Anti Clockwise spiral traversal of a binary tree

Given a binary tree, the task is to print the nodes of the tree in an anti-clockwise spiral manner. 

Examples:

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

Input:
      1
     /  \
    2    3
   /    / \
  4    5   6
 / \  /   / \
7   8 9  10  11
Output: 1 7 8 9 10 11 3 2 4 5 6

Approach: The idea is to use two variables i initialized to 1 and j initialized to the height of tree and run a while loop which wont break until i becomes greater than j. We will use another variable flag and initialize it to 0. Now in the while loop we will check a condition that if flag is equal to 0 we will traverse the tree from right to left and mark flag as 1 so that next time we traverse the tree from left to right and then increment the value of i so that next time we visit the level just below the current level. Also when we will traverse the level from bottom we will mark flag as 0 so that next time we traverse the tree from left to right and then decrement the value of j so that next time we visit the level just above the current level. Repeat the whole process until the binary tree is completely traversed.

Below is the implementation of the above approach:  

C++

// C++ implementation of the approach
#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)
{
    // Base condition
    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(1 + lheight, 1 + rheight);
}

// 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 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 anti clockwise spiral
// traversal of a binary tree
void antiClockWiseSpiral(struct Node* root)
{
    int i = 1;
    int j = height(root);

    // Flag to mark a change in the direction
    // of printing nodes
    int flag = 0;
    while (i <= j) {

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

            // Set the value of flag as zero
            // so that nodes are next time
            // printed from left to right
            flag = 1;

            // Increment i
            i++;
        }

        // If flag is one print nodes
        // from left to right
        else {
            leftToRight(root, j);

            // Set the value of flag as zero
            // so that nodes are next time
            // printed from right to left
            flag = 0;

            // Decrement j
            j--;
        }
    }
}

// Driver code
int main()
{
    struct Node* root = new Node(1);
    root->left = new Node(2);
    root->right = new Node(3);
    root->left->left = new Node(4);
    root->right->left = new Node(5);
    root->right->right = new Node(7);
    root->left->left->left = new Node(10);
    root->left->left->right = new Node(11);
    root->right->right->left = new Node(8);

    antiClockWiseSpiral(root);

    return 0;
}

Java

// Java implementation of the approach
class GfG
{

// 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)
{
    // Base condition
    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(1 + lheight, 1 + rheight);
}

// 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 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 anti clockwise spiral
// traversal of a binary tree
static void antiClockWiseSpiral(Node root)
{
    int i = 1;
    int j = height(root);

    // Flag to mark a change in the direction
    // of printing nodes
    int flag = 0;
    while (i <= j)
    {

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

            // Set the value of flag as zero
            // so that nodes are next time
            // printed from left to right
            flag = 1;

            // Increment i
            i++;
        }

        // If flag is one print nodes
        // from left to right
        else {
            leftToRight(root, j);

            // Set the value of flag as zero
            // so that nodes are next time
            // printed from right to left
            flag = 0;

            // Decrement j
            j--;
        }
    }
}

// Driver code
public static void main(String[] args)
{
    Node root = new Node(1);
    root.left = new Node(2);
    root.right = new Node(3);
    root.left.left = new Node(4);
    root.right.left = new Node(5);
    root.right.right = new Node(7);
    root.left.left.left = new Node(10);
    root.left.left.right = new Node(11);
    root.right.right.left = new Node(8);

    antiClockWiseSpiral(root);
}
}

// This code is contributed by Prerna Saini.

