另一个使用堆栈将递归函数转换为迭代函数的简单而完整的示例。

#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;
}

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。

在谷歌中搜索“延续传递样式”。有一个转换为尾部递归样式的一般过程;还有一个将尾部递归函数转换为循环的通用过程。

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)))
            ...
}

一个系统如何接受任何递归函数并使用堆栈执行它的粗略描述:

这是为了在没有细节的情况下展示想法。考虑这个函数，它将打印出图的节点:

function show(node)
0. if isleaf(node):
1.  print node.name
2. else:
3.  show(node.left)
4.  show(node)
5.  show(node.right)

例如图表: A - B > C - > show(A)将打印B, A, C

函数调用意味着保存本地状态和延续点，以便返回，然后跳转到要调用的函数。

例如，假设show(A)开始运行。函数调用在第3行。显示(B)的意思 -将项目添加到堆栈，意思是“你需要在第2行继续使用本地变量状态node=A” —到第0行，节点为B。

为了执行代码，系统运行指令。当遇到函数调用时，系统将需要的信息推回到原来的位置，运行函数代码，当函数完成时，弹出关于需要继续执行的位置的信息。

实际上，最常见的方法是保留自己的堆栈。下面是一个C语言的递归快速排序函数:

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);
}

以下是我们如何通过保持自己的堆栈来实现迭代:

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;
    }
}

显然，这个例子没有检查堆栈边界……实际上，你可以根据最坏的情况来确定堆栈的大小。但你懂的。