.left = new Node(8);
tree.root.left.left.right = new Node(9);
tree.root.left.right.left = new Node(10);
tree.root.left.right.right = new Node(11);
tree.root.right.left.left = new Node(12);
tree.root.right.left.right = new Node(13);
tree.root.right.right.left = new Node(14);
tree.root.right.right.right = new Node(15);
tree.root.left.left.left.left = new Node(16);
tree.root.left.left.left.right = new Node(17);
tree.root.left.left.right.left = new Node(18);
tree.root.left.left.right.right = new Node(19);
tree.root.left.right.left.left = new Node(20);
tree.root.left.right.left.right = new Node(21);
tree.root.left.right.right.left = new Node(22);
tree.root.left.right.right.right = new Node(23);
tree.root.right.left.left.left = new Node(24);
tree.root.right.left.left.right = new Node(25);
tree.root.right.left.right.left = new Node(26);
tree.root.right.left.right.right = new Node(27);
tree.root.right.right.left.left = new Node(28);
tree.root.right.right.left.right = new Node(29);
tree.root.right.right.right.left = new Node(30);
tree.root.right.right.right.right = new Node(31);
System.out.println("Specific Level Order traversal of binary"
+"tree is ");
tree.printSpecificLevelOrder(tree.root);
}
}
// This code has been contributed by Mayank Jaiswal
# Python program for special order traversal
# A binary tree node
class Node:
# A constructor for making a new node
def __init__(self, key):
self.data = key
self.left = None
self.right = None
# Given a perfect binary tree print its node in
# specific order
def printSpecificLevelOrder(root):
if root is None:
return
# Let us print root and next level first
print (root.data,end=" ")
# Since it is perfect Binary tree,
# one of the node is needed to be checked
if root.left is not None :
print (root.left.data,end=" ")
print (root.right.data,end=" ")
# Do anything more if there are nodes at next level
# in given perfect Binary Tree
if root.left.left is None:
return
# Create a queue and enqueue left and right
# children of root
q = []
q.append(root.left)
q.append(root.right)
# We process two nodes at a time, so we need
# two variables to store two front items of queue
first = None
second = None
# Traversal loop
while(len(q) > 0):
# Pop two items from queue
first = q.pop(0)
second = q.pop(0)
# Print children of first and second in reverse order
print (first.left.data,end=" ")
print (second.right.data,end=" ")
print (first.right.data,end=" ")
print (second.left.data,end=" ")
# If first and second have grandchildren,
# enqueue them in reverse order
if first.left.left is not None:
q.append(first.left)
q.append(second.right)
q.append(first.right)
q.append(second.left)
# Driver program to test above function
# Perfect Binary Tree of Height 4
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)
root.right.right = Node(7)
root.left.left.left = Node(8)
root.left.left.right = Node(9)
root.left.right.left = Node(10)
root.left.right.right = Node(11)
root.right.left.left = Node(12)
root.right.left.right = Node(13)
root.right.right.left = Node(14)
root.right.right.right = Node(15)
root.left.left.left.left = Node(16)
root.left.left.left.right = Node(17)
root.left.left.right.left = Node(18)
root.left.left.right.right = Node(19)
root.left.right.left.left = Node(20)
root.left.right.left.right = Node(21)
root.left.right.right.left = Node(22)
root.left.right.right.right = Node(23)
root.right.left.left.left = Node(24)
root.right.left.left.right = Node(25)
root.right.left.right.left = Node(26)
root.right.left.right.right = Node(27)
root.right.right.left.left = Node(28)
root.right.right.left.right = Node(29)
root.right.right.right.left = Node(30)
root.right.right.right.right = Node(31)
print ("Specific Level Order traversal of binary tree is")
printSpecificLevelOrder(root);
# This code is contributed by Nikhil Kumar Singh(nickzuck_007)
// C# program for special level
// order traversal
using System;
using System.Collections.Generic;
/* Class containing left and right
child of current node and key value*/
public class Node
{
public int data;
public Node left, right;
public Node(int item)
{
data = item;
left = right = null;
}
}
class GFG
{
public Node root;
/* Given a perfect binary tree,
print its nodes in specific
level order */
public virtual void printSpecificLevelOrder(Node node)
{
if (node == null)
{
return;
}
// Let us print root and next level first
Console.Write(node.data);
// Since it is perfect Binary Tree,
// right is not checked
if (node.left != null)
{
Console.Write(" " + node.left.data +
" " + node.right.data);
}
// Do anything more if there
// are nodes at next level in
// given perfect Binary Tree
if (node.left.left == null)
{
return;
}
// Create a queue and enqueue left
// and right children of root
LinkedList<Node> q = new LinkedList<Node>();
q.AddLast(node.left);
q.AddLast(node.right);
// We process two nodes at a time,
// so we need two variables to
// store two front items of queue
Node first = null, second = null;
// traversal loop
while (q.Count > 0)
{
// Pop two items from queue
first = q.First.Value;
q.RemoveFirst();
second = q.First.Value;
q.RemoveFirst();
// Print children of first and
// second in reverse order
Console.Write(" " + first.left.data +
" " + second.right.data);
