Skip to content
Related Articles

Related Articles

Improve Article
Save Article
Like Article

Clockwise Spiral Traversal of Binary Tree | Set – 2

  • Difficulty Level : Expert
  • Last Updated : 11 Aug, 2021

Given a Binary Tree. The task is to print the circular clockwise spiral order traversal of the given binary tree.
Examples: 
 

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

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

 

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.  To complete your preparation from learning a language to DS Algo and many more,  please refer Complete Interview Preparation Course.

In case you wish to attend live classes with experts, please refer DSA Live Classes for Working Professionals and Competitive Programming Live for Students.

We have already discussed Clockwise spiral traversal of binary tree using a 2D array. Here we will discuss another approach which will not use 2D array.
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 left to right and mark flag as 1 so that next time we traverse the tree from right to left 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 right to left 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 clockwise spiral
// traversal of a binary tree without using 2D array
void ClockWiseSpiral(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 left to right
        if (flag == 0) {
            leftToRight(root, i);
 
            // Set the value of flag as zero
            // so that nodes are next time
            // printed from right to left
            flag = 1;
 
            // Increment i
            i++;
        }
 
        // If flag is one print nodes
        // from right to left
        else {
            rightToLeft(root, j);
 
            // Set the value of flag as zero
            // so that nodes are next time
            // printed from left to right
            flag = 0;
 
            // Decrement j
            j--;
        }
    }
}
 
// Driver code
int main()
{
    struct Node* root = new Node(10);
 
    root->left = new Node(12);
    root->right = new Node(13);
 
    root->right->left = new Node(14);
    root->right->right = new Node(15);
 
    root->right->left->left = new Node(21);
    root->right->left->right = new Node(22);
    root->right->right->left = new Node(23);
    root->right->right->right = new Node(24);
 
    ClockWiseSpiral(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 clockwise spiral
// traversal of a binary tree without using 2D array
static void ClockWiseSpiral(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 left to right
        if (flag == 0)
        {
            leftToRight(root, i);
 
            // Set the value of flag as zero
            // so that nodes are next time
            // printed from right to left
            flag = 1;
 
            // Increment i
            i++;
        }
 
        // If flag is one print nodes
        // from right to left
        else
        {
            rightToLeft(root, j);
 
            // Set the value of flag as zero
            // so that nodes are next time
            // printed from left to right
            flag = 0;
 
            // Decrement j
            j--;
        }
    }
}
 
// Driver code
public static void main(String[] args)
{
    Node root = new Node(10);
 
    root.left = new Node(12);
    root.right = new Node(13);
 
    root.right.left = new Node(14);
    root.right.right = new Node(15);
 
    root.right.left.left = new Node(21);
    root.right.left.right = new Node(22);
    root.right.right.left = new Node(23);
    root.right.right.right = new Node(24);
 
    ClockWiseSpiral(root);
}
}
 
// This code is contributed by Prerna Saini

Python3




# Python3 implementation of the approach
 
# Binary tree
class Node:
     
    # Constructor to create a new node
    def __init__(self, data):
        self.data = data
        self.left = self.right = None
 
# 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 clockwise spiral
# traversal of a binary tree without using 2D array
def ClockWiseSpiral(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 left to right
        if (flag == 0) :
            leftToRight(root, i)
             
            # Set the value of flag as zero
            # so that nodes are next time
            # printed from right to left
            flag = 1
             
            # Increment i
            i += 1
         
        # If flag is one print nodes
        # from right to left
        else:
            rightToLeft(root, j)
             
            # Set the value of flag as zero
            # so that nodes are next time
            # printed from left to right
            flag = 0
             
            # Decrement j
            j -= 1
 
# Driver code
 
root = Node(10)
 
root.left = Node(12)
root.right = Node(13)
 
root.right.left = Node(14)
root.right.right = Node(15)
 
root.right.left.left = Node(21)
root.right.left.right = Node(22)
root.right.right.left = Node(23)
root.right.right.right = Node(24)
 
ClockWiseSpiral(root)
 
# This code is contributed by SHUBHAMSINGH10

C#




// C# implementation of the approach
using System;
using System.Collections.Generic;
 
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 clockwise spiral
// traversal of a binary tree without using 2D array
static void ClockWiseSpiral(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 left to right
        if (flag == 0)
        {
            leftToRight(root, i);
 
            // Set the value of flag as zero
            // so that nodes are next time
            // printed from right to left
            flag = 1;
 
            // Increment i
            i++;
        }
 
        // If flag is one print nodes
        // from right to left
        else
        {
            rightToLeft(root, j);
 
            // Set the value of flag as zero
            // so that nodes are next time
            // printed from left to right
            flag = 0;
 
            // Decrement j
            j--;
        }
    }
}
 
// Driver code
public static void Main(String[] args)
{
    Node root = new Node(10);
 
    root.left = new Node(12);
    root.right = new Node(13);
 
    root.right.left = new Node(14);
    root.right.right = new Node(15);
 
    root.right.left.left = new Node(21);
    root.right.left.right = new Node(22);
    root.right.right.left = new Node(23);
    root.right.right.right = new Node(24);
 
    ClockWiseSpiral(root);
}
}
 
/* This code contributed by PrinciRaj1992 */

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 clockwise spiral
    // traversal of a binary tree without using 2D array
    function ClockWiseSpiral(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 left to right
            if (flag == 0)
            {
                leftToRight(root, i);
 
                // Set the value of flag as zero
                // so that nodes are next time
                // printed from right to left
                flag = 1;
 
                // Increment i
                i++;
            }
 
            // If flag is one print nodes
            // from right to left
            else
            {
                rightToLeft(root, j);
 
                // Set the value of flag as zero
                // so that nodes are next time
                // printed from left to right
                flag = 0;
 
                // Decrement j
                j--;
            }
        }
    }
     
    let root = new Node(10);
   
    root.left = new Node(12);
    root.right = new Node(13);
   
    root.right.left = new Node(14);
    root.right.right = new Node(15);
   
    root.right.left.left = new Node(21);
    root.right.left.right = new Node(22);
    root.right.right.left = new Node(23);
    root.right.right.right = new Node(24);
   
    ClockWiseSpiral(root);
 
</script>
Output: 
10 24 23 22 21 12 13 15 14

 

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




My Personal Notes arrow_drop_up
Recommended Articles
Page :