最近需要实现一个功能,需要把一个树形结构的数据打散成包含最小分支数目的数组结构。描述起来别扭,还是举栗子吧:
以 JavaScript 为例:其中,id 为主键,parent 为该对象对应的父级元素ID
let tree = [
{
id:1,
parent:0,
name:'hello',
children:[
{
id:5,
parent:1,
name:'bill gate',
children:[]
},
{
id:6,
parent:1,
name:'jack ma',
children:[]
}
]
},
{
id:2,
parent:0,
name:'world',
children:[
{
id:3,
parent:2,
name:'foo',
children:[]
},
{
id:4,
parent:2,
name:'bar',
children:[]
}
]
}
];
上面的数组比较简单,因为每个元素的子元素个数和深度是一样的,但实际上的需求是变动是非常大的,最明显的使用场景是权限树,权限树涉及到的数据层级是不确定的,相邻的两个元素的子元素层级都可能会不同,这样的场景会比上面的例子复杂得多,其实这里可以考虑递归,复杂的数据结构如下:
let tree = [
{
id: 1,
parent: 0,
name: 'hello',
children: [
{
id: 5,
parent: 1,
name: 'bill gate',
children: []
},
{
id: 6,
parent: 1,
name: 'jack ma',
children: []
}
]
},
{
id: 2,
parent: 0,
name: 'world',
children: [
{
id: 3,
parent: 2,
name: 'foo',
children: [
{
id: 7,
parent: 3,
name: 'tom',
children: []
},
{
id: 8,
parent: 3,
name: 'jerry',
children: [
{
id: 9,
name: 'peter'
}
]
}
]
},
{
id: 4,
parent: 2,
name: 'bar',
children: []
}
]
}
];
上面的两个样本数组,对其进行处理后我们期望得到的结果应该分别如下:可以理解为对一个族谱进行拆分,每一个分支一个数组,这样,一个族谱有多少个分支,最终生成的数组就有多少个元素。
例子一:
[
['hello','bill gate'],
['hello','jack ma'],
['world','foo'],
['world','bar']
]
例子二:
[
['hello','bill gate'],
['hello','jack ma'],
['world','foo','tom'],
['world','foo','jerry','peter'],
['world','bar']
]
要实现这样的需求有什么比较好的算法呢?
3 回答
![?](http://img1.sycdn.imooc.com/545866c40001561502200220-100-100.jpg)
白板的微信
TA贡献1883条经验 获得超3个赞
递归遍历
let tree = [
{
id: 1,
parent: 0,
name: 'hello',
children: [
{
id: 5,
parent: 1,
name: 'bill gate',
children: []
},
{
id: 6,
parent: 1,
name: 'jack ma',
children: []
}
]
},
{
id: 2,
parent: 0,
name: 'world',
children: [
{
id: 3,
parent: 2,
name: 'foo',
children: [
{
id: 7,
parent: 3,
name: 'tom',
children: []
},
{
id: 8,
parent: 3,
name: 'jerry',
children: [
{
id: 9,
parent: 8,
name: 'peter',
children: []
}
]
}
]
},
{
id: 4,
parent: 2,
name: 'bar',
children: []
}
]
}
];
function treeToPath(tree, path, currentPath) {
var currentPath = currentPath || [];
var path = path || [];
for(let i = 0; i < tree.length; i++) {
if(i !== 0) {
currentPath.pop();
}
currentPath.push(tree[i].name);
if(tree[i].children.length) {
treeToPath(tree[i].children, path, currentPath);
}else {
path.push(currentPath.slice(0));
}
}
currentPath.pop();
return path;
}
console.log(treeToPath(tree));
![?](http://img1.sycdn.imooc.com/54584d080001566902200220-100-100.jpg)
森栏
TA贡献1810条经验 获得超5个赞
我理解为需要 依次 输出从 顶点 到 每一个 终点节点 的路径
$json = "[{\"id\":1,\"name\":\"hello\",\"children\":[{\"id\":5,\"name\":\"bill gate\"},{\"id\":6,\"name\":\"jack ma\"}]},{\"id\":2,\"name\":\"world\",\"children\":[{\"id\":3,\"name\":\"foor\"},{\"id\":4,\"name\":\"bar\"}]}]";
$arrays = json_decode($json, true);
define("NAME", "name");
$result = []; //存放結果
function tree($array){
static $path = [];
global $result;
array_push($path, $array[NAME]);
foreach ($array['children'] as $index => $item){
if(isset($item['children']) && count($item['children']) > 0){
//如果非终点就继续递归执行
tree($item);
}else{
array_push($path, $item[NAME]);
$result[] = $path;
array_pop($path);
}
}
array_pop($path);
}
foreach ($arrays as $index => $array){
tree($array);
}
var_dump($result);
- 3 回答
- 0 关注
- 445 浏览
添加回答
举报
0/150
提交
取消