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

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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 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 

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 

Output:

1 10 11 8 3 2 4 5 7

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My Personal Notes

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

C++

Java

Python3

C#

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