我有一堆扁平结构的物体。这些对象具有一个ID和一个ParentID属性，因此它们可以排列在树中。它们没有特定的顺序。每个ParentID属性不一定与ID结构中的 an 匹配。因此，它们可能是从这些物体中出现的几棵树。
您将如何处理这些对象以创建结果树？
我离解决方案不远，但我确信它远非最佳......
我需要创建这些树，然后以正确的顺序将数据插入数据库。
没有循环引用。当 ParentID == null 或在其他对象中找不到 ParentID 时，节点是 RootNode
Costo
问问题
115375 次
19 回答
133
将对象的 ID 存储在映射到特定对象的哈希表中。枚举所有对象并找到它们的父对象（如果存在）并相应地更新其父指针。
class MyObject
{ // The actual object
public int ParentID { get; set; }
public int ID { get; set; }
}
class Node
{
public List<Node> Children = new List<Node>();
public Node Parent { get; set; }
public MyObject AssociatedObject { get; set; }
}
IEnumerable<Node> BuildTreeAndGetRoots(List<MyObject> actualObjects)
{
Dictionary<int, Node> lookup = new Dictionary<int, Node>();
actualObjects.ForEach(x => lookup.Add(x.ID, new Node { AssociatedObject = x }));
foreach (var item in lookup.Values) {
Node proposedParent;
if (lookup.TryGetValue(item.AssociatedObject.ParentID, out proposedParent)) {
item.Parent = proposedParent;
proposedParent.Children.Add(item);
}
}
return lookup.Values.Where(x => x.Parent == null);
}
于 2009-01-14T19:17:58.547 回答
46
这是一个简单的 JavaScript 算法，用于将平面表解析为 N 次运行的父/子树结构：
var table = [
{parent_id: 0, id: 1, children: []},
{parent_id: 0, id: 2, children: []},
{parent_id: 0, id: 3, children: []},
{parent_id: 1, id: 4, children: []},
{parent_id: 1, id: 5, children: []},
{parent_id: 1, id: 6, children: []},
{parent_id: 2, id: 7, children: []},
{parent_id: 7, id: 8, children: []},
{parent_id: 8, id: 9, children: []},
{parent_id: 3, id: 10, children: []}
];
var root = {id:0, parent_id: null, children: []};
var node_list = { 0 : root};
for (var i = 0; i < table.length; i++) {
node_list[table[i].id] = table[i];
node_list[table[i].parent_id].children.push(node_list[table[i].id]);
}
console.log(root);
于 2011-11-23T21:03:46.703 回答
40
根据 Mehrdad Afshari 的回答和 Andrew Hanlon 对加速的评论，这是我的看法。
与原始任务的重要区别：根节点的 ID==parentID。
class MyObject
{ // The actual object
public int ParentID { get; set; }
public int ID { get; set; }
}
class Node
{
public List<Node> Children = new List<Node>();
public Node Parent { get; set; }
public MyObject Source { get; set; }
}
List<Node> BuildTreeAndGetRoots(List<MyObject> actualObjects)
{
var lookup = new Dictionary<int, Node>();
var rootNodes = new List<Node>();
foreach (var item in actualObjects)
{
// add us to lookup
Node ourNode;
if (lookup.TryGetValue(item.ID, out ourNode))
{ // was already found as a parent - register the actual object
ourNode.Source = item;
}
else
{
ourNode = new Node() { Source = item };
lookup.Add(item.ID, ourNode);
}
// hook into parent
if (item.ParentID == item.ID)
{ // is a root node
rootNodes.Add(ourNode);
}
else
{ // is a child row - so we have a parent
Node parentNode;
if (!lookup.TryGetValue(item.ParentID, out parentNode))
{ // unknown parent, construct preliminary parent
parentNode = new Node();
