從樹形結構中獲取所有“名稱”

Question

這里有無數的樹問題，但是對我來說，如果有人可以幫助我解決這個特定問題，那將是很好的。 我參加了新兵訓練營的“模擬面試”，這是一個問題。 我沒有一個很好的方法來解決它。

const people = {
    name: "Robin",
    children: [
        {
            name: "Alberto",
            children: [
                {
                    name: "Quinn",
                    children: [
                        {
                            name: "Conner",
                            children: []
                        },
                        {
                            name: "Lila",
                            children: []
                        }
                    ]
                }
            ]
        },
        {
            name: "Charlie",
            children: []
        }
    ]
}

// Write a function called getNames that returns a string "Robin, Alberto, Quinn, Conner, Lila, Charlie

當您嘗試學習這些東西時，可能會有些不知所措，尤其是當您在篩選具有不同角度解決問題的大量問題時。 因此，非常感謝您對此特定的幫助！

Answer 1

您可以使用reduce方法創建將返回字符串的遞歸函數。

 const people = {"name":"Robin","children":[{"name":"Alberto","children":[{"name":"Quinn","children":[{"name":"Conner","children":[]},{"name":"Lila","children":[]}]}]},{"name":"Charlie","children":[]}]} function getNames(data, name = "") { return data.name + (data.children ? data.children.reduce((r, e) => { return r + ", " + getNames(e) }, "") : "") } console.log(getNames(people)) 

Answer 2

簡單，只需使用遞歸即可。

 const people = { name: "Robin", children: [ { name: "Alberto", children: [ { name: "Quinn", children: [ { name: "Conner", children: [] }, { name: "Lila", children: [] } ] } ] }, { name: "Charlie", } ]}; function getNames(tree) { var names = []; for (var i = 0; i < tree.length; i++) { names.push(tree[i].name) if (tree[i].children) { names = names.concat(getNames(tree[i].children)) } } return names; } var names = getNames([people]); console.log(names); 

Answer 3

可以使用新的flatMap輕松地轉換遞歸數據結構

 const people = { name: "Robin", children: [ { name: "Alberto", children: [ { name: "Quinn", children: [ { name: "Conner", children: [] }, { name: "Lila", children: [] } ] } ] }, { name: "Charlie", } ]}; const getNames = ({ name, children = [] }) => [ name, ...children.flatMap(getNames) ] console.log(getNames(people)) // [ "Robin", "Alberto", "Quinn", "Conner", "Lila", "Charlie" ] 

如果您的環境未定義，則可以編寫自己的flatMap

 const flatMap = (f, xs = [], context = null) => xs.reduce ( (acc, x, i) => acc.concat (f.call (context, x, i, xs)) , [] ) const people = { name: "Robin", children: [ { name: "Alberto", children: [ { name: "Quinn", children: [ { name: "Conner", children: [] }, { name: "Lila", children: [] } ] } ] }, { name: "Charlie", } ]}; const getNames = ({ name, children = [] }) => [ name, ...flatMap (getNames, children) ] console.log(getNames(people)) // [ "Robin", "Alberto", "Quinn", "Conner", "Lila", "Charlie" ] 

節點11支持flatMap ，但從Babel 7開始不再包含在babel-polyfill中，現在必須手動將其導入

// Node 10 or Babel 7
import 'core-js/fn/array/flat-map'

您可以手動填充

 // manual polyfill Array.prototype.flatMap = function (f, context = null) { return this.reduce ( (acc, x, i) => acc.concat (f.call (context, x, i, this)) , [] ) } const people = { name: "Robin", children: [ { name: "Alberto", children: [ { name: "Quinn", children: [ { name: "Conner", children: [] }, { name: "Lila", children: [] } ] } ] }, { name: "Charlie", } ]}; const getNames = ({ name, children = [] }) => [ name, ...children.flatMap(getNames) ] console.log(getNames(people)) // [ "Robin", "Alberto", "Quinn", "Conner", "Lila", "Charlie" ]