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
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 nodestruct 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 treeint 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 rightvoid 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 leftvoid 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 arrayvoid 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 codeint 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).
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