Specific Level Order Traversal of Binary Tree

Given a Binary Tree, the task is to perform a Specific Level Order Traversal of the tree such that at each level print 1st element then the last element, then 2nd element and 2nd last element, until all elements of that level is printed and so on.

Examples:

Input: Below is the given tree: 

Output: 1 8 3 4 7 5 6 
Explanation:
1st level:    1(root) 
2nd level:   8(left),    3(right) 
3rd level:    4(left),     7(right),    5(left),    6(right)

Input: Below is the given tree: 

Output: 20 8 22 4 12 10 14 
Explanation:
1st level:    20(root) 
2nd level:   8(left),    22(right) 
3rd level:    4(left),     12(right) 
4th level:    10(left),    14(right) 

Approach: The idea is to the Level Order Traversal of the given Binary Tree and converts this tree into a Perfect Binary Tree. While traversing if any node whose child Node doesn’t exist then append NULL node as the child node in the queue. Below are the steps: 

1. Perform the level order traversal of the given tree.
2. During traversal, if any node whose child Node doesn’t exist then append NULL node as the child node in the queue
3. During traversal for each level do the following: 
   • Maintain two queues to store the nodes for the two half of each level.
   • For the first half of the level store nodes in a left to right manner.
   • For the second half of the level store nodes in a right to left manner.
   • Iterate the above two queues formed and print in an alternate manner for specific level order traversal.

Below is the implementation of the above approach: 

C++

// C++ Program for the above approach
#include <bits/stdc++.h>
using namespace std;

struct Node {
    int data;
    Node* left;
    Node* right;
};

// Creates and initialize a new node
Node* newNode(int ch)
{
    // Allocating memory to a new node
    Node* n = new Node();
    n->data = ch;
    n->left = NULL;
    n->right = NULL;
    return n;
}

// Function to find the height of tree
int Height(Node* root)
{
    if (root == NULL)
        return 0;
    return 1 + max(Height(root->left), Height(root->right));
}

// Given a perfect binary tree
// print its node in specific order
void printSpecificLevelOrder(queue<Node*> A, queue<Node*> B,
                             int height)
{
    while (height != 0) {
        // Get each one front
        // element of both queue
        Node* X = A.front();
        A.pop();
        Node* Y = B.front();
        B.pop();

        // check if X exist or not
        if (X == NULL) {
            // Assume is has and put
            // their both child as none
            A.push(NULL);
            A.push(NULL);
        }
        else {
            // print the data and store
            // their child first left
            // then right
            cout << X->data << " ";
            A.push(X->left);
            A.push(X->right);
        }

        // checkY exist or not
        if (Y == NULL) {
            // Assume is has and put
            // their both child as none
            B.push(NULL);
            B.push(NULL);
        }
        else {
            // print the data and store
            // their child first left
            // then right
            cout << Y->data << " ";
            B.push(Y->right);
            B.push(Y->left);
        }

        // Decrease by 1 unit
        height -= 1;
    }
}

int main()
{
    // Given Tree
    Node* 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(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->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->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);

    // Initialise Queue
    queue<Node*> A;
    queue<Node*> B;

    int height = 0;

    // check top root manually
    if (root != NULL) {
        cout << root->data << " ";
        A.push(root->left);
        B.push(root->right);

        height = Height(root);
        height = pow(2, (height - 1)) - 1;
    }

    // Function call
    printSpecificLevelOrder(A, B, height);
}

// This code is contributed by Yash Agarwal(yashagarwal2852002)

Java

// Java program for the above approach
import java.util.*;

class GFG{

// Node structure
static class node
{
    int data;
    node left = null;
    node right = null;
}

// Creates and initialize a new node
static node newNode(int ch)
{

    // Allocating memory to a new node
    node n = new node();
    n.data = ch;
    n.left = null;
    n.right = null;
    return n;
}

// Function to find the height of tree
static int Height(node root)
{
    if (root == null)
        return 0;

    return 1 + Math.max(Height(root.left),
                        Height(root.right));
}

// Given a perfect binary tree
// print its node in Specific order
static void printSpecificLevelOrder(Queue<node> A,
                                    Queue<node> B,
                                    int height)
{
    while (height != 0)
    {

        // Get each one front
        // element of both queue
        node  X = A.poll();
        node  Y = B.poll();

        // Check if X exist or not
        if (X == null)
        {

            // Assume is has and put
            // their both child as none
            A.add(null);
            A.add(null);
        }
        else
        {

            // print the data and store
            // their child first left
            // then right
            System.out.print(X.data + " ");
            A.add(X.left);
            A.add(X.right);
        }

        // Check Y exist or not
        if (Y == null)
        {

            // Assume is has and put
            // their both child as none
            B.add(null);
            B.add(null);
        }
        else
        {

            // Print the data and store their
            // child first left then right
            System.out.print(Y.data + " ");
            B.add(Y.right);
            B.add(Y.left);
        }

        // Decrease by 1 unit
        height -= 1;
    }
}

// Driver Code
public static void main (String[] args)
{

    // Given tree
    node 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(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.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.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);

    // Initialise Queue
    Queue<node> A = new LinkedList<>();
    Queue<node> B = new LinkedList<>();

    int height = 0;

    // Check top root manually
    if (root != null)
    {
        System.out.print(root.data + " ");

        A.add(root.left);
        B.add(root.right);

        height = Height(root);
        height = (int)Math.pow(2, (height - 1)) - 1;
    }

    // Function Call
    printSpecificLevelOrder(A, B, height);
}
}

// This code is contributed by offbeat

Python3

# Python3 program for the above approach
from queue import Queue

# 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

# Function to find the height of tree
def Height(root):
    if root == None:
        return 0
    return 1 + max(Height(root.left), Height(root.right))


