Okay, this was the bonus question in CMPS 280's test at southeastern Louisiana Univ. Print singly linked list in reverse in three lines. Any ideas?
C implementation of your bonus question, in three lines:
#include <stdio.h>
struct node {
int data;
struct node* next;
};
void ReversePrint(struct node* head) {
if(head == NULL) return;
ReversePrint(head->next);
printf("%d ", head->data);
}
int main()
{
struct node first;
struct node second;
struct node third;
first.data = 1;
second.data = 2;
third.data = 3;
first.next = &second;
second.next = &third;
ReversePrint(&first); // Should print: 3 2 1
printf("\n");
return 0;
}
If you are allowed to use another Data Structure, then use a Stack.
Step 1: Traverse the linked list from the head node and put the key into the stack, till you reach the last node. This will take O(n) time.
Step 2 : Pop the elements out from the stack. This will take O(1) time.
Therefore, code will be
while(head != null){
stack.push(head.val);
head = head.next;
}
while(stack.isEmpty() != true){
stack.pop();
}
Generally, when asking for help on SO, you should always try to do the problem yourself first. Then, if you are stuck, come here with what you have done so far and clearly show what your problem is to maximize your chances of getting help.
How to ask a good question on SO
However, since it is a past exam question and my answer wont help you cheat :), here's pseudo-code how you can do it recursively:
reversePrint(head)
if head is null then return // Line 1: base case
reversePrint(head.next) // Line 2: print the list after head
print(head.data) // Line 3: print head
Below are the different ways of doing it. Complete source code can be at found the hyperlinks below.
1) Printing using extra memory : https://www.geeksforgeeks.org/print-reverse-linked-list-using-stack/
2) Printing using recursion : https://www.geeksforgeeks.org/print-reverse-of-a-linked-list-without-actually-reversing/
3) Printing by modifying original list - i.e. first reverse the list and then print from start.
Source Code for reversing list : https://www.geeksforgeeks.org/reverse-a-linked-list/
4) Printing without using extra space or modifying original list : https://www.geeksforgeeks.org/print-reverse-linked-list-without-extra-space-modifications/
5) Printing using Carriage return (“r”) : https://www.geeksforgeeks.org/an-interesting-method-to-print-reverse-of-a-linked-list/
Below is my iterative Java solution without stacks. I suppose the space complexity is still O(n) because the length of the StringBuilder grows linearly with the number of items in the list. However, we can get away without using a stack (either the data structure or the recursive call stack), neither of which is necessary if all we're doing is printing the elements to console. Also, the iterative stack solutions use two loops whereas this method only requires one. Finally, the time complexity is still O(n), which is much better than the quadratic algorithm by Geeks for Geeks referenced in another answer.
void printInReverse(ListNode head) {
StringBuilder sb = new StringBuilder();
ListNode n = head;
while (n != null) {
sb.insert(0, "<-" + n.data);
n = n.next;
}
System.out.println(sb.toString());
}
I've used recursion quite a lot on my many years of programming to solve simple problems, but I'm fully aware that sometimes you need iteration due to memory/speed problems.
So, sometime in the very far past I went to try and find if there existed any "pattern" or text-book way of transforming a common recursion approach to iteration and found nothing. Or at least nothing that I can remember it would help.
Are there general rules?
Is there a "pattern"?
Usually, I replace a recursive algorithm by an iterative algorithm by pushing the parameters that would normally be passed to the recursive function onto a stack. In fact, you are replacing the program stack by one of your own.
var stack = [];
stack.push(firstObject);
// while not empty
while (stack.length) {
// Pop off end of stack.
obj = stack.pop();
// Do stuff.
// Push other objects on the stack as needed.
...
}
Note: if you have more than one recursive call inside and you want to preserve the order of the calls, you have to add them in the reverse order to the stack:
foo(first);
foo(second);
has to be replaced by
stack.push(second);
stack.push(first);
Edit: The article Stacks and Recursion Elimination (or Article Backup link) goes into more details on this subject.
Really, the most common way to do it is to keep your own stack. Here's a recursive quicksort function in C:
void quicksort(int* array, int left, int right)
{
if(left >= right)
return;
int index = partition(array, left, right);
quicksort(array, left, index - 1);
quicksort(array, index + 1, right);
}
Here's how we could make it iterative by keeping our own stack:
void quicksort(int *array, int left, int right)
{
int stack[1024];
int i=0;
stack[i++] = left;
stack[i++] = right;
while (i > 0)
{
right = stack[--i];
left = stack[--i];
if (left >= right)
continue;
int index = partition(array, left, right);
stack[i++] = left;
stack[i++] = index - 1;
stack[i++] = index + 1;
stack[i++] = right;
}
}
Obviously, this example doesn't check stack boundaries... and really you could size the stack based on the worst case given left and and right values. But you get the idea.