Console.Write(" " + first.right.data +
" " + second.left.data);
// If first and second have grandchildren,
// enqueue them in reverse order
if (first.left.left != null)
{
q.AddLast(first.left);
q.AddLast(second.right);
q.AddLast(first.right);
q.AddLast(second.left);
}
}
}
// Driver Code
public static void Main(string[] args)
{
GFG tree = new GFG();
tree.root = new Node(1);
tree.root.left = new Node(2);
tree.root.right = new Node(3);
tree.root.left.left = new Node(4);
tree.root.left.right = new Node(5);
tree.root.right.left = new Node(6);
tree.root.right.right = new Node(7);
tree.root.left.left.left = new Node(8);
tree.root.left.left.right = new Node(9);
tree.root.left.right.left = new Node(10);
tree.root.left.right.right = new Node(11);
tree.root.right.left.left = new Node(12);
tree.root.right.left.right = new Node(13);
tree.root.right.right.left = new Node(14);
tree.root.right.right.right = new Node(15);
tree.root.left.left.left.left = new Node(16);
tree.root.left.left.left.right = new Node(17);
tree.root.left.left.right.left = new Node(18);
tree.root.left.left.right.right = new Node(19);
tree.root.left.right.left.left = new Node(20);
tree.root.left.right.left.right = new Node(21);
tree.root.left.right.right.left = new Node(22);
tree.root.left.right.right.right = new Node(23);
tree.root.right.left.left.left = new Node(24);
tree.root.right.left.left.right = new Node(25);
tree.root.right.left.right.left = new Node(26);
tree.root.right.left.right.right = new Node(27);
tree.root.right.right.left.left = new Node(28);
tree.root.right.right.left.right = new Node(29);
tree.root.right.right.right.left = new Node(30);
tree.root.right.right.right.right = new Node(31);
Console.WriteLine("Specific Level Order " +
"traversal of binary" + "tree is ");
tree.printSpecificLevelOrder(tree.root);
}
}
// This code is contributed by Shrikant13
<script>
// JavaScript program for special level
// order traversal
/* Class containing left and right
child of current node and key value*/
class Node
{
constructor(item)
{
this.data = item;
this.left = null;
this.right = null;
}
}
var root = null;
/* Given a perfect binary tree,
print its nodes in specific
level order */
function printSpecificLevelOrder(node)
{
if (node == null)
{
return;
}
// Let us print root and next level first
document.write(node.data);
// Since it is perfect Binary Tree,
// right is not checked
if (node.left != null)
{
document.write(" " + node.left.data +
" " + node.right.data);
}
// Do anything more if there
// are nodes at next level in
// given perfect Binary Tree
if (node.left.left == null)
{
return;
}
// Create a queue and enqueue left
// and right children of root
var q = [];
q.push(node.left);
q.push(node.right);
// We process two nodes at a time,
// so we need two variables to
// store two front items of queue
var first = null, second = null;
// traversal loop
while (q.length > 0)
{
// Pop two items from queue
first = q[0];
q.shift();
second = q[0];
q.shift();
// Print children of first and
// second in reverse order
document.write(" " + first.left.data +
" " + second.right.data);
document.write(" " + first.right.data +
" " + second.left.data);
// If first and second have grandchildren,
// enqueue them in reverse order
if (first.left.left != null)
{
q.push(first.left);
q.push(second.right);
q.push(first.right);
q.push(second.left);
}
}
}
// Driver Code
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);
root.right.right = new Node(7);
root.left.left.left = new Node(8);
root.left.left.right = new Node(9);
root.left.right.left = new Node(10);
root.left.right.right = new Node(11);
root.right.left.left = new Node(12);
root.right.left.right = new Node(13);
root.right.right.left = new Node(14);
root.right.right.right = new Node(15);
root.left.left.left.left = new Node(16);
root.left.left.left.right = new Node(17);
root.left.left.right.left = new Node(18);
root.left.left.right.right = new Node(19);
root.left.right.left.left = new Node(20);
root.left.right.left.right = new Node(21);
root.left.right.right.left = new Node(22);
root.left.right.right.right = new Node(23);
root.right.left.left.left = new Node(24);
root.right.left.left.right = new Node(25);
root.right.left.right.left = new Node(26);
root.right.left.right.right = new Node(27);
root.right.right.left.left = new Node(28);
root.right.right.left.right = new Node(29);
root.right.right.right.left = new Node(30);
root.right.right.right.right = new Node(31);
document.write("Specific Level Order " +
"traversal of binary" + "tree is <br>");
printSpecificLevelOrder(root);
</script>
OutputSpecific Level Order traversal of binary tree is
1 2 3 4 7 5 6 8 15 9 14 10 13 11 12 16 31 17 30 18 29 19 28 20 27 21 26 22 25 23 24
Time Complexity: O(n) where n is the total number of nodes in the tree.
Auxiliary Space : O(n)
Followup Questions:
- The above code prints specific level order from TOP to BOTTOM. How will you do specific level order traversal from BOTTOM to TOP (Amazon Interview | Set 120 - Round 1 Last Problem)
- What if tree is not perfect, but complete.
- What if tree is neither perfect, nor complete. It can be any general binary tree.
Perfect Binary Tree Specific Level Order Traversal
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