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回复讨论(解决方案)

php递归问题

Jun 23, 2016 pm 02:24 PM

本帖最后由 qq43599939 于 2013-08-28 12:42:25 编辑

<?php    $area = array(          0=>array('id'=>1,'name'=>'中国','pid'=>0),          1=>array('id'=>2,'name'=>'北京','pid'=>1),      );      print_r(catsort($area));     function catsort($cate, $pid = 0){        $arr = array();        foreach($cate as $v){            if ($v['pid'] == $pid){                $arr[] = $v;                $arr = array_merge($arr, catsort($cate ,$v['id']));                //$arr[] = $v;            }        }        return $arr;        }    ?>
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这个问题让我很困惑,为什么 $arr[] = $v放在$arr = array_merge($arr, catsort($cate ,$v['id']))前面返回的先是中国,然后才是北京。而放到$arr = array_merge($arr, catsort($cate ,$v['id']))后面的时候却是北京先,后中国呢?请帮忙详细解释下,谢谢


回复讨论(解决方案)

你至少应该阅读过关于书的遍历吧?
递归就是遍历一颗树

根据访问结点操作发生位置命名:
① NLR:前序遍历(PreorderTraversal亦称(先序遍历))
??访问根结点的操作发生在遍历其左右子树之前。
② LNR:中序遍历(InorderTraversal)
??访问根结点的操作发生在遍历其左右子树之中(间)。
③ LRN:后序遍历(PostorderTraversal)
??访问根结点的操作发生在遍历其左右子树之后。

$arr[] = $v放在$arr = array_merge($arr, catsort($cate ,$v['id']))前面
是前序遍历
放到$arr = array_merge($arr, catsort($cate ,$v['id']))后面
是后序遍历

那为什么放在前面就是前序,后面就是后序,不是很理解,能在说清楚点吗

        foreach($cate as $v){            if ($v['pid'] == $pid){                $arr[] = $v;                $arr = array_merge($arr, catsort($cate ,$v['id']));                //$arr[] = $v;            }        }
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放前面的时候,中国先加到$arr中的没什么好说的。

        foreach($cate as $v){            if ($v['pid'] == $pid){                //$arr[] = $v;                $arr = array_merge($arr, catsort($cate ,$v['id'])); //这里catsort进入递归                $arr[] = $v;            }        }
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放后面的时候,外层的array_merge函数处理第二个参数"catsort($cate,$v['id'])"时,递归进入里层,在里层执行了"$arr[]=$v",把北京先加入到了数组中,然后才返回外层,执行外层的"$arr[]=$v".

搞懂了,thank you

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