It seems nobody has addressed where the recursive function calls itself more than once in the body, and handles returning to a specific point in the recursion (i.e. not primitive-recursive). It is said that every recursion can be turned into iteration, so it appears that this should be possible.
I just came up with a C# example of how to do this. Suppose you have the following recursive function, which acts like a postorder traversal, and that AbcTreeNode is a 3-ary tree with pointers a, b, c.
public static void AbcRecursiveTraversal(this AbcTreeNode x, List<int> list) {
if (x != null) {
AbcRecursiveTraversal(x.a, list);
AbcRecursiveTraversal(x.b, list);
AbcRecursiveTraversal(x.c, list);
list.Add(x.key);//finally visit root
}
}
The iterative solution:
int? address = null;
AbcTreeNode x = null;
x = root;
address = A;
stack.Push(x);
stack.Push(null)
while (stack.Count > 0) {
bool #return = x == null;
if (#return == false) {
switch (address) {
case A://
stack.Push(x);
stack.Push(B);
x = x.a;
address = A;
break;
case B:
stack.Push(x);
stack.Push(C);
x = x.b;
address = A;
break;
case C:
stack.Push(x);
stack.Push(null);
x = x.c;
address = A;
break;
case null:
list_iterative.Add(x.key);
#return = true;
break;
}
}
if (#return == true) {
address = (int?)stack.Pop();
x = (AbcTreeNode)stack.Pop();
}
}
Strive to make your recursive call Tail Recursion (recursion where the last statement is the recursive call). Once you have that, converting it to iteration is generally pretty easy.
Well, in general, recursion can be mimicked as iteration by simply using a storage variable. Note that recursion and iteration are generally equivalent; one can almost always be converted to the other. A tail-recursive function is very easily converted to an iterative one. Just make the accumulator variable a local one, and iterate instead of recurse. Here's an example in C++ (C were it not for the use of a default argument):
// tail-recursive
int factorial (int n, int acc = 1)
{
if (n == 1)
return acc;
else
return factorial(n - 1, acc * n);
}
// iterative
int factorial (int n)
{
int acc = 1;
for (; n > 1; --n)
acc *= n;
return acc;
}
Knowing me, I probably made a mistake in the code, but the idea is there.
Even using stack will not convert a recursive algorithm into iterative. Normal recursion is function based recursion and if we use stack then it becomes stack based recursion. But its still recursion.
For recursive algorithms, space complexity is O(N) and time complexity is O(N).
For iterative algorithms, space complexity is O(1) and time complexity is O(N).
But if we use stack things in terms of complexity remains same. I think only tail recursion can be converted into iteration.
The stacks and recursion elimination article captures the idea of externalizing the stack frame on heap, but does not provide a straightforward and repeatable way to convert. Below is one.
While converting to iterative code, one must be aware that the recursive call may happen from an arbitrarily deep code block. Its not just the parameters, but also the point to return to the logic that remains to be executed and the state of variables which participate in subsequent conditionals, which matter. Below is a very simple way to convert to iterative code with least changes.
Consider this recursive code:
struct tnode
{
tnode(int n) : data(n), left(0), right(0) {}
tnode *left, *right;
int data;
};
void insertnode_recur(tnode *node, int num)
{
if(node->data <= num)
{
if(node->right == NULL)
node->right = new tnode(num);
else
insertnode(node->right, num);
}
else
{
if(node->left == NULL)
node->left = new tnode(num);
else
insertnode(node->left, num);
}
}
Iterative code:
// Identify the stack variables that need to be preserved across stack
// invocations, that is, across iterations and wrap them in an object
struct stackitem
{
stackitem(tnode *t, int n) : node(t), num(n), ra(0) {}
tnode *node; int num;
int ra; //to point of return
};
void insertnode_iter(tnode *node, int num)
{
vector<stackitem> v;
//pushing a stackitem is equivalent to making a recursive call.
v.push_back(stackitem(node, num));
while(v.size())
{
// taking a modifiable reference to the stack item makes prepending
// 'si.' to auto variables in recursive logic suffice
// e.g., instead of num, replace with si.num.