lookup.Add(item.ParentID, parentNode);
}
parentNode.Children.Add(ourNode);
ourNode.Parent = parentNode;
}
}
return rootNodes;
}
于 2015-04-29T11:17:59.403 回答
20
Python解决方案
def subtree(node, relationships):
return {
v: subtree(v, relationships)
for v in [x[0] for x in relationships if x[1] == node]
}
例如：
# (child, parent) pairs where -1 means no parent
flat_tree = [
(1, -1),
(4, 1),
(10, 4),
(11, 4),
(16, 11),
(17, 11),
(24, 17),
(25, 17),
(5, 1),
(8, 5),
(9, 5),
(7, 9),
(12, 9),
(22, 12),
(23, 12),
(2, 23),
(26, 23),
(27, 23),
(20, 9),
(21, 9)
]
subtree(-1, flat_tree)
产生：
{
"1": {
"4": {
"10": {},
"11": {
"16": {},
"17": {
"24": {},
"25": {}
}
}
},
"5": {
"8": {},
"9": {
"20": {},
"12": {
"22": {},
"23": {
"2": {},
"27": {},
"26": {}
}
},
"21": {},
"7": {}
}
}
}
}
于 2017-05-02T00:16:35.407 回答
14
JS 版本，它返回一个根或一组根，每个根都有一个包含相关子项的 Children 数组属性。不依赖于有序输入，相当紧凑，并且不使用递归。享受！
// creates a tree from a flat set of hierarchically related data
var MiracleGrow = function(treeData, key, parentKey)
{
var keys = [];
treeData.map(function(x){
x.Children = [];
keys.push(x[key]);
});
var roots = treeData.filter(function(x){return keys.indexOf(x[parentKey])==-1});
var nodes = [];
roots.map(function(x){nodes.push(x)});
while(nodes.length > 0)
{
var node = nodes.pop();
var children = treeData.filter(function(x){return x[parentKey] == node[key]});
children.map(function(x){
node.Children.push(x);
nodes.push(x)
});
}
if (roots.length==1) return roots[0];
return roots;
}
// demo/test data
var treeData = [
{id:9, name:'Led Zep', parent:null},
{id:10, name:'Jimmy', parent:9},
{id:11, name:'Robert', parent:9},
{id:12, name:'John', parent:9},
{id:8, name:'Elec Gtr Strings', parent:5},
{id:1, name:'Rush', parent:null},
{id:2, name:'Alex', parent:1},
{id:3, name:'Geddy', parent:1},
{id:4, name:'Neil', parent:1},
{id:5, name:'Gibson Les Paul', parent:2},
{id:6, name:'Pearl Kit', parent:4},
{id:7, name:'Rickenbacker', parent:3},
{id:100, name:'Santa', parent:99},
{id:101, name:'Elf', parent:100},
];
var root = MiracleGrow(treeData, "id", "parent")
console.log(root)
于 2016-12-14T15:04:31.587 回答
5
对于任何对 Eugene 解决方案的 C# 版本感兴趣的人，请注意node_list作为地图访问，因此请改用字典。
请记住，此解决方案仅在表按parent_id排序时才有效。
var table = new[]
{
new Node { parent_id = 0, id = 1 },
new Node { parent_id = 0, id = 2 },
new Node { parent_id = 0, id = 3 },
new Node { parent_id = 1, id = 4 },
new Node { parent_id = 1, id = 5 },
new Node { parent_id = 1, id = 6 },
new Node { parent_id = 2, id = 7 },
new Node { parent_id = 7, id = 8 },
new Node { parent_id = 8, id = 9 },
new Node { parent_id = 3, id = 10 },
};
var root = new Node { id = 0 };
var node_list = new Dictionary<int, Node>{
{ 0, root }
};
foreach (var item in table)
{
node_list.Add(item.id, item);
node_list[item.parent_id].children.Add(node_list[item.id]);
}
节点定义如下。
class Node
{
public int id { get; set; }
