Reverse Level Order traversal in spiral form

Given a binary tree, the task is to print the reverse level order in spiral form.

Examples:

Input: 
     1
   /  \
  2    3
 / \  / \
4   5 6  7
Output: 4 5 6 7 3 2 1

Input:
     5
   /   \
  6     4
 / \   / 
7   1 8
 \     \
  3     10
Output: 3 10 8 1 7 6 4 5

Approach: The idea is to traverse the tree in a Reverse Level Order manner but with a slight modification. We will use a variable flag and initially set it’s value to one. As we complete the reverse level order traversal of the tree, from left to right we will set the value of flag to zero, so that next time we traverse the Tree from right to left and as we complete the traversal we set it’s value back to one. We will repeat this whole step until we have traversed the Binary Tree completely.

Below is the implementation of the above approach:  

C++

// C++ program to print reverse level order
// traversal of a binary tree in spiral form
#include <bits/stdc++.h>
using namespace std;

// Binary tree Node
struct Node {
    struct Node* left;
    struct Node* right;
    int data;

    Node(int data)
    {
        this->data = data;
        this->left = NULL;
        this->right = NULL;
    }
};

// Recursive Function to find height
// of binary tree
int height(struct Node* root)
{
    if (root == NULL)
        return 0;

    // Compute the height of each subtree
    int lheight = height(root->left);
    int rheight = height(root->right);

    // Return the maximum of two
    return max(lheight + 1, rheight + 1);
}

// Function to Print Nodes from right to left
void rightToLeft(struct Node* root, int level)
{
    if (root == NULL)
        return;

    if (level == 1)
        cout << root->data << " ";

    else if (level > 1) {
        rightToLeft(root->right, level - 1);
        rightToLeft(root->left, level - 1);
    }
}

// Function to Print Nodes from left to right
void leftToRight(struct Node* root, int level)
{
    if (root == NULL)
        return;

    if (level == 1)
        cout << root->data << " ";

    else if (level > 1) {
        leftToRight(root->left, level - 1);
        leftToRight(root->right, level - 1);
    }
}

// Function to print Reverse level order traversal
// of a Binary tree in spiral form
void reverseSpiral(struct Node* root)
{
    // Flag is used to mark the change
    // in level
    int flag = 1;

    // Height of tree
    int h = height(root);

    for (int i = h; i >= 1; i--) {

        // If flag value is one print Nodes
        // from left to right
        if (flag == 1) {
            leftToRight(root, i);

            // Mark flag as zero so that next time
            // tree is traversed from right to left
            flag = 0;
        }

        // If flag is zero print Nodes
        // from right to left
        else if (flag == 0) {
            rightToLeft(root, i);

            // Mark flag as one so that next time
            // Nodes are printed from left to right
            flag = 1;
        }
    }
}

// Driver code
int main()
{
    struct Node* root = new Node(5);
    root->left = new Node(9);
    root->right = new Node(3);
    root->left->left = new Node(6);
    root->right->right = new Node(4);
    root->left->left->left = new Node(8);
    root->left->left->right = new Node(7);

    reverseSpiral(root);

    return 0;
}

Java

// Java program to print reverse level order
// traversal of a binary tree in spiral form
class Solution {

    // Binary tree Node
    static class Node {
        Node left;
        Node right;
        int data;

        Node(int data)
        {
            this.data = data;
            this.left = null;
            this.right = null;
        }
    };

    // Recursive Function to find height
    // of binary tree
    static int height(Node root)
    {
        if (root == null)
            return 0;

        // Compute the height of each subtree
        int lheight = height(root.left);
        int rheight = height(root.right);

        // Return the maximum of two
        return Math.max(lheight + 1, rheight + 1);
    }

    // Function to Print Nodes from right to left
    static void rightToLeft(Node root, int level)
    {
        if (root == null)
            return;

        if (level == 1)
            System.out.print(root.data + " ");

        else if (level > 1) {
            rightToLeft(root.right, level - 1);
            rightToLeft(root.left, level - 1);
        }
    }