stackitem &si = v.back();
switch(si.ra)
{
// this jump simulates resuming execution after return from recursive
// call
case 1: goto ra1;
case 2: goto ra2;
default: break;
}
if(si.node->data <= si.num)
{
if(si.node->right == NULL)
si.node->right = new tnode(si.num);
else
{
// replace a recursive call with below statements
// (a) save return point,
// (b) push stack item with new stackitem,
// (c) continue statement to make loop pick up and start
// processing new stack item,
// (d) a return point label
// (e) optional semi-colon, if resume point is an end
// of a block.
si.ra=1;
v.push_back(stackitem(si.node->right, si.num));
continue;
ra1: ;
}
}
else
{
if(si.node->left == NULL)
si.node->left = new tnode(si.num);
else
{
si.ra=2;
v.push_back(stackitem(si.node->left, si.num));
continue;
ra2: ;
}
}
v.pop_back();
}
}
Notice how the structure of the code still remains true to the recursive logic and modifications are minimal, resulting in less number of bugs. For comparison, I have marked the changes with ++ and --. Most of the new inserted blocks except v.push_back, are common to any converted iterative logic
void insertnode_iter(tnode *node, int num)
{
+++++++++++++++++++++++++
vector<stackitem> v;
v.push_back(stackitem(node, num));
while(v.size())
{
stackitem &si = v.back();
switch(si.ra)
{
case 1: goto ra1;
case 2: goto ra2;
default: break;
}
------------------------
if(si.node->data <= si.num)
{
if(si.node->right == NULL)
si.node->right = new tnode(si.num);
else
{
+++++++++++++++++++++++++
si.ra=1;
v.push_back(stackitem(si.node->right, si.num));
continue;
ra1: ;
-------------------------
}
}
else
{
if(si.node->left == NULL)
si.node->left = new tnode(si.num);
else
{
+++++++++++++++++++++++++
si.ra=2;
v.push_back(stackitem(si.node->left, si.num));
continue;
ra2: ;
-------------------------
}
}
+++++++++++++++++++++++++
v.pop_back();
}
-------------------------
}
Search google for "Continuation passing style." There is a general procedure for converting to a tail recursive style; there is also a general procedure for turning tail recursive functions into loops.
Just killing time... A recursive function
void foo(Node* node)
{
if(node == NULL)
return;
// Do something with node...
foo(node->left);
foo(node->right);
}
can be converted to
void foo(Node* node)
{
if(node == NULL)
return;
// Do something with node...
stack.push(node->right);
stack.push(node->left);
while(!stack.empty()) {
node1 = stack.pop();
if(node1 == NULL)
continue;
// Do something with node1...
stack.push(node1->right);
stack.push(node1->left);
}
}
Thinking of things that actually need a stack:
If we consider the pattern of recursion as:
if(task can be done directly) {
return result of doing task directly
} else {
split task into two or more parts
solve for each part (possibly by recursing)
return result constructed by combining these solutions
}
For example, the classic Tower of Hanoi
if(the number of discs to move is 1) {
just move it
} else {
move n-1 discs to the spare peg
move the remaining disc to the target peg
move n-1 discs from the spare peg to the target peg, using the current peg as a spare
}
This can be translated into a loop working on an explicit stack, by restating it as:
place seed task on stack
while stack is not empty
take a task off the stack
if(task can be done directly) {
Do it
} else {
Split task into two or more parts
Place task to consolidate results on stack
Place each task on stack
}
}
For Tower of Hanoi this becomes:
stack.push(new Task(size, from, to, spare));
while(! stack.isEmpty()) {
task = stack.pop();
if(task.size() = 1) {
just move it
} else {
stack.push(new Task(task.size() -1, task.spare(), task,to(), task,from()));
stack.push(new Task(1, task.from(), task.to(), task.spare()));
stack.push(new Task(task.size() -1, task.from(), task.spare(), task.to()));
}
}
There is considerable flexibility here as to how you define your stack. You can make your stack a list of Command objects that do sophisticated things. Or you can go the opposite direction and make it a list of simpler types (e.g. a "task" might be 4 elements on a stack of int, rather than one element on a stack of Task).
All this means is that the memory for the stack is in the heap rather than in the Java execution stack, but this can be useful in that you have more control over it.
Generally the technique to avoid stack overflow is for recursive functions is called trampoline technique which is widely adopted by Java devs.
