javascript - 判断两个树形数组是否相同(list1旧数组,list2新数组), 不同返回 不同的属性 属性值 求大佬指点 返回true或false也可以 顺序可以打乱?

Question

var list1 = {
  transactionYear: 0,
  transactionMonth: 0,
  transactionDay: 1,
  groupId: "428CB095-A67C-4A88-9CD1-E472E496446E",
  // 数组
  accessoryList: [
    {
      id: "CF0D7DEF-5702-4CBD-B8AF-51F499586A28",
      createTime: "2023-06-16 11:05:41",
    },
  ],
  // 数组
  relationList: [
    {
      id: "29406050-AE9D-40EA-9EDA-F8084D2F798A",
      createTime: "2023-06-16 11:05:41",
    },
  ],
  // 数组
  employeeList: [
    {
      reportEmployee: false,
      employeePerformance: 101.0,
    },
  ],
}

var list2 = {
  transactionYear: 0,
  transactionMonth: 0,
  transactionDay: 1,
  groupId: "428CB095-A67C-4A88-9CD1-E472E496446E",
  // 数组
  accessoryList: [
    {
      id: "CF0D7DEF-5702-4CBD-B8AF-51F499586A28",
      createTime: "2023-06-16 11:05:41",
    },
  ],
  // 数组
  relationList: [
    {
      id: "29406050-AE9D-40EA-9EDA-F8084D2F798A",
      createTime: "2023-06-16 11:05:41",
    },
  ],
  // 数组
  employeeList: [
    {
      reportEmployee: false,
      employeePerformance: 101.0,
    },
  ],
}

Answer 1

function isSame(a,b) {
  const type = v => ({}).toString.call(v);
  // 是否传入相同对象
  if(a === b) return true;
  // 是否类型一致
  if(type(a) !== type(b)) return false;
  // 空值判断
  // 因为类型判断在前，故而空值直接返回true
  if(a === null || a === void 0) return true;
  // 数组优先判断长度，长度一致再遍历对比
  if(Array.isArray(a)) return a.length === b.length && a.every((v,i) => isSame(v, b[i]));
  // 对象同数组，优先判断key数量
  if(typeof a == 'object') return Object.keys(a).length == Object.keys(b).length && Object.entries(a).every(([k,v]) => isSame(v, b[k]))
  // 最后判断值是否相等
  return a === b;
}

Answer 2

function isEqual(a, b) {
    if (Object.is(a, b)) return true;
    if (
        typeof a !== "object" ||
        a === null ||
        typeof b !== "object" ||
        b === null
    ) {
        return false;
    }
    const keysA = Object.keys(a);
    const keysB = Object.keys(b);
    if (keysA.length !== keysB.length) return false;
    for (let i = 0; i < keysA.length; ++i) {
        if (
            !Object.hasOwn(b, keysA[i]) ||
            !isEqual(a[keysA[i]], b[keysB[i]])
        ) {
            return false;
        }
    }
    return true;
}
console.log(isEqual(list1, list2));