    // Function to Print Nodes from left to right
    static void leftToRight(Node root, int level)
    {
        if (root == null)
            return;

        if (level == 1)
            System.out.print(root.data + " ");

        else if (level > 1) {
            leftToRight(root.left, level - 1);
            leftToRight(root.right, level - 1);
        }
    }

    // Function to print Reverse level order traversal
    // of a Binary tree in spiral form
    static void reverseSpiral(Node root)
    {
        // Flag is used to mark the change
        // in level
        int flag = 1;

        // Height of tree
        int h = height(root);

        for (int i = h; i >= 1; i--) {

            // If flag value is one print Nodes
            // from left to right
            if (flag == 1) {
                leftToRight(root, i);

                // Mark flag as zero so that next time
                // tree is traversed from right to left
                flag = 0;
            }

            // If flag is zero print Nodes
            // from right to left
            else if (flag == 0) {
                rightToLeft(root, i);

                // Mark flag as one so that next time
                // Nodes are printed from left to right
                flag = 1;
            }
        }
    }

    // Driver code
    public static void main(String args[])
    {
        Node root = new Node(5);
        root.left = new Node(9);
        root.right = new Node(3);
        root.left.left = new Node(6);
        root.right.right = new Node(4);
        root.left.left.left = new Node(8);
        root.left.left.right = new Node(7);

        reverseSpiral(root);
    }
}

// This code is contributed by Arnab Kundu

Python3

# Python3 program to print reverse level order
# traversal of a binary tree in spiral form

class newNode:

    # Construct to create a newNode
    def __init__(self, key):
        self.data = key
        self.left = None
        self.right = None

# Recursive Function to find height
# of binary tree
def height(root):

    if (root == None):
        return 0

    # Compute the height of each subtree
    lheight = height(root.left)
    rheight = height(root.right)

    # Return the maximum of two
    return max(lheight + 1, rheight + 1)

# Function to Print Nodes from right to left
def rightToLeft(root, level):
    if (root == None):
        return

    if (level == 1):
        print(root.data, end =" ")

    elif (level > 1):
        rightToLeft(root.right, level - 1)
        rightToLeft(root.left, level - 1)

# Function to Print Nodes from left to right
def leftToRight(root, level):
    if (root == None):
        return

    if (level == 1) :
        print(root.data, end = " ")

    elif (level > 1):
        leftToRight(root.left, level - 1)
        leftToRight(root.right, level - 1)

# Function to print Reverse level order traversal
# of a Binary tree in spiral form
def reverseSpiral(root):

    # Flag is used to mark the
    # change in level
    flag = 1

    # Height of tree
    h = height(root)

    for i in range(h, 0, -1):

        # If flag value is one print Nodes
        # from left to right
        if (flag == 1):
            leftToRight(root, i)

            # Mark flag as zero so that next time
            # tree is traversed from right to left
            flag = 0

        # If flag is zero print Nodes
        # from right to left
        elif (flag == 0):
            rightToLeft(root, i)

            # Mark flag as one so that next time
            # Nodes are printed from left to right
            flag = 1

# Driver Code
if __name__ == '__main__':
    root = newNode(5)
    root.left = newNode(9)
    root.right = newNode(3)
    root.left.left = newNode(6)
    root.right.right = newNode(4)
    root.left.left.left = newNode(8)
    root.left.left.right = newNode(7)

    reverseSpiral(root)

# This code is contributed
# by SHUBHAMSINGH10

C#

// C# program to print reverse level order
// traversal of a binary tree in spiral form
using System;

class GFG {

    // Binary tree Node
    public class Node {
        public Node left;
        public Node right;
        public int data;

        public Node(int data)
        {
            this.data = data;
            this.left = null;
            this.right = null;
        }
    };

    // Recursive Function to find height
    // of binary tree
    static int height(Node root)
    {
        if (root == null)
            return 0;

        // Compute the height of each subtree
        int lheight = height(root.left);
        int rheight = height(root.right);

        // Return the maximum of two
        return Math.Max(lheight + 1, rheight + 1);
    }