However, for C# there is a little helper method here that turns your recursive function to iterative without requiring to change logic or make the code in-comprehensible. C# is such a nice language that amazing stuff is possible with it.
It works by wrapping parts of the method by a helper method. For example the following recursive function:
int Sum(int index, int[] array)
{
//This is the termination condition
if (int >= array.Length)
//This is the returning value when termination condition is true
return 0;
//This is the recursive call
var sumofrest = Sum(index+1, array);
//This is the work to do with the current item and the
//result of recursive call
return array[index]+sumofrest;
}
Turns into:
int Sum(int[] ar)
{
return RecursionHelper<int>.CreateSingular(i => i >= ar.Length, i => 0)
.RecursiveCall((i, rv) => i + 1)
.Do((i, rv) => ar[i] + rv)
.Execute(0);
}
One pattern to look for is a recursion call at the end of the function (so called tail-recursion). This can easily be replaced with a while. For example, the function foo:
void foo(Node* node)
{
if(node == NULL)
return;
// Do something with node...
foo(node->left);
foo(node->right);
}
ends with a call to foo. This can be replaced with:
void foo(Node* node)
{
while(node != NULL)
{
// Do something with node...
foo(node->left);
node = node->right;
}
}
which eliminates the second recursive call.
A question that had been closed as a duplicate of this one had a very specific data structure:
The node had the following structure:
typedef struct {
int32_t type;
int32_t valueint;
double valuedouble;
struct cNODE *next;
struct cNODE *prev;
struct cNODE *child;
} cNODE;
The recursive deletion function looked like:
void cNODE_Delete(cNODE *c) {
cNODE*next;
while (c) {
next=c->next;
if (c->child) {
cNODE_Delete(c->child)
}
free(c);
c=next;
}
}
In general, it is not always possible to avoid a stack for recursive functions that invoke itself more than one time (or even once). However, for this particular structure, it is possible. The idea is to flatten all the nodes into a single list. This is accomplished by putting the current node's child at the end of the top row's list.
void cNODE_Delete (cNODE *c) {
cNODE *tmp, *last = c;
while (c) {
while (last->next) {
last = last->next; /* find last */
}
if ((tmp = c->child)) {
c->child = NULL; /* append child to last */
last->next = tmp;
tmp->prev = last;
}
tmp = c->next; /* remove current */
free(c);
c = tmp;
}
}
This technique can be applied to any data linked structure that can be reduce to a DAG with a deterministic topological ordering. The current nodes children are rearranged so that the last child adopts all of the other children. Then the current node can be deleted and traversal can then iterate to the remaining child.
Recursion is nothing but the process of calling of one function from the other only this process is done by calling of a function by itself. As we know when one function calls the other function the first function saves its state(its variables) and then passes the control to the called function. The called function can be called by using the same name of variables ex fun1(a) can call fun2(a).
When we do recursive call nothing new happens. One function calls itself by passing the same type and similar in name variables(but obviously the values stored in variables are different,only the name remains same.)to itself. But before every call the function saves its state and this process of saving continues. The SAVING IS DONE ON A STACK.
NOW THE STACK COMES INTO PLAY.
So if you write an iterative program and save the state on a stack each time and then pop out the values from stack when needed, you have successfully converted a recursive program into an iterative one!
The proof is simple and analytical.
In recursion the computer maintains a stack and in iterative version you will have to manually maintain the stack.
Think over it, just convert a depth first search(on graphs) recursive program into a dfs iterative program.
All the best!
TLDR
You can compare the source code below, before and after to intuitively understand the approach without reading this whole answer.
I ran into issues with some multi-key quicksort code I was using to process very large blocks of text to produce suffix arrays. The code would abort due to the extreme depth of recursion required. With this approach, the termination issues were resolved. After conversion the maximum number of frames required for some jobs could be captured, which was between 10K and 100K, taking from 1M to 6M memory. Not an optimum solution, there are more effective ways to produce suffix arrays. But anyway, here's the approach used.
The approach
A general way to convert a recursive function to an iterative solution that will apply to any case is to mimic the process natively compiled code uses during a function call and the return from the call.
Taking an example that requires a somewhat involved approach, we have the multi-key quicksort algorithm. This function has three successive recursive calls, and after each call, execution begins at the next line.
The state of the function is captured in the stack frame, which is pushed onto the execution stack. When sort() is called from within itself and returns, the stack frame present at the time of the call is restored. In that way all the variables have the same values as they did before the call - unless they were modified by the call.