public int parent_id { get; set; }
public List<Node> children = new List<Node>();
}
于 2017-12-12T06:46:20.790 回答
4
在这里找到了一个很棒的 JavaScript 版本：http: //oskarhane.com/create-a-nested-array-recursively-in-javascript/
假设你有一个这样的数组：
const models = [
{id: 1, title: 'hello', parent: 0},
{id: 2, title: 'hello', parent: 0},
{id: 3, title: 'hello', parent: 1},
{id: 4, title: 'hello', parent: 3},
{id: 5, title: 'hello', parent: 4},
{id: 6, title: 'hello', parent: 4},
{id: 7, title: 'hello', parent: 3},
{id: 8, title: 'hello', parent: 2}
];
你想让对象像这样嵌套：
const nestedStructure = [
{
id: 1, title: 'hello', parent: 0, children: [
{
id: 3, title: 'hello', parent: 1, children: [
{
id: 4, title: 'hello', parent: 3, children: [
{id: 5, title: 'hello', parent: 4},
{id: 6, title: 'hello', parent: 4}
]
},
{id: 7, title: 'hello', parent: 3}
]
}
]
},
{
id: 2, title: 'hello', parent: 0, children: [
{id: 8, title: 'hello', parent: 2}
]
}
];
这是一个使它发生的递归函数。
function getNestedChildren(models, parentId) {
const nestedTreeStructure = [];
const length = models.length;
for (let i = 0; i < length; i++) { // for-loop for perf reasons, huge difference in ie11
const model = models[i];
if (model.parent == parentId) {
const children = getNestedChildren(models, model.id);
if (children.length > 0) {
model.children = children;
}
nestedTreeStructure.push(model);
}
}
return nestedTreeStructure;
}
用法：
const models = [
{id: 1, title: 'hello', parent: 0},
{id: 2, title: 'hello', parent: 0},
{id: 3, title: 'hello', parent: 1},
{id: 4, title: 'hello', parent: 3},
{id: 5, title: 'hello', parent: 4},
{id: 6, title: 'hello', parent: 4},
{id: 7, title: 'hello', parent: 3},
{id: 8, title: 'hello', parent: 2}
];
const nestedStructure = getNestedChildren(models, 0);
于 2018-03-20T14:51:56.583 回答
4
这是 Mehrdad Afshari 的答案的 java 解决方案。
import java.util.ArrayList;
import java.util.HashMap;
import java.util.Iterator;
import java.util.List;
import java.util.Map;
public class Tree {
Iterator<Node> buildTreeAndGetRoots(List<MyObject> actualObjects) {
Map<Integer, Node> lookup = new HashMap<>();
actualObjects.forEach(x -> lookup.put(x.id, new Node(x)));
//foreach (var item in lookup.Values)
lookup.values().forEach(item ->
{
Node proposedParent;
if (lookup.containsKey(item.associatedObject.parentId)) {
proposedParent = lookup.get(item.associatedObject.parentId);
item.parent = proposedParent;
proposedParent.children.add(item);
}
}
);
//return lookup.values.Where(x =>x.Parent ==null);
return lookup.values().stream().filter(x -> x.parent == null).iterator();
}
}
class MyObject { // The actual object
public int parentId;
public int id;
}
class Node {
public List<Node> children = new ArrayList<Node>();
public Node parent;
public MyObject associatedObject;
public Node(MyObject associatedObject) {
this.associatedObject = associatedObject;
}
}