    // Function to Print Nodes from right to left
    static void rightToLeft(Node root, int level)
    {
        if (root == null)
            return;

        if (level == 1)
            Console.Write(root.data + " ");

        else if (level > 1) {
            rightToLeft(root.right, level - 1);
            rightToLeft(root.left, level - 1);
        }
    }

    // Function to Print Nodes from left to right
    static void leftToRight(Node root, int level)
    {
        if (root == null)
            return;

        if (level == 1)
            Console.Write(root.data + " ");

        else if (level > 1) {
            leftToRight(root.left, level - 1);
            leftToRight(root.right, level - 1);
        }
    }

    // Function to print Reverse level order traversal
    // of a Binary tree in spiral form
    static void reverseSpiral(Node root)
    {
        // Flag is used to mark the change
        // in level
        int flag = 1;

        // Height of tree
        int h = height(root);

        for (int i = h; i >= 1; i--) {

            // If flag value is one print Nodes
            // from left to right
            if (flag == 1) {
                leftToRight(root, i);

                // Mark flag as zero so that next time
                // tree is traversed from right to left
                flag = 0;
            }

            // If flag is zero print Nodes
            // from right to left
            else if (flag == 0) {
                rightToLeft(root, i);

                // Mark flag as one so that next time
                // Nodes are printed from left to right
                flag = 1;
            }
        }
    }

    // Driver code
    public static void Main(String[] args)
    {
        Node root = new Node(5);
        root.left = new Node(9);
        root.right = new Node(3);
        root.left.left = new Node(6);
        root.right.right = new Node(4);
        root.left.left.left = new Node(8);
        root.left.left.right = new Node(7);

        reverseSpiral(root);
    }
}

/* This code is contributed by PrinciRaj1992 */

Javascript

<script>

// Javascript program to print reverse level order
// traversal of a binary tree in spiral form

// Binary tree Node
class Node
{
    constructor(data)
    {
        this.data = data;
        this.left = null;
        this.right = null;
    }
}

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

    // Compute the height of each subtree
    let lheight = height(root.left);
    let rheight = height(root.right);

    // Return the maximum of two
    return Math.max(lheight + 1, rheight + 1);
}

// Function to Print Nodes from right to left
function rightToLeft(root, level)
{
    if (root == null)
        return;

    if (level == 1)
        document.write(root.data + " ");

    else if (level > 1)
    {
        rightToLeft(root.right, level - 1);
        rightToLeft(root.left, level - 1);
    }
}

// Function to Print Nodes from left to right
function leftToRight(root, level)
{
    if (root == null)
        return;

    if (level == 1)
        document.write(root.data + " ");

    else if (level > 1)
    {
        leftToRight(root.left, level - 1);
        leftToRight(root.right, level - 1);
    }
}

// Function to print Reverse level order traversal
// of a Binary tree in spiral form
function reverseSpiral(root)
{

    // Flag is used to mark the change
    // in level
    let flag = 1;

    // Height of tree
    let h = height(root);

    for(let i = h; i >= 1; i--)
    {

        // If flag value is one print Nodes
        // from left to right
        if (flag == 1)
        {
            leftToRight(root, i);

            // Mark flag as zero so that next time
            // tree is traversed from right to left
            flag = 0;
        }

        // If flag is zero print Nodes
        // from right to left
        else if (flag == 0)
        {
            rightToLeft(root, i);

            // Mark flag as one so that next time
            // Nodes are printed from left to right
            flag = 1;
        }
    }
}

// Driver code
let root = new Node(5);
root.left = new Node(9);
root.right = new Node(3);
root.left.left = new Node(6);
root.right.right = new Node(4);
root.left.left.left = new Node(8);
root.left.left.right = new Node(7);

reverseSpiral(root);

// This code is contributed by unknown2108

</script>

Output:

8 7 4 6 9 3 5

Time Complexity Analysis : The best case time complexity of reverseSpiral() is O(n+n/2+n/4…) = O(n) in case of a perfect binary tree. And the worst case time complexity is O(n²) [O((n)+(n-1)+(n-2)+…] in case of a skewed tree taken as input.