Recursive function
def sort(a: list_view, d: int):
if len(a) <= 1:
return
p = pivot(a, d)
i, j = partition(a, d, p)
sort(a[0:i], d)
sort(a[i:j], d + 1)
sort(a[j:len(a)], d)
Taking this model, and mimicking it, a list is set up to act as the stack. In this example tuples are used to mimic frames. If this were encoded in C, structs could be used. The data can be contained within a data structure instead of just pushing one value at a time.
Reimplemented as "iterative"
# Assume `a` is view-like object where slices reference
# the same internal list of strings.
def sort(a: list_view):
stack = []
stack.append((LEFT, a, 0)) # Initial frame.
while len(stack) > 0:
frame = stack.pop()
if len(frame[1]) <= 1: # Guard.
continue
stage = frame[0] # Where to jump to.
if stage == LEFT:
_, a, d = frame # a - array/list, d - depth.
p = pivot(a, d)
i, j = partition(a, d, p)
stack.append((MID, a, i, j, d)) # Where to go after "return".
stack.append((LEFT, a[0:i], d)) # Simulate function call.
elif stage == MID: # Picking up here after "call"
_, a, i, j, d = frame # State before "call" restored.
stack.append((RIGHT, a, i, j, d)) # Set up for next "return".
stack.append((LEFT, a[i:j], d + 1)) # Split list and "recurse".
elif stage == RIGHT:
_, a, _, j, d = frame
stack.append((LEFT, a[j:len(a)], d)
else:
pass
When a function call is made, information on where to begin execution after the function returns is included in the stack frame. In this example, if/elif/else blocks represent the points where execution begins after return from a call. In C this could be implemented as a switch statement.
In the example, the blocks are given labels; they're arbitrarily labeled by how the list is partitioned within each block. The first block, "LEFT" splits the list on the left side. The "MID" section represents the block that splits the list in the middle, etc.
With this approach, mimicking a call takes two steps. First a frame is pushed onto the stack that will cause execution to resume in the block following the current one after the "call" "returns". A value in the frame indicates which if/elif/else section to fall into on the loop that follows the "call".
Then the "call" frame is pushed onto the stack. This sends execution to the first, "LEFT", block in most cases for this specific example. This is where the actual sorting is done regardless which section of the list was split to get there.
Before the looping begins, the primary frame pushed at the top of the function represents the initial call. Then on each iteration, a frame is popped. The "LEFT/MID/RIGHT" value/label from the frame is used to fall into the correct block of the if/elif/else statement. The frame is used to restore the state of the variables needed for the current operation, then on the next iteration the return frame is popped, sending execution to the subsequent section.
Return values
If the recursive function returns a value used by itself, it can be treated the same way as other variables. Just create a field in the stack frame for it. If a "callee" is returning a value, it checks the stack to see if it has any entries; and if so, updates the return value in the frame on the top of the stack. For an example of this you can check this other example of this same approach to recursive to iterative conversion.
Conclusion
Methods like this that convert recursive functions to iterative functions, are essentially also "recursive". Instead of the process stack being utilized for actual function calls, another programmatically implemented stack takes its place.
What is gained? Perhaps some marginal improvements in speed. Or it could serve as a way to get around stack limitations imposed by some compilers and/or execution environments (stack pointer hitting the guard page). In some cases, the amount of data pushed onto the stack can be reduced. Do the gains offset the complexity introduced in the code by mimicking something that we get automatically with the recursive implementation?
In the case of the sorting algorithm, finding a way to implement this particular one without a stack could be challenging, plus there are so many iterative sorting algorithms available that are much faster. It's been said that any recursive algorithm can be implemented iteratively. Sure... but some algorithms don't convert well without being modified to such a degree that they're no longer the same algorithm.
It may not be such a great idea to convert recursive algorithms just for the sake of converting them. Anyway, for what it's worth, the above approach is a generic way of converting that should apply to just about anything.
If you find you really need an iterative version of a recursive function that doesn't use a memory eating stack of its own, the best approach may be to scrap the code and write your own using the description from a scholarly article, or work it out on paper and then code it from scratch, or other ground up approach.
There is a general way of converting recursive traversal to iterator by using a lazy iterator which concatenates multiple iterator suppliers (lambda expression which returns an iterator). See my Converting Recursive Traversal to Iterator.
Another simple and complete example of turning the recursive function into iterative one using the stack.
