Level order traversal in spiral form | Using Deque
Given a Binary Tree, the task is to print spiral order traversal of the given tree. For below tree, the function should print 1, 2, 3, 4, 5, 6, 7.
Examples:
Input: 1 / \ 3 2 Output : 1 3 2 Input : 10 / \ 20 30 / \ 40 60 Output : 10 20 30 60 40
We have seen recursive and iterative solutions using two stacks and an approach using one stack and one queue. In this post, a solution with one deque is discussed. The idea is to use a direction variable and decide whether to pop elements from the front or from the rear based on the value of this direction variable.
Below is the implementation of the above approach:
C++
// C++ program to print level order traversal // in spiral form using one deque. #include <bits/stdc++.h> using namespace std; class Node { public : int data; Node *left, *right; Node( int val) { data = val; left = NULL; right = NULL; } }; void spiralOrder(Node* root) { deque<Node*> d; // Push root d.push_back(root); // Direction 0 shows print right to left // and for Direction 1 left to right int dir = 0; while (!d.empty()) { int size = d.size(); while (size--) { // One whole level // will be print in this loop if (dir == 0) { Node* temp = d.back(); d.pop_back(); if (temp->right) d.push_front(temp->right); if (temp->left) d.push_front(temp->left); cout << temp->data << " " ; } else { Node* temp = d.front(); d.pop_front(); if (temp->left) d.push_back(temp->left); if (temp->right) d.push_back(temp->right); cout << temp->data << " " ; } } cout << endl; // Direction change dir = 1 - dir; } } int main() { // Build the Tree Node* root = new Node(10); root->left = new Node(20); root->right = new Node(30); root->left->left = new Node(40); root->left->right = new Node(60); // Call the Function spiralOrder(root); return 0; } |
Java
// Java program to print level order traversal // in spiral form using one deque. import java.util.*; class GFG { static class Node { int data; Node left, right; Node( int val) { data = val; left = null ; right = null ; } }; static void spiralOrder(Node root) { Deque<Node> d = new LinkedList<Node>(); // Push root d.addLast(root); // Direction 0 shows print right to left // and for Direction 1 left to right int dir = 0 ; while (d.size() > 0 ) { int size = d.size(); while (size--> 0 ) { // One whole level // will be print in this loop if (dir == 0 ) { Node temp = d.peekLast(); d.pollLast(); if (temp.right != null ) d.addFirst(temp.right); if (temp.left != null ) d.addFirst(temp.left); System.out.print(temp.data + " " ); } else { Node temp = d.peekFirst(); d.pollFirst(); if (temp.left != null ) d.addLast(temp.left); if (temp.right != null ) d.addLast(temp.right); System.out.print(temp.data + " " ); } } System.out.println(); // Direction change dir = 1 - dir; } } // Driver code public static void main(String args[]) { // Build the Tree Node root = new Node( 10 ); root.left = new Node( 20 ); root.right = new Node( 30 ); root.left.left = new Node( 40 ); root.left.right = new Node( 60 ); // Call the Function spiralOrder(root); } } // This code is contributed by Arnab Kundu |
Python3
# Python program to print level order traversal # in spiral form using one deque. class Node : def __init__( self ,val) : self .data = val; self .left = None ; self .right = None ; def spiralOrder(root) : d = []; # Push root d.append(root); # Direction 0 shows print right to left # and for Direction 1 left to right direct = 0 ; while ( len (d) ! = 0 ) : size = len (d); while (size) : size - = 1 ; # One whole level # will be print in this loop if (direct = = 0 ) : temp = d.pop(); if (temp.right) : d.insert( 0 , temp.right); if (temp.left) : d.insert( 0 , temp.left); print (temp.data, end = " " ); else : temp = d[ 0 ]; d.pop( 0 ); if (temp.left) : d.append(temp.left); if (temp.right) : d.append(temp.right); print (temp.data ,end = " " ); print () # Direction change direct = 1 - direct; if __name__ = = "__main__" : # Build the Tree root = Node( 10 ); root.left = Node( 20 ); root.right = Node( 30 ); root.left.left = Node( 40 ); root.left.right = Node( 60 ); # Call the Function spiralOrder(root); # This code is contributed by AnkitRai01 |
C#
//C# code for the above approach using System; using System.Collections.Generic; class GFG { class Node { public int data; public Node left, right; public Node( int val) { data = val; left = null ; right = null ; } } static void SpiralOrder(Node root) { Queue<Node> d = new Queue<Node>(); // Push root d.Enqueue(root); // Direction 0 shows print right to left // and for Direction 1 left to right int dir = 0; while (d.Count > 0) { int size = d.Count; while (size-- > 0) { // One whole level // will be print in this loop if (dir == 1) { Node temp = d.Peek(); d.Dequeue(); if (temp.right != null ) d.Enqueue(temp.right); if (temp.left != null ) d.Enqueue(temp.left); Console.Write(temp.data + " " ); } else { Node temp = d.Peek(); d.Dequeue(); if (temp.left != null ) d.Enqueue(temp.left); if (temp.right != null ) d.Enqueue(temp.right); Console.Write(temp.data + " " ); } } Console.WriteLine(); // Direction change dir = 1 - dir; } } // Driver code // Driver code static void Main( string [] args) { // Build the Tree Node root = new Node(10); root.left = new Node(20); root.right = new Node(30); root.left.left = new Node(40); root.left.right = new Node(60); // Call the Function SpiralOrder(root); } } //This code is contributed by Potta Lokesh |
Javascript
<script> // JavaScript program to print level order traversal // in spiral form using one deque. class Node { constructor(val) { this .data = val; this .left = null ; this .right = null ; } }; function spiralOrder(root) { var d = []; // Push root d.push(root); // Direction 0 shows print right to left // and for Direction 1 left to right var dir = 0; while (d.length!=0) { var size = d.length; while (size-- >0) { // One whole level // will be print in this loop if (dir == 0) { var temp = d[d.length-1]; d.pop(); if (temp.right!= null ) d.unshift(temp.right); if (temp.left!= null ) d.unshift(temp.left); document.write( temp.data + " " ); } else { var temp = d[0]; d.shift(); if (temp.left != null ) d.push(temp.left); if (temp.right!= null ) d.push(temp.right); document.write( temp.data + " " ); } } document.write( "<br>" ); // Direction change dir = 1 - dir; } } // Build the Tree var root = new Node(10); root.left = new Node(20); root.right = new Node(30); root.left.left = new Node(40); root.left.right = new Node(60); // Call the Function spiralOrder(root); </script> |
Output:
10 20 30 60 40
Time Complexity: O(N)
Space Complexity: O(N)
where N is the number of Nodes
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