想想那些真正需要堆栈的东西:

如果我们考虑递归的模式为:

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
}

例如，经典的河内塔

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
}

这可以转化为一个循环工作在一个显式的堆栈，通过重申它为:

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
   }
}

对于《河内塔》来说，这就变成了:

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()));
    }
}

在如何定义堆栈方面，这里有相当大的灵活性。你可以让你的堆栈成为一个Command对象列表，这些对象可以做一些复杂的事情。或者你可以走相反的方向，让它成为一个简单类型的列表(例如，一个“task”可能是一个int堆栈上的4个元素，而不是一个task堆栈上的一个元素)。

这意味着堆栈的内存在堆中，而不是在Java执行堆栈中，但这可能很有用，因为您可以更好地控制它。

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)))
            ...
}

要寻找的一种模式是函数末尾的递归调用(所谓的尾部递归)。这很容易用一段时间来代替。例如，函数foo:

void foo(Node* node)
{
    if(node == NULL)
       return;
    // Do something with node...
    foo(node->left);
    foo(node->right);
}

以调用foo结束。这可以替换为:

void foo(Node* node)
{
    while(node != NULL)
    {
        // Do something with node...
        foo(node->left);
        node = node->right;
     }
}

这消除了第二次递归调用。

堆栈和递归消除文章抓住了将堆栈框架外部化到堆上的思想，但没有提供直接和可重复的转换方法。下面是一个。

在转换为迭代代码时，必须意识到递归调用可能发生在任意深度的代码块中。它不仅是参数，而且是返回到仍然要执行的逻辑的点，以及参与后续条件的变量的状态，这很重要。下面是一种转换为迭代代码的非常简单的方法。

考虑下面的递归代码:

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);
    }    
}

迭代代码:

// 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();
    }
}

请注意，代码的结构仍然保持忠于递归逻辑，并且修改是最小的，从而减少了错误的数量。为了便于比较，我用++和——标记了更改。除了v.push_back之外，大多数新插入的块对于任何转换的迭代逻辑都是通用的

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();
    }

-------------------------

}

我的例子是用Clojure编写的，但是应该很容易翻译成任何语言。

给定这个函数，当n值较大时StackOverflows:

(defn factorial [n]
  (if (< n 2)
    1
    (*' n (factorial (dec n)))))

我们可以用以下方式定义一个使用自己堆栈的版本:

(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)))))

其中return定义为:

(defn return
  [v stack]
  (reduce (fn [acc f]
            (f acc))
          v
          stack))

这也适用于更复杂的函数，例如阿克曼函数:

(defn ackermann [m n]
  (cond
    (zero? m)
    (inc n)

    (zero? n)
    (recur (dec m) 1)

    :else
    (recur (dec m)
           (ackermann m (dec n)))))

可以转化为:

(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)))))

似乎没有人指出递归函数在主体中调用自己超过一次的位置，并处理返回递归中的特定点(即不是原始递归)。据说每一个递归都可以转化为迭代，所以这似乎是可能的。

我刚刚想出了一个如何做到这一点的c#示例。假设您有以下递归函数，它的作用类似于poststorder遍历，AbcTreeNode是一个带有指针a、b、c的3元树。

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
        }
}

迭代解:

        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();
            }


        }