遍历数组对象

前提条件：数组对象中的id唯一

 1 const tree = [
 2     {id: 1},
 3     {id:2, 
 4         children:[
 5             {id: 3, 
 6                 children:[
 7                     {id: 4, 
 8                          children: [
 9                              {id: 5}
10                          ]
11                      }
12                  ]
13               }
14           ]
15      }
16  ]

1. 面试题-通过id，在上述数据中找出id所在的对象
   

    1 function getItem(data, searchId){
    2     let result;
    3     if(Array.isArray(data)){
    4         for(let item of data) {
    5             const {id, children} = item;
    6             if (searchId === id) {
    7                 result = item;
    8                 break;
    9             }
   10             if(Array.isArray(children)){
   11                 result = getItem(children, searchId)
   12             }
   13         }
   14     } else {
   15         console.log('没有找到');
   16         return null;
   17     }
   18     return result;
   19 }

2. 面试题-通过id，获取父节点id
   

    1 function getParentId(data, searchId, parentId) {
    2     let result;
    3     if(Array.isArray(data)){
    4         for (let item of data) {
    5             const {id, children} = item;
    6             if (id === searchId) {
    7                 result = parentId;
    8                 break;
    9             }
   10             if(Array.isArray(children)){
   11                 result = getParentId(children, searchId, id);
   12             }
   13         }
   14     } else {
   15         return 'null'
   16     }
   17 
   18     return result;
   19 }

3. 面试题-通过id，获取所有的上级节点id
   

    1 function getAllParentIds(data, searchId, parentIds = []) {
    2     let result;
    3     if(Array.isArray(data)){
    4         for (let item of data) {
    5             const {id, children} = item;
    6             if (id === searchId) {
    7                 result = parentIds;
    8                 break;
    9             }
   10             if(Array.isArray(children)){
   11                 result = getAllParentIds(children, searchId, parentIds.concat(id));
   12             }
   13         }
   14     } else {
   15         return 'null'
   16     }
   17 
   18     return result;
   19 }