如何为每个嵌套孩子存储所有父母的参考？

Question

Fiddle : http://jsfiddle.net/p35cobgx/3/

Below is my tree structure :

Node
   Node-1
     Node-1-1......

Now I want to store back reference of all parents for each of the nodes like below :

Node

Node-1 (should store back reference of only Node because this is last parent for Node-1)

Node-1-1(should store back reference of all parents ie Node-1 and Node because Node is the last parent for Node-1-1).

But problem with my code is i am unable to store reference of Node for Node-1-1 .

Answer 1

I'm not sure if that's exactly the answer you wanted, but here it is:

You can access a parent node by using:

node.parentNode

This is a property and holds the reference for the parent node.

Also, it's sometimes better to access parent element, instead of node, because whitespace can also be a node, so you can use this property:

node.parentElement

Answer 2

The algorithm I would use is to store only the immediate parent in the node.

So if A is the parent to B and B of C

That makes it A->B->C

C.parent = B; B.parent = A;

If I wanted all parents of C, I would do the below

    var parentArray = getParent(C).slice();
    this.parents = [];

    //use parentArray as you like now!


    function getParent(obj){

    if(obj.parent){
    this.parents.push(obj.parent) //this.parents is an array
    getParent(obj.parent);
    }else{
    return this.parents;
    }
    }

So if you make a call getParent(C); you get all its parents B and A. If you make a call getParent(B); you get only A. If you make a call getParent(A); you get no parents as it doesn't have any parents.

Hope that helps!

Answer 3

使用此条件来分配父项：

obj.parent = data.parent || null;