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Reverse Anti Clockwise Spiral Traversal of a Binary Tree

  • Difficulty Level : Hard
  • Last Updated : 11 Aug, 2021
Geek Week

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

Examples: 

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

Input :
           20
         /   \
        8     22
      /   \  /   \
     5     3 4    25
          / \
         10  14
Output : 10 14 20 5 3 4 25 22 8

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 1. Now in the while loop we will check a condition that if flag is equal to 1 we will traverse the tree from left to right and mark flag as 0 so that next time we traverse the tree from right to left and then decrement the value of j so that next time we visit the level just above the current level. Also when we will traverse the level from top we will 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. 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 Reverse anti clockwise spiral
// traversal of a binary tree
void ReverseAntiClockWiseSpiral(struct Node* root)
{
    int i = 1;
    int j = height(root);
 
    // Flag to mark a change in the direction
    // of printing nodes
    int flag = 1;
    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(20);
    root->left = new Node(8);
    root->right = new Node(22);
    root->left->left = new Node(5);
    root->left->right = new Node(3);
    root->right->left = new Node(4);
    root->right->right = new Node(25);
    root->left->right->left = new Node(10);
    root->left->right->right = new Node(14);
 
    ReverseAntiClockWiseSpiral(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 Reverse anti clockwise spiral
// traversal of a binary tree
static void ReverseAntiClockWiseSpiral(Node root)
{
    int i = 1;
    int j = height(root);
 
    // Flag to mark a change in the direction
    // of printing nodes
    int flag = 1;
    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(20);
    root.left = new Node(8);
    root.right = new Node(22);
    root.left.left = new Node(5);
    root.left.right = new Node(3);
    root.right.left = new Node(4);
    root.right.right = new Node(25);
    root.left.right.left = new Node(10);
    root.left.right.right = new Node(14);
 
    ReverseAntiClockWiseSpiral(root);
 
}
}
 
// This code is contributed by Prerna Saini.

Python3




# Python3 implementation of the approach
  
# Binary tree node
class Node:
     
    def __init__(self, data):
         
        self.left = None
        self.right = None
        self.data = data
         
# 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 Reverse anti clockwise
# spiral traversal of a binary tree
def ReverseAntiClockWiseSpiral(root):
 
    i = 1
    j = height(root)
  
    # Flag to mark a change in the
    # direction of printing nodes
    flag = 1;
     
    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 = Node(20)
    root.left = Node(8)
    root.right = Node(22)
    root.left.left = Node(5)
    root.left.right = Node(3)
    root.right.left = Node(4)
    root.right.right = Node(25)
    root.left.right.left = Node(10)
    root.left.right.right = Node(14)
  
    ReverseAntiClockWiseSpiral(root)
 
# This code is contributed by rutvik_56

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 Reverse anti clockwise spiral
// traversal of a binary tree
static void ReverseAntiClockWiseSpiral(Node root)
{
    int i = 1;
    int j = height(root);
 
    // Flag to mark a change in the direction
    // of printing nodes
    int flag = 1;
    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(20);
    root.left = new Node(8);
    root.right = new Node(22);
    root.left.left = new Node(5);
    root.left.right = new Node(3);
    root.right.left = new Node(4);
    root.right.right = new Node(25);
    root.left.right.left = new Node(10);
    root.left.right.right = new Node(14);
 
    ReverseAntiClockWiseSpiral(root);
 
}
}
 
// This code has been contributed by 29AjayKumar

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 Reverse anti clockwise spiral
    // traversal of a binary tree
    function ReverseAntiClockWiseSpiral(root)
    {
        let i = 1;
        let j = height(root);
 
        // Flag to mark a change in the direction
        // of printing nodes
        let flag = 1;
        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(20);
    root.left = new Node(8);
    root.right = new Node(22);
    root.left.left = new Node(5);
    root.left.right = new Node(3);
    root.right.left = new Node(4);
    root.right.right = new Node(25);
    root.left.right.left = new Node(10);
    root.left.right.right = new Node(14);
  
    ReverseAntiClockWiseSpiral(root);
 
</script>
Output: 



10 14 20 5 3 4 25 22 8    

 

Time Complexity: O(N^2), where N is the total number of nodes in the binary tree. 
Auxiliary Space: O(N)

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