Question

假设我有一个可以容纳其他实例的类：

function Thing() {
    this.subThings = [];
}

我将n subThings添加到最重要的内容：

var rootThing = new Thing();

rootThing.subThings.push(new Thing());
rootThing.subThings.push(new Thing());
rootThing.subThings.push(new Thing());

然后我将n subThings添加到rootThing subThings的某些rootThing.subThings[0].subThings.push(new Thing()); rootThing.subThings[0].subThings.push(new Thing()); rootThing.subThings[0].subThings.push(new Thing());中：

rootThing
|
+--subThing
|  |
|  +--subThing
|  |
|  +--subThing
|  |
|  +--subThing
|
+--subThing
|
+--subThing

此时，结构如下所示：

subThing

那么，如何在subThing中获取所有rootThing和所有 function getAllChildren(beginNode) { var allChildren = []; beginNode.subThings.forEach(function(childNode) { allChildren.push(childNode); // allChildren = allChildren.concat(childNode.subThings); // something along these lines }); return allChildren; } console.log(getAllChildren(rootThing));的列表？

{{1}}

Answer 1

感谢@zerkms，这是一个有效的实现：

function getAllChildren(beginNode) {
    var allChildren = beginNode.subThings;

    beginNode.subThings.forEach(function(childNode) {
        allChildren = allChildren.concat(getAllChildren(childNode));
    });

    return allChildren;
}

Answer 2

你在找这样的东西吗？

var GetAllThings = function(Thing thing) {
    var result = [];

    //Assume you want to remove the subthings as you flatten
    var subthings = thing.subthings;
    thing.subThings = null;

    result.push(thing);

    for(var i = 0; i < subthings i++) {
        result.concat(GetAllThings(subthings[i]));
    }

    return result;
};

这基本上会使树变平，但如果您只是想查看/处理它，上面的注释是正确的，因为您可以将其作为json对象来处理。

如何递归获取嵌套对象中所有子项的列表？

2 个答案: