Javascript 在具有完整路径的树中查找匹配节点

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: []
        }]
    }]
}]

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

每个人可以有多个孩子，每个孩子可以有任意数量的孩子，以此类推

我希望保留包含匹配子项的完整路径，并排除不匹配子项

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

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

---

您需要生成一个只包含相关部分的新对象

该建议适用于单级迭代，适用于子级递归

函数getNodes（数组，cb）{ 返回数组.reduce（函数iter（r，a）{ 儿童； 如果（cb（a））{ 返回r.concat（a）； } if（Array.isArray（a.children））{ children=a.children.reduce（iter，[]）； } if（儿童长度）{ 返回r.concat（{name:a.name，children:children}）； } 返回r； }, []); } var数据=[{name:“Peter”，children:[{name:“John”，children:[[]}]，{name:“Joseph”，children:[]}]}，{name:“Joseph”，children:[{name:“Sam”，children:[{name:“JohnXX”，children:[]}]}]}]，{name:“Carl”，children:[{name:“Sam”，children:[{name:“Sam”，children； log（getNodes（数据，函数（o）{returno.name.indexOf（'John'）！==-1；}））

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.as控制台包装{max height:100%！important；top:0；}

此代码返回整个树，而不是删除不匹配的子项。请将代码放入问题中。