# Given a perfect binary tree
# print its node in Specific order
def printSpecificLevelOrder(A, B, height):

    while height != 0:

        # Get each one front
        # element of both queue
        X = A.get()
        Y = B.get()

        # Check if X exist or not
        if X == None:
            # Assume is has and put
            # their both child as none
            A.put(None)
            A.put(None)
        else:
            # print the data and store
            # their child first left
            # then right
            print(X.data, end =" ")
            A.put(X.left)
            A.put(X.right)

        # Check Y exist or not
        if Y == None:
            # Assume is has and put
            # their both child as none
            B.put(None)
            B.put(None)
        else:
            # print the data and store their
            # child first left then right
            print(Y.data, end =" ")
            B.put(Y.right)
            B.put(Y.left)
        # Decrease by 1 unit
        height-= 1


# Driver Code

# Given Tree
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(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.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.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)

# Initialise Queue
A = Queue(100)
B = Queue(100)

# Check top root manually
if root != None:
    print(root.data, end =" ")

    A.put(root.left)
    B.put(root.right)

    height = Height(root)
    height = 2**(height-1)-1

# Function Call
    printSpecificLevelOrder(A, B, height)

C#

// C# program for the above approach
using System;
using System.Collections.Generic;

class GFG{

// Node structure
class node
{
    public int data;
    public node left = null;
    public node right = null;
}

// Creates and initialize a new node
static node newNode(int ch)
{

    // Allocating memory to a new node
    node n = new node();
    n.data = ch;
    n.left = null;
    n.right = null;
    return n;
}

// Function to find the height of tree
static int Height(node root)
{
    if (root == null)
        return 0;

    return 1 + Math.Max(Height(root.left),
                        Height(root.right));
}

// Given a perfect binary tree
// print its node in Specific order
static void printSpecificLevelOrder(Queue<node> A,
                                    Queue<node> B,
                                    int height)
{
    while (height != 0)
    {

        // Get each one front
        // element of both queue
        node X = A.Peek();
        A.Dequeue();
        node Y = B.Peek();
        B.Dequeue();

        // Check if X exist or not
        if (X == null)
        {

            // Assume is has and put
            // their both child as none
            A.Enqueue(null);
            A.Enqueue(null);
        }
        else
        {

            // print the data and store
            // their child first left
            // then right
            Console.Write(X.data + " ");
            A.Enqueue(X.left);
            A.Enqueue(X.right);
        }

        // Check Y exist or not
        if (Y == null)
        {

            // Assume is has and put
            // their both child as none
            B.Enqueue(null);
            B.Enqueue(null);
        }
        else
        {

            // Print the data and store their
            // child first left then right
            Console.Write(Y.data + " ");
            B.Enqueue(Y.right);
            B.Enqueue(Y.left);
        }

        // Decrease by 1 unit
        height -= 1;
    }
}

// Driver Code
public static void Main()
{

    // Given tree
    node 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(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.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.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);

    // Initialise Queue
    Queue<node> A = new Queue<node>();
    Queue<node> B = new Queue<node>();

    int height = 0;

    // Check top root manually
    if (root != null)
    {
        Console.Write(root.data + " ");

        A.Enqueue(root.left);
        B.Enqueue(root.right);

        height = Height(root);
        height = (int)Math.Pow(2, (height - 1)) - 1;
    }

    // Function Call
    printSpecificLevelOrder(A, B, height);
}
}

// This code is contributed by ipg2016107

Javascript

<script>
    // Javascript program for the above approach

    // Node structure
    class node
    {
        constructor(ch) {
           this.data = ch;
           this.left = this.right = null;
        }
    }

    // Creates and initialize a new node
    function newNode(ch)
    {

        // Allocating memory to a new node
        let n = new node(ch);
        return n;
    }

    // Function to find the height of tree
    function Height(root)
    {
        if (root == null)
            return 0;

        return 1 + Math.max(Height(root.left),
                            Height(root.right));
    }

    // Given a perfect binary tree
    // print its node in Specific order
    function printSpecificLevelOrder(A, B, height)
    {
        while (height != 0)
        {

            // Get each one front
            // element of both queue
            let X = A[0];
            A.shift();
            let Y = B[0];
            B.shift();

            // Check if X exist or not
            if (X == null)
            {

                // Assume is has and put
                // their both child as none
                A.push(null);
                A.push(null);
            }
            else
            {

                // print the data and store
                // their child first left
                // then right
                document.write(X.data + " ");
                A.push(X.left);
                A.push(X.right);
            }

            // Check Y exist or not
            if (Y == null)
            {

                // Assume is has and put
                // their both child as none
                B.push(null);
                B.push(null);
            }
            else
            {

                // Print the data and store their
                // child first left then right
                document.write(Y.data + " ");
                B.push(Y.right);
                B.push(Y.left);
            }

            // Decrease by 1 unit
            height -= 1;
        }
    }

    // Given tree
    let 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(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.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.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);

    // Initialise Queue
    let A = [];
    let B = [];

    let height = 0;

    // Check top root manually
    if (root != null)
    {
        document.write(root.data + " ");

        A.push(root.left);
        B.push(root.right);

        height = Height(root);
        height = Math.pow(2, (height - 1)) - 1;
    }

    // Function Call
    printSpecificLevelOrder(A, B, height);

</script>

Output:
1 2 3 4 7 5 8 15 9 14 10 11 16 31 17 30 18 29 19 28 20 21 22 23 

Time Complexity: O(N), N is the number of nodes 
Auxiliary Space: O(N), N is the number of nodes