#include <iostream>
#include <stack>
using namespace std;
int GCD(int a, int b) { return b == 0 ? a : GCD(b, a % b); }
struct Par
{
int a, b;
Par() : Par(0, 0) {}
Par(int _a, int _b) : a(_a), b(_b) {}
};
int GCDIter(int a, int b)
{
stack<Par> rcstack;
if (b == 0)
return a;
rcstack.push(Par(b, a % b));
Par p;
while (!rcstack.empty())
{
p = rcstack.top();
rcstack.pop();
if (p.b == 0)
continue;
rcstack.push(Par(p.b, p.a % p.b));
}
return p.a;
}
int main()
{
//cout << GCD(24, 36) << endl;
cout << GCDIter(81, 36) << endl;
cin.get();
return 0;
}
My examples are in Clojure, but should be fairly easy to translate to any language.
Given this function that StackOverflows for large values of n:
(defn factorial [n]
(if (< n 2)
1
(*' n (factorial (dec n)))))
we can define a version that uses its own stack in the following manner:
(defn factorial [n]
(loop [n n
stack []]
(if (< n 2)
(return 1 stack)
;; else loop with new values
(recur (dec n)
;; push function onto stack
(cons (fn [n-1!]
(*' n n-1!))
stack)))))
where return is defined as:
(defn return
[v stack]
(reduce (fn [acc f]
(f acc))
v
stack))
This works for more complex functions too, for example the ackermann function:
(defn ackermann [m n]
(cond
(zero? m)
(inc n)
(zero? n)
(recur (dec m) 1)
:else
(recur (dec m)
(ackermann m (dec n)))))
can be transformed into:
(defn ackermann [m n]
(loop [m m
n n
stack []]
(cond
(zero? m)
(return (inc n) stack)
(zero? n)
(recur (dec m) 1 stack)
:else
(recur m
(dec n)
(cons #(ackermann (dec m) %)
stack)))))
A rough description of how a system takes any recursive function and executes it using a stack:
This intended to show the idea without details. Consider this function that would print out nodes of a graph:
function show(node)
0. if isleaf(node):
1. print node.name
2. else:
3. show(node.left)
4. show(node)
5. show(node.right)
For example graph:
A->B
A->C
show(A) would print B, A, C
Function calls mean save the local state and the continuation point so you can come back, and then jump the the function you want to call.
For example, suppose show(A) begins to run. The function call on line 3. show(B) means
- Add item to the stack meaning "you'll need to continue at line 2 with local variable state node=A"
- Goto line 0 with node=B.
To execute code, the system runs through the instructions. When a function call is encountered, the system pushes information it needs to come back to where it was, runs the function code, and when the function completes, pops the information about where it needs to go to continue.
This link provides some explanation and proposes the idea of keeping "location" to be able to get to the exact place between several recursive calls:
However, all these examples describe scenarios in which a recursive call is made a fixed amount of times. Things get trickier when you have something like:
function rec(...) {
for/while loop {
var x = rec(...)
// make a side effect involving return value x
}
}
This is an old question but I want to add a different aspect as a solution. I'm currently working on a project in which I used the flood fill algorithm using C#. Normally, I implemented this algorithm with recursion at first, but obviously, it caused a stack overflow. After that, I changed the method from recursion to iteration. Yes, It worked and I was no longer getting the stack overflow error. But this time, since I applied the flood fill method to very large structures, the program was going into an infinite loop. For this reason, it occurred to me that the function may have re-entered the places it had already visited. As a definitive solution to this, I decided to use a dictionary for visited points. If that node(x,y) has already been added to the stack structure for the first time, that node(x,y) will be saved in the dictionary as the key. Even if the same node is tried to be added again later, it won't be added to the stack structure because the node is already in the dictionary. Let's see on pseudo-code:
startNode = pos(x,y)
Stack stack = new Stack();
Dictionary visited<pos, bool> = new Dictionary();
stack.Push(startNode);
while(stack.count != 0){
currentNode = stack.Pop();
if "check currentNode if not available"
continue;
if "check if already handled"
continue;
else if "run if it must be wanted thing should be handled"
// make something with pos currentNode.X and currentNode.X
// then add its neighbor nodes to the stack to iterate
// but at first check if it has already been visited.
if(!visited.Contains(pos(x-1,y)))
visited[pos(x-1,y)] = true;
stack.Push(pos(x-1,y));
if(!visited.Contains(pos(x+1,y)))
...
if(!visited.Contains(pos(x,y+1)))
...
if(!visited.Contains(pos(x,y-1)))
...
}