只要迭代一遍，如果找到了一个子元素，就从子元素中获取串联的数组。

function flat(array) {
    var result = [];
    array.forEach(function (a) {
        result.push(a);
        if (Array.isArray(a.children)) {
            result = result.concat(flat(a.children));
        }
    });
    return result;
}

var data = [{ id: "acc.1260446672222.11", type: "EXPENSES_FOLDER", name: "Expense Group", balance: 3418.11, children: [{ id: "acc.1260446672238.27", type: "EXPENSE", name: "Advertising, Promotion and Entertainment Account", balance: 0, children: [] }, { id: "acc.9a2492ba-0d82-4f4a-a1b4-14868f1e1a39", type: "EXPENSES_FOLDER", name: "Premises Costs", balance: 0, children: [{ id: "acc.287ba5b6-5536-428b-950f-d71d2af73ccc", type: "EXPENSE", name: "Use of Home - Gas", balance: 0, children: [] }, { id: "acc.7091ee15-3f02-4bd1-94e5-5918cf986969", type: "EXPENSE", name: "Hire of Venue, Studios, Teaching Rooms", balance: 0, children: [] }] }, { id: "acc.827ec446-edeb-4f2b-8032-d306292d2d83", type: "EXPENSES_FOLDER", name: "Administrative Expenses", balance: 558.61, children: [{ id: "acc.0ed5fc81-7734-4452-86a9-db22a6b0f8e8", type: "EXPENSE", name: "Bank Charges", balance: 15, children: [] }, { id: "acc.e2cdb2c0-8565-4991-a35a-d4596b0ddf45", type: "EXPENSE", name: "Software & Computer Peripherals", balance: 417.13, children: [] }, { id: "acc.96d5d00e-43f4-4d3a-b97b-fdf258c65514", type: "EXPENSE", name: "Printing, photocopying etc", balance: 55.93, children: [] }, { id: "acc.494dd64a-4fb3-42b8-be3e-8f3b59a2ef59", type: "EXPENSE", name: "Artists Administration Service", balance: 0, children: [] }, { id: "acc.1260446672238.35", type: "EXPENSE", name: "Stationery", balance: 0, children: [] }, { id: "acc.96d89d0d-5465-488b-b37f-d41ca114c5e6", type: "EXPENSE", name: "Mobile Telephone", balance: 41.19, children: [] }, { id: "acc.1260446672238.33", type: "EXPENSE", name: "Home Telephone", balance: 0, children: [] }, { id: "acc.1260446672238.38", type: "EXPENSE", name: "Postage/delivery", balance: 29.36, children: [] }] }, { id: "acc.b9c9bbc7-43df-472e-9ac8-c7c76f08f49a", type: "EXPENSES_FOLDER", name: "Instruments, Equipment Maintenance etc", balance: 1002.48, children: [{ id: "acc.1260446672238.32", type: "OTHER_EXPENSES", name: "Instrument Insurance", balance: 157.48, children: [] }, { id: "acc.2a1cca15-2868-4770-a3e7-d43a6268c6a1", type: "EXPENSE", name: "Instrument Repairs & Maintenance", balance: 845, children: [] }, { id: "acc.a908aee0-84fb-450a-916b-4cec25265aef", type: "EXPENSE", name: "Accessories & Replacement Parts", balance: 0, children: [] }] }, { id: "acc.a42cdd86-0d9e-4f3f-af0d-7c4525374731", type: "EXPENSES_FOLDER", name: "Motor Vehicle", balance: 0, children: [{ id: "acc.cb325e7e-0ce4-4c78-9cb4-20659df733a6", type: "EXPENSE", name: "Fuel and Oil", balance: 0, children: [] }] }, { id: "acc.4bdd9e26-ce64-4e7f-b46a-82ec9de06ded", type: "EXPENSES_FOLDER", name: "Other Travel", balance: 132.1, children: [{ id: "acc.77dd2142-f2de-4a2c-9247-061d0661bc0a", type: "EXPENSE", name: "Taxis", balance: 24.5, children: [] }, { id: "acc.2b54abdd-7ef5-43cd-bdb9-c8c981b59ff2", type: "EXPENSE", name: "Public Transport", balance: 107.6, children: [] }] }, { id: "acc.e4695b70-31fa-4e23-afd0-97335dcd5b9e", type: "EXPENSE", name: "Subsitence", balance: 