javascript - 在具有完整路径的树中查找匹配节点

Question

我有一个像这样的对象数组

[{
    name: "Peter",
    children: [{
        name: "John",
        children: [{
            name: "Joseph",
            children: []
        }]
    }, {
        name: "Shawn",
        children: [{
            name: "Joseph",
            children: []
        }]
    }]
}, {
    name: "Carl",
    children: [{
        name: "Sam",
        children: [{
            name: "JohnXX",
            children: []
        }]
    }]
}]

每个人可以有多个 child ，每个 child 可以有任意数量的 child ，依此类推。

我想保留具有匹配子项的完整路径并排除不匹配子项。例如，如果我搜索 John 输出应该是这样的

[{
    name: "Peter",
    children: [{
        name: "John",
        children: [{
            name: "Joseph",
            children: []
        }]
    }]
}, {
    name: "Carl",
    children: [{
        name: "Sam",
        children: [{
            name: "JohnXX",
            children: []
        }]
    }]
}]

Answer 1

您需要生成一个仅包含相关部分的新对象。

该提案对于单个级别进行迭代，对于子级进行递归。

function getNodes(array, cb) {
    return array.reduce(function iter(r, a) {
        var children;
        if (cb(a)) {
            return r.concat(a);
        }
        if (Array.isArray(a.children)) {
            children = a.children.reduce(iter, []);
        }
        if (children.length) {
            return r.concat({ name: a.name, children: children });
        }
        return r;
    }, []);
}

var data = [{ name: "Peter", children: [{ name: "John", children: [{ name: "Joseph", children: [] }] }, { name: "Shawn", children: [{ name: "Joseph", children: [] }] }] }, { name: "Carl", children: [{ name: "Sam", children: [{ name: "JohnXX", children: [] }] }] }];

console.log(getNodes(data, function (o) { return o.name.indexOf('John') !== -1; }));

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