于 2018-09-29T11:27:58.977 回答
3
我根据@Mehrdad Afshari 的回答用 C# 写了一个通用的解决方案：
void Example(List<MyObject> actualObjects)
{
List<TreeNode<MyObject>> treeRoots = actualObjects.BuildTree(obj => obj.ID, obj => obj.ParentID, -1);
}
public class TreeNode<T>
{
public TreeNode(T value)
{
Value = value;
Children = new List<TreeNode<T>>();
}
public T Value { get; private set; }
public List<TreeNode<T>> Children { get; private set; }
}
public static class TreeExtensions
{
public static List<TreeNode<TValue>> BuildTree<TKey, TValue>(this IEnumerable<TValue> objects, Func<TValue, TKey> keySelector, Func<TValue, TKey> parentKeySelector, TKey defaultKey = default(TKey))
{
var roots = new List<TreeNode<TValue>>();
var allNodes = objects.Select(overrideValue => new TreeNode<TValue>(overrideValue)).ToArray();
var nodesByRowId = allNodes.ToDictionary(node => keySelector(node.Value));
foreach (var currentNode in allNodes)
{
TKey parentKey = parentKeySelector(currentNode.Value);
if (Equals(parentKey, defaultKey))
{
roots.Add(currentNode);
}
else
{
nodesByRowId[parentKey].Children.Add(currentNode);
}
}
return roots;
}
}
于 2016-01-13T17:35:30.553 回答
1
在我看来这个问题很模糊，我可能会创建一个从 ID 到实际对象的映射。在伪java中（我没有检查它是否工作/编译），它可能是这样的：
Map<ID, FlatObject> flatObjectMap = new HashMap<ID, FlatObject>();
for (FlatObject object: flatStructure) {
flatObjectMap.put(object.ID, object);
}
并查找每个父母：
private FlatObject getParent(FlatObject object) {
getRealObject(object.ParentID);
}
private FlatObject getRealObject(ID objectID) {
flatObjectMap.get(objectID);
}
通过重用getRealObject(ID)并执行从对象到对象集合（或它们的 ID）的映射，您也可以得到一个父->子映射。
于 2009-01-14T19:27:52.037 回答
1
假设 Dictionary 类似于 TreeMap，我可以在 4 行代码和 O(n log n) 时间内完成此操作。
dict := Dictionary new.
ary do: [:each | dict at: each id put: each].
ary do: [:each | (dict at: each parent) addChild: each].
root := dict at: nil.
编辑：好的，现在我读到一些 parentID 是假的，所以忘记上面的，然后这样做：
dict := Dictionary new.
dict at: nil put: OrderedCollection new.
ary do: [:each | dict at: each id put: each].
ary do: [:each |
(dict at: each parent ifAbsent: [dict at: nil])
add: each].
roots := dict at: nil.
于 2009-01-14T20:03:13.353 回答
1
大多数答案都假设您希望在数据库之外执行此操作。如果您的树本质上是相对静态的，并且您只需要以某种方式将树映射到数据库中，您可能需要考虑在数据库端使用嵌套集表示。查看 Joe Celko 的书籍（或此处
查看 Celko 的概述）。
如果仍然绑定到 Oracle dbs，请查看他们的 CONNECT BY 以获得直接的 SQL 方法。
无论采用哪种方法，您都可以在将数据加载到数据库之前完全跳过树的映射。只是想我会提供这个作为替代方案，它可能完全不适合您的特定需求。原始问题的整个“正确顺序”部分在某种程度上暗示您出于某种原因需要数据库中的“正确”顺序？这也可能促使我去处理那里的树木。
于 2009-01-14T20:31:12.813 回答
1
这与提问者所寻找的并不完全相同，但我很难理解这里提供的措辞模棱两可的答案，我仍然认为这个答案符合标题。
我的答案是将一个平面结构映射到一个直接在对象上的树，你所拥有的只是ParentID每个对象上的一个。ParentID是null或者0如果它是一个根。与提问者相反，我假设所有 valid 都ParentID指向列表中的其他内容：
var rootNodes = new List<DTIntranetMenuItem>();
var dictIntranetMenuItems = new Dictionary<long, DTIntranetMenuItem>();
//Convert the flat database items to the DTO's,
//that has a list of children instead of a ParentID.