Python3

# Python3 implementation of the approach

# Binary tree node
class newNode:

    # Constructor to create a newNode
    def __init__(self, data):
        self.data = data
        self.left = None
        self.right = None
        self.visited = False

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

    # Base condition
    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(1 + lheight, 1 + rheight)

# 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 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 anti clockwise spiral
# traversal of a binary tree
def antiClockWiseSpiral(root):

    i = 1
    j = height(root)

    # Flag to mark a change in the
    # direction of printing nodes
    flag = 0
    while (i <= j):

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

            # Set the value of flag as zero
            # so that nodes are next time
            # printed from left to right
            flag = 1

            # Increment i
            i += 1

        # If flag is one print nodes
        # from left to right
        else:
            leftToRight(root, j)

            # Set the value of flag as zero
            # so that nodes are next time
            # printed from right to left
            flag = 0

            # Decrement j
            j-=1

# Driver Code
if __name__ == '__main__':
    root = newNode(1)
    root.left = newNode(2)
    root.right = newNode(3)
    root.left.left = newNode(4)
    root.right.left = newNode(5)
    root.right.right = newNode(7)
    root.left.left.left = newNode(10)
    root.left.left.right = newNode(11)
    root.right.right.left = newNode(8)

    antiClockWiseSpiral(root)

# This code is contributed by
# SHUBHAMSINGH10

C#

// C# implementation of the approach
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)
{
    // Base condition
    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(1 + lheight, 1 + rheight);
}

// 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 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 anti clockwise spiral
// traversal of a binary tree
static void antiClockWiseSpiral( Node root)
{
    int i = 1;
    int j = height(root);

    // Flag to mark a change in the direction
    // of printing nodes
    int flag = 0;
    while (i <= j)
    {

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

            // Set the value of flag as zero
            // so that nodes are next time
            // printed from left to right
            flag = 1;

            // Increment i
            i++;
        }

        // If flag is one print nodes
        // from left to right
        else
        {
            leftToRight(root, j);

            // Set the value of flag as zero
            // so that nodes are next time
            // printed from right to left
            flag = 0;

            // Decrement j
            j--;
        }
    }
}

// Driver code
public static void Main(String []args)
{
    Node root = new Node(1);
    root.left = new Node(2);
    root.right = new Node(3);
    root.left.left = new Node(4);
    root.right.left = new Node(5);
    root.right.right = new Node(7);
    root.left.left.left = new Node(10);
    root.left.left.right = new Node(11);
    root.right.right.left = new Node(8);

    antiClockWiseSpiral(root);

}
}

//This code is contributed by Arnab Kundu

Javascript

<script>

    // JavaScript implementation of the approach

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

    // Recursive Function to find height
    // of binary tree
    function height(root)
    {
        // Base condition
        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(1 + lheight, 1 + rheight);
    }

    // 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 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 anti clockwise spiral
    // traversal of a binary tree
    function antiClockWiseSpiral(root)
    {
        let i = 1;
        let j = height(root);

        // Flag to mark a change in the direction
        // of printing nodes
        let flag = 0;
        while (i <= j)
        {

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

                // Set the value of flag as zero
                // so that nodes are next time
                // printed from left to right
                flag = 1;

                // Increment i
                i++;
            }

            // If flag is one print nodes
            // from left to right
            else {
                leftToRight(root, j);

                // Set the value of flag as zero
                // so that nodes are next time
                // printed from right to left
                flag = 0;

                // Decrement j
                j--;
            }
        }
    }

    let root = new Node(1);
    root.left = new Node(2);
    root.right = new Node(3);
    root.left.left = new Node(4);
    root.right.left = new Node(5);
    root.right.right = new Node(7);
    root.left.left.left = new Node(10);
    root.left.left.right = new Node(11);
    root.right.right.left = new Node(8);

    antiClockWiseSpiral(root);

</script>

Output:

1 10 11 8 3 2 4 5 7 

Another Approach:
The above approach have O(n^2) worst case complexity due to calling the print level everytime. An improvement over it can be storing the level wise nodes and use it to print. 