Auxiliary Space: O(n) for call stack since using recursion.

Iterative Implementation – 
The above problem can be solved with the help of queue and stack. Just perform a normal spiral traversal, use stack to use reverse the elements alternatively. Below is the implementation of the same – 

C++14

#include <bits/stdc++.h>
using namespace std;

// Node structure
struct Node {
    int data;
    Node* left;
    Node* right;
    Node(int val)
    {
        data = val;
        left = right = NULL;
    }
};

// function to perform
// reversal spiral traversal
void reverseSpiral(Node* root)
{
    if (root == NULL) {
        return;
    }
    queue<Node*> q;
    stack<int> s;
    bool reverse = true;
    q.push(root);
    // iterate until queue is empty
    while (!q.empty()) {
        // calculate size of level
        int count = q.size();

        // iterate over level nodes
        for (int i = 0; i < count; i++) {
            Node* curr = q.front();
            q.pop();
            s.push(curr->data);
            if (reverse) {
                if (curr->right) {
                    q.push(curr->right);
                }
                if (curr->left) {
                    q.push(curr->left);
                }
            }
            else {
                if (curr->left) {
                    q.push(curr->left);
                }
                if (curr->right) {
                    q.push(curr->right);
                }
            }
        }
        reverse = !reverse;
    }
    // pop and print reversed nodes
    while (!s.empty()) {
        int curr = s.top();
        cout << curr << " ";
        s.pop();
    }
}

// Driver code
int main()
{
    Node* root = new Node(5);
    root->left = new Node(9);
    root->right = new Node(3);
    root->left->left = new Node(6);
    root->left->right = new Node(4);
    root->left->left->left = new Node(8);
    root->left->left->right = new Node(7);
    reverseSpiral(root);
    return 0;
}

Java

// Java code for the above approach

import java.io.*;
import java.util.*;

class GFG {

    static class Node {
        int data;
        Node left, right;
        Node(int val)
        {
            this.data = val;
            left = right = null;
        }
    }

    // Function to perform reversal spiral traversal
    static void reverseSpiral(Node root)
    {
        if (root == null) {
            return;
        }
        Queue<Node> q = new LinkedList<>();
        Stack<Integer> s = new Stack<>();
        boolean reverse = true;
        q.add(root);

        // Iterate until queue is empty
        while (!q.isEmpty()) {
            // Calculate size of level
            int count = q.size();

            // Iterate over level nodes
            for (int i = 0; i < count; i++) {
                Node curr = q.peek();
                q.remove();
                s.push(curr.data);
                if (reverse) {
                    if (curr.right != null) {
                        q.add(curr.right);
                    }
                    if (curr.left != null) {
                        q.add(curr.left);
                    }
                }
                else {
                    if (curr.left != null) {
                        q.add(curr.left);
                    }
                    if (curr.right != null) {
                        q.add(curr.right);
                    }
                }
            }
            reverse = !reverse;
        }
        // pop and print reversed nodes
        while (!s.isEmpty()) {
            int curr = s.peek();
            System.out.print(curr + " ");
            s.pop();
        }
    }

    public static void main(String[] args)
    {
        Node root = new Node(5);
        root.left = new Node(9);
        root.right = new Node(3);
        root.left.left = new Node(6);
        root.left.right = new Node(4);
        root.left.left.left = new Node(8);
        root.left.left.right = new Node(7);

        reverseSpiral(root);
    }
}

// This code is contributed by lokeshmvs21.