0, children: [] }, { id: "acc.02d222bf-4dff-4308-afe9-69b93f412ada", type: "EXPENSE", name: "Hotel and Accomodation", balance: 0, children: [] }, { id: "acc.d61cd5b4-2c80-4ab8-93d0-9d5726bd253b", type: "EXPENSES_FOLDER", name: "Fees and Commission Paid", balance: 0, children: [{ id: "acc.1262189019758.7", type: "EXPENSE", name: "Pupils exam entry fees", balance: 0, children: [] }, { id: "acc.a7d7efd3-d0da-4704-babb-079b6077f3fe", type: "EXPENSE", name: "Audition, competition entry fees", balance: 0, children: [] }, { id: "acc.3b91ee4e-40a8-46d8-aa05-3afa5974b3ef", type: "EXPENSE", name: "Deputies, Other Musicians", balance: 0, children: [] }] }, { id: "acc.250d6872-6023-4599-a0b6-b7159eebbfa1", type: "EXPENSES_FOLDER", name: "Other Professional Expenses", balance: 1739.42, children: [{ id: "acc.b7315228-f85a-4ffb-9199-d1128a409e5f", type: "EXPENSE", name: "Promotion & Publicity", balance: 138.6, children: [] }, { id: "acc.69ca2005-d7a0-448b-b70c-dafb128a48ae", type: "EXPENSE", name: "Other Expenses", balance: 364.5, children: [] }, { id: "acc.dcd999d2-4e18-41be-b9cc-218d4034b88e", type: "EXPENSE", name: "Office Equipment, Furniture", balance: 0, children: [] }, { id: "acc.e0460706-d5c9-4c40-9d1e-0d2058864b92", type: "EXPENSE", name: "CDs, Dowloads etc", balance: 67.57, children: [] }, { id: "acc.1866df79-9e44-459a-a978-727904987469", type: "EXPENSE", name: "Professional Books, Magazines", balance: 104.01, children: [] }, { id: "acc.24c1651d-e7ae-48bc-a32d-311427e0fcea", type: "EXPENSE", name: "Professional Associations", balance: 272.17, children: [] }, { id: "acc.289ab0ac-b9d3-435e-ac82-9da9702b7d4b", type: "EXPENSE", name: "Tuition", balance: 470, children: [] }, { id: "acc.f24cf99b-6291-4b9f-821e-425f4909d4e1", type: "EXPENSE", name: "Scores, Manuscript Paper etc", balance: 215.32, children: [] }, { id: "acc.1af95953-56f0-455e-9d0a-7c4e0477cf0d", type: "EXPENSE", name: "Performance Clothing", balance: 0, children: [] }, { id: "acc.c0585577-535a-4ae2-a02b-e5b249f67c67", type: "EXPENSE", name: "Concerts, Shows etc", balance: 107.25, children: [] }] }, { id: "acc.1260446672222.24", type: "ADMIN", name: "Administrative Expenses", balance: 0, children: [] }, { id: "acc.1260446672238.26", type: "TRAVEL", name: "Travel and Subsistence Account", balance: -14.5, children: [] }, { id: "acc.1260446672238.28", type: "LEGAL", name: "Legal and Professional Costs Account", balance: 0, children: [] }, { id: "acc.1260446672238.36", type: "OTHER_EXPENSES", name: "Rent/Rates", balance: 0, children: [] }, { id: "acc.1262191376548.37", type: "EXPENSE", name: "Research", balance: 0, children: [] }, { id: "acc.1262191388329.38", type: "EXPENSE", name: "Professional Development", balance: 0, children: [] }, { id: "acc.1262192291558.52", type: "EXPENSE", name: "Professional Presentation", balance: 0, children: [] }, { id: "acc.1262193596634.72", type: "EXPENSE", name: "Subscriptions", balance: 0, children: [] }, { id: "acc.1262265941130.16", type: "EXPENSE", name: "Piano accompaniment", balance: 0, children: [] }, { id: "acc.1267370824329.1", type: "EXPENSE", name: "Cost of Sales", balance: 0, children: [] }] }],
    result = flat(data);

document.write('<pre>' + JSON.stringify(result, 0, 4) + '</pre>');