C++

// C++ implementation of the above approach

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

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

void antiClockWiseSpiral(struct Node* root)
{
    // Initialize the queue
    queue<Node*> q;

    // Add the root node
    q.push(root);

    //  Initialize the vector
    vector<Node*> topone;

    // Until queue is not empty
    while (!q.empty()) {
        int len = q.size();

        // len is greater than zero
        while (len > 0) {
            Node* nd = q.front();
            q.pop();
            if (nd != NULL) {
                topone.push_back(nd);
                if (nd->right != NULL)
                    q.push(nd->right);
                if (nd->left != NULL)
                    q.push(nd->left);
            }
            len--;
        }
        topone.push_back(NULL);
    }
    bool top = true;
    int l = 0, r = (int)topone.size() - 2;

    while (l < r) {
        if (top) {
            while (l < (int)topone.size()) {
                Node* nd = topone[l++];
                if (nd == NULL) {
                    break;
                }
                cout << nd->data << " ";
            }
        }
        else {
            while (r >= l) {
                Node* nd = topone[r--];
                if (nd == NULL)
                    break;
                cout << nd->data << " ";
            }
        }
        top = !top;
    }
}

// Build Tree
int main()
{
    /*
               1
            2     3
          4     5   7
         10 11     8
   */

    struct Node* root = new Node(1);
    root->left = new Node(2);
    root->right = new Node(3);
    root->left->left = new Node(4);
    root->right->left = new Node(5);
    root->right->right = new Node(7);
    root->left->left->left = new Node(10);
    root->left->left->right = new Node(11);
    root->right->right->left = new Node(8);

    antiClockWiseSpiral(root);

    return 0;
}

Java

// Java implementation of the above approach

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

class GFG {

    // Structure of each node
    class Node {
        int val;
        Node left, right;
        Node(int val)
        {
            this.val = val;
            this.left = this.right = null;
        }
    }

    private void work(Node root)
    {
        // Initialize queue
        Queue<Node> q = new LinkedList<>();

        // Add the root node
        q.add(root);

        // Initialize the vector
        Vector<Node> topone = new Vector<>();

        // Until queue is not empty
        while (!q.isEmpty()) {
            int len = q.size();

            // len is greater than zero
            while (len > 0) {
                Node nd = q.poll();
                if (nd != null) {
                    topone.add(nd);
                    if (nd.right != null)
                        q.add(nd.right);
                    if (nd.left != null)
                        q.add(nd.left);
                }
                len--;
            }
            topone.add(null);
        }
        boolean top = true;
        int l = 0, r = topone.size() - 2;

        while (l < r) {
            if (top) {
                while (l < topone.size()) {
                    Node nd = topone.get(l++);
                    if (nd == null) {
                        break;
                    }
                    System.out.print(nd.val + " ");
                }
            }
            else {
                while (r >= l) {
                    Node nd = topone.get(r--);
                    if (nd == null)
                        break;
                    System.out.print(nd.val + " ");
                }
            }
            top = !top;
        }
    }

    // Build Tree
    public void solve()
    {
        /*
                            1
                         2     3
                       4     5   7
                      10 11     8
                    */

        Node root = new Node(1);
        root.left = new Node(2);
        root.right = new Node(3);
        root.left.left = new Node(4);
        root.right.left = new Node(5);
        root.right.right = new Node(7);
        root.left.left.left = new Node(10);
        root.left.left.right = new Node(11);
        root.right.right.left = new Node(8);

        // Function call
        work(root);
    }

    // Driver Code
    public static void main(String[] args)
    {
        GFG t = new GFG();
        t.solve();
    }
}

Python3

# Python 3 implementation of the above approach

from collections import  deque as dq

class Node:
    def __init__(self, data):

        self.data = data
        self.left = None
        self.right = None

def antiClockWiseSpiral(root):

    # Initialize the queue
    q=dq([root])

    # Initialize the list
    topone=[]

    # Until queue is not empty
    while (q):
        l = len(q)

        # l is greater than zero
        while (l > 0):
            nd = q.popleft()
            if (nd != None):
                topone.append(nd)
                if (nd.right != None):
                    q.append(nd.right)
                if (nd.left != None):
                    q.append(nd.left)
            l-=1

        topone.append(None)

    top = True
    l = 0; r = len(topone) - 2

    while (l < r):
        if (top):
            while (l < len(topone)):
                nd = topone[l]
                l+=1
                if (nd == None):
                    break
                print(nd.data,end=" ")
        else:
            while (r >= l):
                nd = topone[r]
                r-=1
                if (nd == None):
                    break
                print(nd.data,end=" ")


        top = not top
    print()

# Build Tree
if __name__ == '__main__':


    #        1
    #     2     3
    #   4     5  7
    # 10 11     8


    root = Node(1)
    root.left = Node(2)
    root.right =  Node(3)
    root.left.left = Node(4)
    root.right.left = Node(5)
    root.right.right = Node(7)
    root.left.left.left = Node(10)
    root.left.left.right = Node(11)
    root.right.right.left = Node(8)

    antiClockWiseSpiral(root)

C#

// C# implementation of the above approach
using System;
using System.Collections;
using System.Collections.Generic;

public class GFG {

  // Structure of each node
  class Node {
    public int val;
    public Node left, right;
    public Node(int val)
    {
      this.val = val;
      this.left = this.right = null;
    }
  }

  private void work(Node root)
  {
    // Initialize queue
    Queue q = new Queue();

    // Add the root node
    q.Enqueue(root);

    // Initialize the vector
    List<Node> topone = new List<Node>();

    // Until queue is not empty
    while (q.Count != 0) {
      int len = q.Count;

      // len is greater than zero
      while (len > 0) {
        Node nd = (Node)q.Dequeue();
        if (nd != null) {
          topone.Add(nd);
          if (nd.right != null)
            q.Enqueue(nd.right);
          if (nd.left != null)
            q.Enqueue(nd.left);
        }
        len--;
      }
      topone.Add(null);
    }
    bool top = true;
    int l = 0, r = topone.Count - 2;

    while (l < r) {
      if (top) {
        while (l < topone.Count) {
          Node nd = topone[l++];
          if (nd == null) {
            break;
          }
          Console.Write(nd.val + " ");
        }
      }
      else {
        while (r >= l) {
          Node nd = topone[r--];
          if (nd == null)
            break;
          Console.Write(nd.val + " ");
        }
      }
      top = !top;
    }
  }

  // Build Tree
  public void solve()
  {
    /*
                            1
                         2     3
                       4     5   7
                      10 11     8
                    */

    Node root = new Node(1);
    root.left = new Node(2);
    root.right = new Node(3);
    root.left.left = new Node(4);
    root.right.left = new Node(5);
    root.right.right = new Node(7);
    root.left.left.left = new Node(10);
    root.left.left.right = new Node(11);
    root.right.right.left = new Node(8);

    // Function call
    work(root);
  }

  static public void Main()
  {

    // Code
    GFG t = new GFG();
    t.solve();
  }
}

// This code is contributed by lokeshmvs21.

Javascript

<script>

// JavaScript code to implement the above approach

class Node {

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

}

function antiClockWiseSpiral(root){

    // Initialize the queue
    let q = [root]

    // Initialize the list
    let topone=[]

    // Until queue is not empty
    while (q.length>0){
        let l = q.length

        // l is greater than zero
        while (l > 0){
            let nd = q.shift()
            if (nd != null){
                topone.push(nd)
                if (nd.right != null)
                    q.push(nd.right)
                if (nd.left != null)
                    q.push(nd.left)
            }
            l-=1
        }

        topone.push(null)
    }

    let top = true
    let l = 0,r = topone.length - 2

    while (l < r){
        if (top){
            while (l < topone.length){
                let nd = topone[l]
                l+=1
                if (nd == null)
                    break
                document.write(nd.data," ")
            }
        }
        else{
            while (r >= l){
                let nd = topone[r]
                r-=1
                if (nd == null)
                    break
                document.write(nd.data," ")
            }
        }

        top = top^1
    }
    document.write("</br>")
}


// driver code
// Build Tree

let root = new Node(1)
root.left = new Node(2)
root.right = new Node(3)
root.left.left = new Node(4)
root.right.left = new Node(5)
root.right.right = new Node(7)
root.left.left.left = new Node(10)
root.left.left.right = new Node(11)
root.right.right.left = new Node(8)

antiClockWiseSpiral(root)
// code is contributed by shinjanpatra

</script>

Output:

1 10 11 8 3 2 4 5 7 

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