Python3

from collections import deque

# Node structure
class Node:
    def __init__(self, x):
        self.data = x
        self.right = None
        self.left = None

# function to perform
# reversal spiral traversal
def reverseSpiral(root):
    if (root == None):
        return

    q =  deque()
    s = []
    reverse = True
    q.append(root)

    # iterate until queue is empty
    while (len(q) > 0):

        # calculate size of level
        count = len(q)

        # iterate over level nodes
        for i in range(count):
            curr = q.popleft()
            s.append(curr.data)
            if (reverse):
                if (curr.right):
                    q.append(curr.right)
                if (curr.left):
                    q.append(curr.left)
            else:
                if (curr.left):
                    q.append(curr.left)
                if (curr.right):
                    q.append(curr.right)

        reverse = not reverse

    # pop and print reversed nodes
    while (len(s) > 0):
        curr = s[-1]
        print(curr, end = " ")
        del s[-1]

# Driver code
if __name__ == '__main__':
    root = Node(5)
    root.left = Node(9)
    root.right = Node(3)
    root.left.left = Node(6)
    root.left.right = Node(4)
    root.left.left.left = Node(8)
    root.left.left.right = Node(7)
    reverseSpiral(root)

# This code is contributed by mohit kumar 29.

C#

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

namespace ConsoleApp
{
    class Program
    {
        // Structure for a binary tree node
        class Node
        {
            public int data;
            public Node left, right;
            public Node(int val)
            {
                this.data = val;
                left = right = null;
            }
        }

        // Function to perform reversal spiral traversal
        static void ReverseSpiral(Node root)
        {
            if (root == null)
            {
                return;
            }
            Queue<Node> q = new Queue<Node>();
            Stack<int> s = new Stack<int>();
            bool reverse = true;
            q.Enqueue(root);

            // Iterate until queue is empty
            while (q.Count > 0)
            {
                // Calculate size of level
                int count = q.Count;

                // Iterate over level nodes
                for (int i = 0; i < count; i++)
                {
                    Node curr = q.Peek();
                    q.Dequeue();
                    s.Push(curr.data);
                    if (reverse)
                    {
                        if (curr.right != null)
                        {
                            q.Enqueue(curr.right);
                        }
                        if (curr.left != null)
                        {
                            q.Enqueue(curr.left);
                        }
                    }
                    else
                    {
                        if (curr.left != null)
                        {
                            q.Enqueue(curr.left);
                        }
                        if (curr.right != null)
                        {
                            q.Enqueue(curr.right);
                        }
                    }
                }
                reverse = !reverse;
            }
            // pop and print reversed nodes
            while (s.Count > 0)
            {
                int curr = s.Peek();
                Console.Write(curr + " ");
                s.Pop();
            }
        }

        static void Main(string[] args)
        {
            Node root = new Node(5);
            root.left = new Node(9);
            root.right = new Node(3);
            root.left.left = new Node(6);
            root.left.right = new Node(4);
            root.left.left.left = new Node(8);
            root.left.left.right = new Node(7);

            ReverseSpiral(root);
        }
    }
}
//This code is contributed by Potta Lokesh

Javascript

<script>

class Node
{
    constructor(data)
    {
        this.data = data;
        this.left = this.right = null;
    }
}

// function to perform
// reversal spiral traversal
function reverseSpiral(root)
{
    if (root == null)
        return
    let q =  [];
    let s = [];
    let reverse = true;

    q.push(root)
    while ((q).length > 0)
    {
        // calculate size of level
        let count = (q).length;

        // iterate over level nodes
        for (let i = 0; i < (count); i++)
        {
            curr = q.shift();
            s.push(curr.data);
            if (reverse)
            {
                if (curr.right)
                    q.push(curr.right);
                if (curr.left)
                    q.push(curr.left);
            }
            else
            {
                if (curr.left)
                    q.push(curr.left);
                if (curr.right)
                    q.push(curr.right);
            }
         }
        reverse = !reverse
    }

    // pop and prreversed nodes
    while ((s).length > 0)
    {
        curr = s.pop();
        document.write(curr+" ");
    }
}

// Driver code
let root = new Node(5)
root.left = new Node(9)
root.right = new Node(3)
root.left.left = new Node(6)
root.left.right = new Node(4)
root.left.left.left = new Node(8)
root.left.left.right = new Node(7)
reverseSpiral(root)

// This code is contributed by avanitrachhadiya2155
</script>

Output:

8 7 4 6 9 3 5

Time Complexity: O(N) 
Auxiliary Space: O(N)

Improved by :HarendraSingh22