予具有相同类型的对象的图。每个对象被链接与0,1或许多其他问题。我需要遍历图中的所有可能的路径。

I have a graph of objects of the same type. Every object is linked with 0, 1 or many others. I need to walk through all possible paths in the graph.

现在我能做到这一点递归，但有一个堆栈溢出的危险。可以有几万人。

Now I could do that with recursion, but there is the danger of a stack overflow. There can be tens of thousands of them.

我听说有比递归，其中一种方法保持了一遍又一遍自称更好的办法。

I've heard there are better ways than recursion where a method keeps calling itself over and over again.

有什么替代方案是什么样子？

What do the alternatives look like?

推荐答案

线沿线的东西：

（我们presume树节点包含属性的NSArray *儿童）

(let's presume TreeNode contains a property NSArray * children)

-(void)iterateOverTree:(TreeNode *)node
{
    NSMutableArray * elements = [NSMutableArray array];
    [elements addObject:node];

    while([elements count])
    {
        TreeNode * current = [elements objectAtIndex:0];
        [self doStuffWithNode:current];
        for(TreeNode * child in current.children)
        {
            [elements addObject:child];
        }

        [elements removeObjectAtIndex:0];
    }
}

当心，未经测试code：）

Beware, untested code :)