从树结构中抓取所有“名字”

Question

这里有无穷无尽的树木问题，但对我来说，如果有人能帮我解决这个问题，那就太好了。我参加了新兵训练营的“模拟面试”，这是个问题。我没有很好的方法来接近它。

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

当你试图学习这些东西的时候，它可能会有点让人难以抗拒，特别是当你在筛选许多有不同角度解决它们的问题时。因此，在这方面的帮助将是非常感谢的！

Answer 1

一种方法是将reduce方法与递归方法结合使用，在这种方法中，您可以将concat到初始的name字符串.f的输出

​

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))

​

另一个递归解决方案是使用map和flatten方法，您可以选择所有名称，然后使用join方法将它们转换为字符串。

​

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


function getString({ name, children }) {
  return [name, ...children.map(getString).flat()].join(', ')
}

const result = getString(people)
console.log(result)

​

Answer 2

简单，只需使用递归。

​

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

使用新的flatMap可以轻松地转换递归数据结构。

​

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" ]

​

如果您的环境没有定义它，您可以编写您自己的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" ]

​

Node 11支持flatMap，但在babel-polyfill 7中不再包括它，现在必须是人工进口。

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

你可以手动填充它

​

// 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" ]

​