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++
// 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 anti clockwise spiral // traversal of a binary treevoid 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 codeint 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 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 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 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 Codeif __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 approachusing System;class GFG{// Binary tree nodepublic 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 treestatic 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 rightstatic 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 leftstatic 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 treestatic 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 codepublic 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 |
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 anti clockwise spiral // traversal of a binary tree function antiClockWiseSpiral(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 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(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); </script> |
Output:
1 10 11 8 3 2 4 5 7
Another Approach:
The above approach have O(n^2) worst case complexity due to calling the print level everytime. An improvement over it can be storing the level wise nodes and use it to print.
C++
// C++ implementation of the above approach#include <bits/stdc++.h>using namespace std;struct Node { struct Node* left; struct Node* right; int data; Node(int data) { this->data = data; this->left = NULL; this->right = NULL; }};void antiClockWiseSpiral(struct Node* root){ // Initialize the queue queue<Node*> q; // Add the root node q.push(root); // Initialize the vector vector<Node*> topone; // Until queue is not empty while (!q.empty()) { int len = q.size(); // len is greater than zero while (len > 0) { Node* nd = q.front(); q.pop(); if (nd != NULL) { topone.push_back(nd); if (nd->right != NULL) q.push(nd->right); if (nd->left != NULL) q.push(nd->left); } len--; } topone.push_back(NULL); } bool top = true; int l = 0, r = (int)topone.size() - 2; while (l < r) { if (top) { while (l < (int)topone.size()) { Node* nd = topone[l++]; if (nd == NULL) { break; } cout << nd->data << " "; } } else { while (r >= l) { Node* nd = topone[r--]; if (nd == NULL) break; cout << nd->data << " "; } } top = !top; }}// Build Treeint main(){ /* 1 2 3 4 5 7 10 11 8 */ 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 above approachimport 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(); }} |
Python3
# Python 3 implementation of the above approachfrom collections import deque as dqclass Node: def __init__(self, data): self.data = data self.left = None self.right = Nonedef antiClockWiseSpiral(root): # Initialize the queue q=dq([root]) # Initialize the list topone=[] # Until queue is not empty while (q): l = len(q) # l is greater than zero while (l > 0): nd = q.popleft() if (nd != None): topone.append(nd) if (nd.right != None): q.append(nd.right) if (nd.left != None): q.append(nd.left) l-=1 topone.append(None) top = True l = 0; r = len(topone) - 2 while (l < r): if (top): while (l < len(topone)): nd = topone[l] l+=1 if (nd == None): break print(nd.data,end=" ") else: while (r >= l): nd = topone[r] r-=1 if (nd == None): break print(nd.data,end=" ") top = not top print()# Build Treeif __name__ == '__main__': # 1 # 2 3 # 4 5 7 # 10 11 8 root = Node(1) root.left = Node(2) root.right = Node(3) root.left.left = Node(4) root.right.left = Node(5) root.right.right = Node(7) root.left.left.left = Node(10) root.left.left.right = Node(11) root.right.right.left = Node(8) antiClockWiseSpiral(root) |
C#
// C# implementation of the above approachusing System;using System.Collections;using System.Collections.Generic;public class GFG { // Structure of each node class Node { public int val; public Node left, right; public Node(int val) { this.val = val; this.left = this.right = null; } } private void work(Node root) { // Initialize queue Queue q = new Queue(); // Add the root node q.Enqueue(root); // Initialize the vector List<Node> topone = new List<Node>(); // Until queue is not empty while (q.Count != 0) { int len = q.Count; // len is greater than zero while (len > 0) { Node nd = (Node)q.Dequeue(); if (nd != null) { topone.Add(nd); if (nd.right != null) q.Enqueue(nd.right); if (nd.left != null) q.Enqueue(nd.left); } len--; } topone.Add(null); } bool top = true; int l = 0, r = topone.Count - 2; while (l < r) { if (top) { while (l < topone.Count) { Node nd = topone[l++]; if (nd == null) { break; } Console.Write(nd.val + " "); } } else { while (r >= l) { Node nd = topone[r--]; if (nd == null) break; Console.Write(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); } static public void Main() { // Code GFG t = new GFG(); t.solve(); }}// This code is contributed by lokeshmvs21. |
Javascript
<script>// JavaScript code to implement the above approachclass Node { constructor(data){ this.data = data; this.left = null,this.right = null; }}function antiClockWiseSpiral(root){ // Initialize the queue let q = [root] // Initialize the list let topone=[] // Until queue is not empty while (q.length>0){ let l = q.length // l is greater than zero while (l > 0){ let nd = q.shift() if (nd != null){ topone.push(nd) if (nd.right != null) q.push(nd.right) if (nd.left != null) q.push(nd.left) } l-=1 } topone.push(null) } let top = true let l = 0,r = topone.length - 2 while (l < r){ if (top){ while (l < topone.length){ let nd = topone[l] l+=1 if (nd == null) break document.write(nd.data," ") } } else{ while (r >= l){ let nd = topone[r] r-=1 if (nd == null) break document.write(nd.data," ") } } top = top^1 } document.write("</br>")}// driver code// Build Treelet 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)// code is contributed by shinjanpatra</script> |
Output:
1 10 11 8 3 2 4 5 7
Time Complexity: O(N)
Auxiliary Space: O(N)
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