Question

我有一个用JavaScript编写的应用程序。在这个应用程序中，我有一个JavaScript对象树。可以在此JSFiddle中看到示例树和下面的代码。

我正在尝试编写一个函数，它将返回一个作为祖先的ID列表。具有特定ID的元素的祖先。目前，我有以下内容：

function getAncestors(childId, branch) {
  var ancestors = [];
  for (var i = 0; i < branch.length; i++) {
    for (var j = 0; j < branch[i].children.length; j++) {
      if (branch[i].children[j].id === childId) {
        ancestors.push(branch[i].id);
        return ancestors;
      } else {
        var _ancestors = getAncestors(childId, branch[i].children);
        for (var k = 0; k < _ancestors.length; k++) {
          if (ancestors.indexOf(_ancestors[k]) === -1) {
            ancestors.push(_ancestors[k]);
          }
        }
      }
    }
  }
  return ancestors;
}

它总是返回第一个父级。但是，它并不能归还所有祖先。例如，在JSFiddle中，我尝试按顺序获取包含：[201,2]的数组。我不确定我做错了什么。我一直盯着这看，看起来很正确。但是，显然，它没有用。

Answer 1

您可以使用迭代和递归方法，对Array#some和实际父母进行回调。

function getParents(id, array) {
    function iter(parents) {
        return function (a) {
            if (a.id === id) {
                result = parents;
                return true;
            }
            return a.children && a.children.some(iter([a.id].concat(parents)));
        };
    }

    var result;
    array.some(iter([]));
    return result;
}

var tree = [{ id: 1, name: 'Left', children: [{ id: 100, name: 'C1', children: [{ id: 1000, name: 'C2A', children: [] }, { id: 1001, name: 'D2A', children: [] }, { id: 1002, name: 'C2B', children: [] }] }, { id: 101, name: 'C2', children: [{ id: 2000, name: 'D7B', children: [] }, { id: 2001, name: 'E2A', children: [] }, { id: 2002, name: 'C2X', children: [] }] }] }, { id: 2, name: 'Middle', children: [{ id: 200, name: 'Z1', children: [{ id: 3000, name: 'R2A', children: [] }, { id: 3001, name: 'DYA', children: [] }, { id: 3002, name: 'Q2B', children: [] }] }, { id: 201, name: 'X2', children: [{ id: 4000, name: 'DMA', children: [] }, { id: 4001, name: 'ELA', children: [] }, { id: 4002, name: 'CRX', children: [] }] }] }, { id: 3, name: 'Right', children: [{ id: 300, name: 'Z1', children: [{ id: 5000, name: 'F7A', children: [] }, { id: 5001, name: 'EW5', children: [] }, { id: 5002, name: 'D5B', children: [] }] }, { id: 301, name: 'X2', children: [{ id: 6000, name: 'OMA', children: [] }, { id: 6001, name: 'NLA', children: [] }, { id: 6002, name: 'MRX', children: [] }] }] }];

console.log(getParents(4001, tree));

版本没有移动数组，但带有实际深度的标记。结果现在逆转了。

function getParents(id, array) {
    function iter(depth) {
        return function (a) {
            result[depth] = a.id;
            if (a.id === id) {
                result.length = depth;
                return true;
            }
            return a.children && a.children.some(iter(depth + 1));
        };
    }

    var result = [];
    return array.some(iter(0)) && result;
}

var tree = [{ id: 1, name: 'Left', children: [{ id: 100, name: 'C1', children: [{ id: 1000, name: 'C2A', children: [] }, { id: 1001, name: 'D2A', children: [] }, { id: 1002, name: 'C2B', children: [] }] }, { id: 101, name: 'C2', children: [{ id: 2000, name: 'D7B', children: [] }, { id: 2001, name: 'E2A', children: [] }, { id: 2002, name: 'C2X', children: [] }] }] }, { id: 2, name: 'Middle', children: [{ id: 200, name: 'Z1', children: [{ id: 3000, name: 'R2A', children: [] }, { id: 3001, name: 'DYA', children: [] }, { id: 3002, name: 'Q2B', children: [] }] }, { id: 201, name: 'X2', children: [{ id: 4000, name: 'DMA', children: [] }, { id: 4001, name: 'ELA', children: [] }, { id: 4002, name: 'CRX', children: [] }] }] }, { id: 3, name: 'Right', children: [{ id: 300, name: 'Z1', children: [{ id: 5000, name: 'F7A', children: [] }, { id: 5001, name: 'EW5', children: [] }, { id: 5002, name: 'D5B', children: [] }] }, { id: 301, name: 'X2', children: [{ id: 6000, name: 'OMA', children: [] }, { id: 6001, name: 'NLA', children: [] }, { id: 6002, name: 'MRX', children: [] }] }] }];

console.log(getParents(4001, tree));

Answer 2

这是工作代码（使用迭代）：

function getAncestors(childId, newbranch) {
    for (var i = 0; i < newbranch.length; i++) { // Search all branches
        var branch = newbranch[i];

        if (branch.id === childId) {
            return [];
        } else {
            if (branch.children && branch.children.length) {
                var x;

                if ((x = getAncestors(childId, branch.children)) !== false) {
                    x.push(branch.id)
                    return x;
                }
            }
        }
    }

    return false;
}

结果：

[201,2]

编辑（更短）

function getAncestors(childId, branch) {
    var i, ancestors;

    for (i = 0; !ancestors && branch && i < branch.length; i++) {
        if (branch[i].id === childId) return [];
        ancestors = getAncestors(childId, branch[i].children);
        if (ancestors) ancestors.push(branch[i].id);
    }
    return ancestors;
}

JavaScript - 递归查找父节点

2 个答案: