Level order traversal in spiral form | Using one stack and one queue
Last Updated :
26 Aug, 2022
Write a function to print spiral order traversal of a tree. For below tree, function should print 1, 2, 3, 4, 5, 6, 7.
You are allowed to use only one stack.
We have seen recursive and iterative solutions using two stacks . In this post, a solution with one stack and one queue is discussed. The idea is to keep on entering nodes like normal level order traversal, but during printing, in alternative turns push them onto the stack and print them, and in other traversals, just print them the way they are present in the queue.
Following is the implementation of the idea.
C++
// CPP program to print level order traversal
// in spiral form using one queue and one stack.
#include <bits/stdc++.h>
using namespace std;
struct Node {
int data;
Node *left, *right;
};
/* Utility function to create a new tree node */
Node* newNode(int val)
{
Node* new_node = new Node;
new_node->data = val;
new_node->left = new_node->right = NULL;
return new_node;
}
/* Function to print a tree in spiral form
using one stack */
void printSpiralUsingOneStack(Node* root)
{
if (root == NULL)
return;
stack<int> s;
queue<Node*> q;
bool reverse = true;
q.push(root);
while (!q.empty()) {
int size = q.size();
while (size) {
Node* p = q.front();
q.pop();
// if reverse is true, push node's
// data onto the stack, else print it
if (reverse)
s.push(p->data);
else
cout << p->data << " ";
if (p->left)
q.push(p->left);
if (p->right)
q.push(p->right);
size--;
}
// print nodes from the stack if
// reverse is true
if (reverse) {
while (!s.empty()) {
cout << s.top() << " ";
s.pop();
}
}
// the next row has to be printed as
// it is, hence change the value of
// reverse
reverse = !reverse;
}
}
// Driver Code
int main()
{
Node* root = newNode(1);
root->left = newNode(2);
root->right = newNode(3);
root->left->left = newNode(7);
root->left->right = newNode(6);
root->right->left = newNode(5);
root->right->right = newNode(4);
printSpiralUsingOneStack(root);
return 0;
}
// Java program to print level order traversal
// in spiral form using one queue and one stack.
import java.util.*;
class GFG
{
static class Node
{
int data;
Node left, right;
};
/* Utility function to create a new tree node */
static Node newNode(int val)
{
Node new_node = new Node();
new_node.data = val;
new_node.left = new_node.right = null;
return new_node;
}
/* Function to print a tree in spiral form
using one stack */
static void printSpiralUsingOneStack(Node root)
{
if (root == null)
return;
Stack<Integer> s = new Stack<Integer>();
Queue<Node> q = new LinkedList<Node>();
boolean reverse = true;
q.add(root);
while (!q.isEmpty())
{
int size = q.size();
while (size > 0)
{
Node p = q.peek();
q.remove();
// if reverse is true, push node's
// data onto the stack, else print it
if (reverse)
s.add(p.data);
else
System.out.print(p.data + " ");
if (p.left != null)
q.add(p.left);
if (p.right != null)
q.add(p.right);
size--;
}
// print nodes from the stack if
// reverse is true
if (reverse)
{
while (!s.empty())
{
System.out.print(s.peek() + " ");
s.pop();
}
}
// the next row has to be printed as
// it is, hence change the value of
// reverse
reverse = !reverse;
}
}
// Driver Code
public static void main(String[] args)
{
Node root = newNode(1);
root.left = newNode(2);
root.right = newNode(3);
root.left.left = newNode(7);
root.left.right = newNode(6);
root.right.left = newNode(5);
root.right.right = newNode(4);
printSpiralUsingOneStack(root);
