Given a stack s, the task is to print the elements of the stack from bottom to top, such that the elements are still present in the stack without their order being changed in the stack.
Examples:
Input :
| 4 |
| 3 |
| 2 |
| 1 |
|________|
Output :1 2 3 4
Approach 1 (Recursion): The idea is to pop the element of the stack and call the recursive function PrintStack. Once the stack becomes empty start printing the element which was popped last and the last element that was popped was the bottom-most element. Thus, elements will be printed from bottom to top. Now push back the element that was printed, this will preserve the order of the elements in the stack.
Below is the implementation of the above approach:
C++
// C++ program to print the elements of a// stack from bottom to top#include <bits/stdc++.h>using namespace std;// Recursive function to print stack elements// from bottom to top without changing// their ordervoid PrintStack(stack<int> s){ // If stack is empty then return if (s.empty()) return; int x = s.top(); // Pop the top element of the stack s.pop(); // Recursively call the function PrintStack PrintStack(s); // Print the stack element starting // from the bottom cout << x << " "; // Push the same element onto the stack // to preserve the order s.push(x);}// Driver codeint main(){ // Stack s stack<int> s; s.push(1); s.push(2); s.push(3); s.push(4); PrintStack(s); return 0;} |
Java
// Java program to print the elements of a // stack from bottom to topimport java.util.*;class GfG { // Recursive function to print stack elements // from bottom to top without changing // their order static void PrintStack(Stack<Integer> s) { // If stack is empty then return if (s.isEmpty()) return; int x = s.peek(); // Pop the top element of the stack s.pop(); // Recursively call the function PrintStack PrintStack(s); // Print the stack element starting // from the bottom System.out.print(x + " "); // Push the same element onto the stack // to preserve the order s.push(x); } // Driver code public static void main(String[] args) { // Stack s Stack<Integer> s = new Stack<Integer> (); s.push(1); s.push(2); s.push(3); s.push(4); PrintStack(s); }} // This code is contributed by Prerna Saini. |
Python3
# Python3 program to print the elements of a # stack from bottom to top # Stack class with all functionality of a stackimport sysclass Stack: def __init__(self): self.s = [] def push(self, data): self.s.append(data) def pop(self): return self.s.pop() def peek(self): return self.s[-1] def count(self): return len(self.s)# Recursive function to print stack elements # from bottom to top without changing # their order def printStack(s): # if stack is empty then simply return if s.count() == 0: return x = s.peek() # pop top most element of the stack s.pop() # recursively call the function printStack printStack(s) # Print the stack element starting # from the bottom print("{} ".format(x), end = "") # Push the same element onto the stack # to preserve the order s.push(x)# Driver codeif __name__=='__main__': s=Stack() s.push(1) s.push(2) s.push(3) s.push(4) printStack(s)# This code is contributed by Vikas Kumar |
C#
// C# program to print the elements of a // stack from bottom to topusing System;using System.Collections.Generic;class GfG { // Recursive function to print stack elements // from bottom to top without changing // their order static void PrintStack(Stack<int> s) { // If stack is empty then return if (s.Count == 0) return; int x = s.Peek(); // Pop the top element of the stack s.Pop(); // Recursively call the function PrintStack PrintStack(s); // Print the stack element starting // from the bottom Console.Write(x + " "); // Push the same element onto the stack // to preserve the order s.Push(x); } // Driver code public static void Main() { // Stack s Stack<int> s = new Stack<int> (); s.Push(1); s.Push(2); s.Push(3); s.Push(4); PrintStack(s); }} /* This code contributed by PrinciRaj1992 */ |
Javascript
<script>// Javascript program to print the elements// of a stack from bottom to top// Recursive function to print stack// elements from bottom to top without // changing their orderfunction PrintStack(s){ // If stack is empty then return if (s.length == 0) return; var x = s[s.length - 1]; // Pop the top element of the stack s.pop(); // Recursively call the function PrintStack PrintStack(s); // Print the stack element starting // from the bottom document.write(x + " "); // Push the same element onto the stack // to preserve the order s.push(x);}// Driver code// Stack svar s = [];s.push(1);s.push(2);s.push(3);s.push(4);PrintStack(s);// This code is contributed by importantly</script> |
Output:
1 2 3 4
Time Complexity: O(N)
Auxiliary Space: O(N)
Approach 2 (Using another stack): The idea is to push every element into another temporary stack and then print elements of the temporary stack.