foreach (var efIntranetMenuItem in flatIntranetMenuItems) //List<tblIntranetMenuItem>
{
//Automapper (nuget)
DTIntranetMenuItem intranetMenuItem =
Mapper.Map<DTIntranetMenuItem>(efIntranetMenuItem);
intranetMenuItem.Children = new List<DTIntranetMenuItem>();
dictIntranetMenuItems.Add(efIntranetMenuItem.ID, intranetMenuItem);
}
foreach (var efIntranetMenuItem in flatIntranetMenuItems)
{
//Getting the equivalent object of the converted ones
DTIntranetMenuItem intranetMenuItem = dictIntranetMenuItems[efIntranetMenuItem.ID];
if (efIntranetMenuItem.ParentID == null || efIntranetMenuItem.ParentID <= 0)
{
rootNodes.Add(intranetMenuItem);
}
else
{
var parent = dictIntranetMenuItems[efIntranetMenuItem.ParentID.Value];
parent.Children.Add(intranetMenuItem);
//intranetMenuItem.Parent = parent;
}
}
return rootNodes;
于 2016-11-11T18:02:13.307 回答
1
这是一个红宝石实现：
它将按属性名称或方法调用的结果进行分类。
CatalogGenerator = ->(depth) do
if depth != 0
->(hash, key) do
hash[key] = Hash.new(&CatalogGenerator[depth - 1])
end
else
->(hash, key) do
hash[key] = []
end
end
end
def catalog(collection, root_name: :root, by:)
method_names = [*by]
log = Hash.new(&CatalogGenerator[method_names.length])
tree = collection.each_with_object(log) do |item, catalog|
path = method_names.map { |method_name| item.public_send(method_name)}.unshift(root_name.to_sym)
catalog.dig(*path) << item
end
tree.with_indifferent_access
end
students = [#<Student:0x007f891d0b4818 id: 33999, status: "on_hold", tenant_id: 95>,
#<Student:0x007f891d0b4570 id: 7635, status: "on_hold", tenant_id: 6>,
#<Student:0x007f891d0b42c8 id: 37220, status: "on_hold", tenant_id: 6>,
#<Student:0x007f891d0b4020 id: 3444, status: "ready_for_match", tenant_id: 15>,
#<Student:0x007f8931d5ab58 id: 25166, status: "in_partnership", tenant_id: 10>]
catalog students, by: [:tenant_id, :status]
# this would out put the following
{"root"=>
{95=>
{"on_hold"=>
[#<Student:0x007f891d0b4818
id: 33999,
status: "on_hold",
tenant_id: 95>]},
6=>
{"on_hold"=>
[#<Student:0x007f891d0b4570 id: 7635, status: "on_hold", tenant_id: 6>,
#<Student:0x007f891d0b42c8
id: 37220,
status: "on_hold",
tenant_id: 6>]},
15=>
{"ready_for_match"=>
[#<Student:0x007f891d0b4020
id: 3444,
status: "ready_for_match",
tenant_id: 15>]},
10=>
{"in_partnership"=>
[#<Student:0x007f8931d5ab58
id: 25166,
status: "in_partnership",
tenant_id: 10>]}}}
于 2018-08-09T20:18:12.330 回答
1
接受的答案对我来说看起来太复杂了，所以我添加了它的 Ruby 和 NodeJS 版本
假设平面节点列表具有以下结构：
nodes = [
{ id: 7, parent_id: 1 },
...
] # ruby
nodes = [
{ id: 7, parentId: 1 },
...
] # nodeJS
将上面的平面列表结构变成树的函数如下所示
对于红宝石：
def to_tree(nodes)
nodes.each do |node|
parent = nodes.find { |another| another[:id] == node[:parent_id] }
next unless parent
node[:parent] = parent
parent[:children] ||= []
parent[:children] << node
end
nodes.select { |node| node[:parent].nil? }
end
对于 NodeJS：
const toTree = (nodes) => {
nodes.forEach((node) => {
const parent = nodes.find((another) => another.id == node.parentId)
if(parent == null) return;
node.parent = parent;
parent.children = parent.children || [];
parent.children = parent.children.concat(node);
});
return nodes.filter((node) => node.parent == null)
};
于 2019-04-17T07:53:23.763 回答
1
一种优雅的方法是将列表中的项目表示为字符串，其中包含一个点分隔的父列表，最后是一个值：
server.port=90
server.hostname=localhost
client.serverport=90
client.database.port=1234
client.database.host=localhost
当组装一棵树时，你最终会得到类似的东西：
server:
