function iterate(instance) {
  for (let child of instance.tree_down) iterate(child);

  instance.children = instance.tree_down;
  delete instance.tree_down;
}

iterate(link_hierarchy);

console.log(link_hierarchy);

我认为最完整(和正确)的方法是:

if (old_key !== new_key) {
    Object.defineProperty(o, new_key,
        Object.getOwnPropertyDescriptor(o, old_key));
    delete o[old_key];
}

此方法确保重命名的属性的行为与原始属性相同。

另外，在我看来，把它包装成一个函数/方法，并把它放入对象的可能性。原型与你的问题无关。

在您最喜欢的编辑器中尝试一下

const obj = {1: 'a', 2: 'b', 3: 'c'}

const OLD_KEY = 1
const NEW_KEY = 10

const { [OLD_KEY]: replaceByKey, ...rest } = obj
const new_obj = {
  ...rest,
  [NEW_KEY]: replaceByKey
}

简单地这么做会有什么问题吗?

someObject = {...someObject, [newKey]: someObject.oldKey}
delete someObject.oldKey

如果愿意，可以将其包装在函数中:

const renameObjectKey = (object, oldKey, newKey) => {
    // if keys are the same, do nothing
    if (oldKey === newKey) return;
    // if old key doesn't exist, do nothing (alternatively, throw an error)
    if (!object.oldKey) return;
    // if new key already exists on object, do nothing (again - alternatively, throw an error)
    if (object.newKey !== undefined) return;

    object = { ...object, [newKey]: object[oldKey] };
    delete object[oldKey];

    return { ...object };
};

// in use
let myObject = {
    keyOne: 'abc',
    keyTwo: 123
};

// avoids mutating original
let renamed = renameObjectKey(myObject, 'keyTwo', 'renamedKey');

console.log(myObject, renamed);
// myObject
/* {
    "keyOne": "abc",
    "keyTwo": 123,
} */

// renamed
/* {
    "keyOne": "abc",
    "renamedKey": 123,
} */

如果你想保持对象的相同顺序

changeObjectKeyName(objectToChange, oldKeyName: string, newKeyName: string){
  const otherKeys = cloneDeep(objectToChange);
  delete otherKeys[oldKeyName];

  const changedKey = objectToChange[oldKeyName];
  return  {...{[newKeyName] : changedKey} , ...otherKeys};

}

使用方法:

changeObjectKeyName ( {'a' : 1}, 'a', 'A');

如果你要改变源对象，ES6可以在一行中完成。

delete Object.assign(o, {[newKey]: o[oldKey] })[oldKey];

如果你想创建一个新对象，可以用两行。

const newObject = {};
delete Object.assign(newObject, o, {[newKey]: o[oldKey] })[oldKey];