C++
// C++ program to print the elements of a// stack from bottom to top#include <bits/stdc++.h>using namespace std;// Recursive function to print stack elements// from bottom to top without changing// their ordervoid PrintStack(stack<int> s){ stack<int> temp; while (s.empty() == false) { temp.push(s.top()); s.pop(); } while (temp.empty() == false) { int t = temp.top(); cout << t << " "; temp.pop(); // To restore contents of // the original stack. s.push(t); }}// Driver codeint main(){ // Stack s stack<int> s; s.push(1); s.push(2); s.push(3); s.push(4); PrintStack(s); return 0;} |
Java
// Java program to print the // elements of a stack from // bottom to topimport java.util.*;class Main{ // Recursive function to print // stack elements from bottom // to top without changing// their orderpublic static void PrintStack(Stack<Integer> s){ Stack<Integer> temp = new Stack<Integer>(); while (s.empty() == false) { temp.push(s.peek()); s.pop(); } while (temp.empty() == false) { int t = temp.peek(); System.out.print(t + " "); temp.pop(); // To restore contents of // the original stack. s.push(t); }}// Driver codepublic static void main(String[] args) { // Stack s Stack<Integer> s = new Stack<Integer>(); s.push(1); s.push(2); s.push(3); s.push(4); PrintStack(s);}}// This code is contributed by divyeshrabadiya07 |
Python3
# Python3 program to print the elements of a # stack from bottom to top # Stack class with all functionality of a stackimport sysclass Stack: def __init__(self): self.s = [] def push(self, data): self.s.append(data) def pop(self): return self.s.pop() def peek(self): return self.s[-1] def count(self): return len(self.s)# Recursive function to print stack elements # from bottom to top without changing # their order def printStack(s): temp = Stack() while(s.count() > 0): temp.push(s.peek()) s.pop() while(temp.count() > 0): t = temp.peek() print("{} " . format(temp.peek()), end = "") temp.pop() # Restore the contents of original stack s.push(t)# Driver codeif __name__=='__main__': s = Stack() s.push(1) s.push(2) s.push(3) s.push(4) printStack(s)# This code is contributed by Vikash Kumar 37 |
C#
// C# program to print the elements of// a stack from bottom to topusing System;using System.Collections;class GFG{ // Recursive function to print stack // elements from bottom to top without// changing their orderstatic void PrintStack(Stack s){ Stack temp = new Stack(); while (s.Count != 0) { temp.Push(s.Peek()); s.Pop(); } while (temp.Count != 0) { int t = (int)temp.Peek(); Console.Write(t + " "); temp.Pop(); // To restore contents of // the original stack. s.Push(t); }}// Driver Code public static void Main(string[] args) { // Stack s Stack s = new Stack(); s.Push(1); s.Push(2); s.Push(3); s.Push(4); PrintStack(s);} } // This code is contributed by rutvik_56 |
Javascript
<script>// JavaScript program to print the elements of a// stack from bottom to top// Recursive function to print stack elements// from bottom to top without changing// their orderfunction PrintStack(s){ var temp = []; while (s.length!=0) { temp.push(s[s.length-1]); s.pop(); } while (temp.length!=0) { var t = temp[temp.length-1]; document.write( t + " "); temp.pop(); // To restore contents of // the original stack. s.push(t); }}// Driver code// Stack svar s = [];s.push(1);s.push(2);s.push(3);s.push(4);PrintStack(s);</script> |
Output:
1 2 3 4
Time Complexity: O(N)
Auxiliary Space: O(N)
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