}
}
// This code is contributed by Princi Singh
# Python program to print level order traversal
# in spiral form using one queue and one stack.
# Utility class to create a new node
class Node:
def __init__(self, key):
self.key = key
self.left = self.right = None
# Utility function to create a new tree node
def newNode(val):
new_node = Node(0)
new_node.data = val
new_node.left = new_node.right = None
return new_node
# Function to print a tree in spiral form
# using one stack
def printSpiralUsingOneStack(root):
if (root == None):
return
s = []
q = []
reverse = True
q.append(root)
while (len(q) > 0) :
size = len(q)
while (size > 0) :
p = q[0]
q.pop(0)
# if reverse is true, push node's
# data onto the stack, else print it
if (reverse):
s.append(p.data)
else:
print( p.data ,end = " ")
if (p.left != None):
q.append(p.left)
if (p.right != None):
q.append(p.right)
size = size - 1
# print nodes from the stack if
# reverse is true
if (reverse) :
while (len(s)) :
print( s[-1],end= " ")
s.pop()
# the next row has to be printed as
# it is, hence change the value of
# reverse
reverse = not reverse
# Driver Code
root = newNode(1)
root.left = newNode(2)
root.right = newNode(3)
root.left.left = newNode(7)
root.left.right = newNode(6)
root.right.left = newNode(5)
root.right.right = newNode(4)
printSpiralUsingOneStack(root)
# This code is contributed by Arnab Kundu
// C# program to print level order traversal
// in spiral form using one queue and one stack.
using System;
using System.Collections.Generic;
class GFG
{
public class Node
{
public int data;
public Node left, right;
};
/* Utility function to create a new tree node */
static Node newNode(int val)
{
Node new_node = new Node();
new_node.data = val;
new_node.left = new_node.right = null;
return new_node;
}
/* Function to print a tree in spiral form
using one stack */
static void printSpiralUsingOneStack(Node root)
{
if (root == null)
return;
Stack<int> s = new Stack<int>();
Queue<Node> q = new Queue<Node>();
Boolean reverse = true;
q.Enqueue(root);
while (q.Count != 0)
{
int size = q.Count;
while (size > 0)
{
Node p = q.Peek();
q.Dequeue();
// if reverse is true, push node's
// data onto the stack, else print it
if (reverse)
s.Push(p.data);
else
Console.Write(p.data + " ");
if (p.left != null)
q.Enqueue(p.left);
if (p.right != null)
q.Enqueue(p.right);
size--;
}
// print nodes from the stack if
// reverse is true
if (reverse)
{
while (s.Count != 0)
{
Console.Write(s.Peek() + " ");
s.Pop();
}
}
// the next row has to be printed as
// it is, hence change the value of
// reverse
reverse = !reverse;
}
}
// Driver Code
public static void Main(String[] args)
{
Node root = newNode(1);
root.left = newNode(2);
root.right = newNode(3);
root.left.left = newNode(7);
root.left.right = newNode(6);
root.right.left = newNode(5);
root.right.right = newNode(4);
printSpiralUsingOneStack(root);
}
}
// This code is contributed by Rajput-Jiv
<script>
// Javascript program to print level
// order traversal in spiral form
// using one queue and one stack.
class Node
{
constructor()
{
this.data = 0;
this.left = null;
this.right = null;
}
};
// Utility function to create
// a new tree node
function newNode(val)
{
var new_node = new Node();
new_node.data = val;
new_node.left = new_node.right = null;
return new_node;
}
// Function to print a tree in spiral form
// using one stack
function printSpiralUsingOneStack(root)
{
if (root == null)
return;
var s = [];
var q = [];
var reverse = true;
q.push(root);
while (q.length != 0)
{
var size = q.length;
while (size > 0)
{
var p = q[0];
q.shift();
// If reverse is true, push node's
// data onto the stack, else print it
if (reverse)
s.push(p.data);
else
document.write(p.data + " ");
if (p.left != null)
q.push(p.left);
if (p.right != null)
q.push(p.right);
size--;
}
// Print nodes from the stack if
// reverse is true
if (reverse)
{
while (s.length != 0)
{
document.write(s[s.length - 1] + " ");
s.pop();
}
}
// The next row has to be printed as
// it is, hence change the value of
// reverse
reverse = !reverse;
}
}
// Driver Code
var root = newNode(1);
root.left = newNode(2);
root.right = newNode(3);
root.left.left = newNode(7);
root.left.right = newNode(6);
root.right.left = newNode(5);
root.right.right = newNode(4);
printSpiralUsingOneStack(root);
// This code is contributed by rutvik_56
</script>
Complexity Analysis:
- Time Complexity : O(n)
- Auxiliary Space : O(n)
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