我试着理解这个版本，包括它的好处和含义

function flatten(into, node){
    if(node == null) return into;
    if(Array.isArray(node)) return node.reduce(flatten, into);
    into.push(node);
    return flatten(into, node.children);
}

var out = flatten([], yourArray);

参数的顺序一开始可能会让人困惑，除非你习惯于简化，或者使用选项优先，数据最后的概念，当涉及到FP和currying或者部分应用时，这是非常方便的。

您可以创建一个递归函数来查看每个帐户，如果它有子帐户，则在子帐户上调用自己，该函数应该返回一个扁平化的帐户数组。
这个例子比实际要复杂一些，因为我假设顺序很重要，父节点应该在子节点之前。

function flattenAccounts(accounts){
  var a = [];
  for(var i=0;i<accounts.length;i++){
    var o = accounts[i];
    if(o.children){
      var c = flattenAccounts(o.children);
      if(c){
          a = a.concat(c);
      }
    }
    a.push(o)
  }
  return a;
}

这是一个更简单的例子，但是子节点会在父节点之前结束。

function flattenAccounts(accounts){
  var a = [];
  for(var i=0;i<accounts.length;i++){
    if(accounts[i].children){
      a = a.concat(flattenAccounts(accounts[i].children) )
    }
    a.push(accounts[i])
  }
  return a;
}

您可以循环遍历数组条目，检查它是否是一个isArray()数组，然后递归地将数组条目扁平化，如下所示：

function deepFlattenArray(arr) {
  const finalArray = [];
  // Loop through the array contents
  for (let i = 0; i < arr.length; i++) {
    if (Array.isArray(arr[i])) {
      // Recursively flatten entries that are arrays and push into the finalArray
      finalArray.push(...deepFlattenArray(arr[i]));
    } else {
      // Push entries that are not arrays
      finalArray.push(arr[i]);
    }
  }
  return finalArray;
}

也可以使用JavaScript递归Map函数，获取嵌套数组的子节点，如下所示：

const dataTree = [//array above];
const getMembers = (members) => {
  let children = [];
  const flattenMembers = members.map(m => {
    if (m.children && m.children.length) {
      children = [...children, ...m.children];
    }
    return m;
  });

  return flattenMembers.concat(children.length ? getMembers(children) : children);
};

getMembers(dataTree);

Source Here

我想指出的是，所有的答案都是通过将子元素复制到根元素，但将原始的子元素保留在原处，从而创建副本-而不是平面数组。
根据数据的使用方式，这可能会导致一些问题，我当然不想被我的团队发现留下一个混乱的数组。
我找不到一个解决方案来删除递归函数中的子函数，所以我添加了第二个“cleanup”函数。
根据Nina Scholz的公认答案：

function flattenArr(arr) {

  var dirtyArr = function(arr) {
    let result = [];

    arr.forEach((arrItem) => {
      result.push(arrItem);

      if (Array.isArray(arrItem.children)) {
        result = result.concat(dirtyArr(arrItem.children));
      }
    });

    return result;
  };

  var cleanArr = function(arr) {
    arr.forEach((arrItem) => {
      if (arrItem.hasOwnProperty('children')) {
        delete arrItem.children;
      }
    });
    return arr;
  };

  return cleanArr(dirtyArr(arr));

}

还有另一种方法..它很简单，但它的工作。

function flattenChildren(arr: any): any {
  return arr.flatMap(({ children, ...o }: { children: any; o: any }) => [
    o,
    ...flattenChildren(children),
  ])
}