js怎样将获取json转换为树形结构

问题：js怎样将获取json转换为树形结构
描述:

比如这样一段json

{
    "name": "1级菜单1",
    "link": "###",
    "isleaf": false,
    "level": 0,
    "children": [
        {
            "name": "2级菜单1",
            "link": "###",
            "isleaf": false,
            "level": 1,
            "children": [
                {
                    "name": "3级菜单1",
                    "link": "###",
                    "isleaf": true,
                    "level": 2,
                    "children": null
                },
                {
                    "name": "3级菜单2",
                    "link": "###",
                    "isleaf": true,
                    "level": 2,
                    "children": null
                }
            ]
        },
        {
            "name": "2级菜单2",
            "link": "###",
            "isleaf": false,
            "level": 1,
            "children": [
                {
                    "name": "3级菜单3",
                    "link": "###",
                    "isleaf": true,
                    "level": 2,
                    "children": null
                }
            ]
        }
    ]
}

我怎样把他转换成

<div>
    <a href="###">一级菜单</a>
    <ul>
        <li>
            <a>2级菜单1</a>
            <ul>
                <li><a href="###">3级菜单1</a></li>
                <li><a href="###">3级菜单2</a></li>
            </ul>
        </li>
        <li>
            <a href="###">2级菜单2</a>
            .......
        </li>
    </ul>
</div>

---

解决方案1:

建议下载个jQuery EASYUI的API里面有关于树的解释，很清楚的
http://www.cnblogs.com/Philoo/archive/2011/11/17/jeasyui_api_index.htm...

解决方案2:

http://stackoverflow.com/questions/8553539/jquery-json-looping-through...

解决方案3:

下面的代码render是主函数，用来render你提供的json数据，它调用了renderMenu函数。

javascriptfunction renderMenu(menu, parent) {
    var el = $('<a href="' + menu.link  +'">' + menu.name + '</a>');
    $(parent).append(el);
    if (menu.isleaf) {
        return parent;
    }

    var ul = $('<ul/>');
    for (var i = 0; i < menu.children.length; ++i) {
        var li = $('<li/>');
        renderMenu(menu.children[i], li);
        ul.append(li);
    }

    $(parent).append(ul);
    return parent;
}

function render(json) {
    var div = $('

<div/>

');
    return renderMenu(json, div);
}

var json = // 你的那段json代码
var menu = render(json);

在浏览器中检查menu显示如下：

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