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Anti Clockwise spiral traversal of a binary tree
  • Difficulty Level : Hard
  • Last Updated : 27 Jan, 2021

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++

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// 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;
}

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Java

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// 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.

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Python3

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# 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 PrNodes 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 PrNodes 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 prnodes
        # 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 prnodes
        # 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

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C#

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// 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

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Output



1 10 11 8 3 2 4 5 7 

Another Approach:
The above approach have O(n^2) worst case complextity due to calling the print level everytime. An imporvement over it can be storing the level wise nodes and use it to print. Below is java code for the same.

Java

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// 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();
    }
}

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Output

1 10 11 8 3 2 4 5 7 

The above code will run in O(n) time complexity with O(n) additional space required

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