port: 90
hostname: localhost
client:
serverport=1234
database:
port: 1234
host: localhost
于 2019-06-15T08:15:20.333 回答
0
您是否只使用这些属性？如果没有，最好创建一个子节点数组，您可以在其中循环遍历所有这些对象以构建此类属性。从那里，选择有子节点但没有父节点的节点，然后从上到下迭代地构建你的树。
于 2009-01-14T19:30:02.423 回答
0
爪哇版
// node
@Data
public class Node {
private Long id;
private Long parentId;
private String name;
private List<Node> children = new ArrayList<>();
}
// flat list to tree
List<Node> nodes = new ArrayList();// load nodes from db or network
Map<Long, Node> nodeMap = new HashMap();
nodes.forEach(node -> {
if (!nodeMap.containsKey(node.getId)) nodeMap.put(node.getId, node);
if (nodeMap.containsKey(node.getParentId)) {
Node parent = nodeMap.get(node.getParentId);
node.setParentId(parent.getId());
parent.getChildren().add(node);
}
});
// tree node
List<Node> treeNode = nodeMap .values().stream().filter(n -> n.getParentId() == null).collect(Collectors.toList());
于 2020-06-10T10:06:52.737 回答
0
请参阅下面带有简单测试程序的完整 Java 8+ 解决方案。
这是对接受多个根的@“Vimal Bhatt”版本进行修改的解决方案
package tree;
import java.util.*;
import java.util.function.Consumer;
import java.util.stream.Stream;
public class Tree {
private <T> void swap(T[] input, int a, int b) {
T tmp = input[a];
input[a] = input[b];
input[b] = tmp;
}
public static void main(String[] args) {
Random r = new Random(8);
MyObject[] myObjects = new MyObject[]{
new MyObject(6, 3),
new MyObject(7, 5),
new MyObject(8, 0),
new MyObject(1, 0),
new MyObject(15, 12),
new MyObject(12, 0),
new MyObject(3, 5),
new MyObject(4, 3),
new MyObject(5, 2),
new MyObject(2, 1),
new MyObject(21, 8),
new MyObject(9, 1)
};
Tree t = new Tree();
// cinco trocas arbitrarias de posição
for (int i = 0; i < 5; i++) {
int a = r.nextInt(7) + 1;
int b = r.nextInt(7) + 1;
t.swap(myObjects, a, b);
}
System.out.println("The list have " + myObjects.length + " objects");
for (MyObject o: myObjects) {
System.out.print(" " + o);
}
Iterator<Node> iterator = t.buildTreeAndGetRoots(Arrays.asList(myObjects));
int counter = 0;
System.out.println("");
while (iterator.hasNext()) {
Node obj = iterator.next();
System.out.println(++counter + "\t" + obj.associatedObject.id + "\t-> " + obj.associatedObject.parentId);
}
}
Iterator<Node> buildTreeAndGetRoots(List<MyObject> actualObjects) {
Node root = null;
Map<Integer, Node> lookup = new HashMap<>();
actualObjects.forEach(x -> lookup.put(x.id, new Node(x)));
Stream<Node> roots = actualObjects.stream()
.filter(x -> x.parentId == 0)
.map(x -> new Node(x));
Consumer<Node> nodeConsumer = item -> {
Node proposedParent;
if (lookup.containsKey(item.associatedObject.parentId)) {
proposedParent = lookup.get(item.associatedObject.parentId);
item.parent = proposedParent;
proposedParent.children.add(item);
}
};
lookup.values().forEach(nodeConsumer);
Stream<Node> s2 = lookup.values().stream().filter(e -> e.associatedObject.parentId != 0);
return Stream.concat(roots, s2).iterator();
}
}
class MyObject { // The actual object
public int parentId;
public int id;
MyObject(int id, int parent) {
this.parentId = parent;
this.id = id;
}
@Override
public String toString() {
return "{ " +
"parent: " + parentId +
", id: " + id +
" }" ;
}
}
class Node {
public List<Node> children = new ArrayList<Node>();
public Node parent;
public MyObject associatedObject;
public Node(MyObject associatedObject) {
this.associatedObject = associatedObject;
}
}
于 2021-06-05T01:21:45.757 回答