Question

我有一堆父/子对，我想尽可能转变为分层树结构。例如，这些可能是配对：

Child : Parent
    H : Ga
    F : G
    G : D
    E : D
    A : E
    B : C
    C : E
    D : NULL
    Z : Y
    Y : X
    X: NULL

需要将其转化为（a）层次结构树：

   D
    ├── E
    │   ├── A
    │   │   └── B
    │   └── C   
    └── G
    |   ├── F
    |   └── H
    |
    X
    |
    └── Y
        |
        └──Z

在Java中，我如何从包含child =＆gt;父对的arrayList转到像那样的Tree？

我需要这个操作的输出是arrayList包含两个元素D和X. 反过来，每个孩子都有孩子的名单，而孩子的名单又包含一个孩子的名单等等

public class MegaMenuDTO {
    private String Id;
    private String name;
    private String parentId;
    private List<MegaMenuDTO> childrenItems=new ArrayList<MegaMenuDTO>();

    public String getId() {
        return Id;
    }
    public void setId(String id) {
        Id = id;
    }
    public String getName() {
        return name;
    }
    public void setName(String name) {
        this.name = name;
    }
    public String getParentId() {
        return parentId;
    }
    public void setParentId(String parentId) {
        this.parentId = parentId;
    }
    public List<MegaMenuDTO> getChildrenItems() {
        return childrenItems;
    }
    public void setChildrenItems(List<MegaMenuDTO> childrenItems) {
        this.childrenItems = childrenItems;
    }
}

我的第一次尝试是

private void arrangeMegaMenuTree(MegaMenuDTO grandParent,
        MegaMenuDTO parent, List<MegaMenuDTO> children) {

    for (MegaMenuDTO child : children) {
        if (child.getParentId().equals(parent.getId())) {
            arrangeMegaMenuTree(parent, child, children);
        }
    }

    if (!grandParent.getId().equals(parent.getId())) {
        grandParent.getChildrenItems().add(parent);
        // children.remove(parent);
    }

}

另一次尝试

private List<MegaMenuDTO> arrangeMegaMenuTree(MegaMenuDTOparent,List<MegaMenuDTO>menuItems) {

    for (MegaMenuDTO child : menuItems) {

        if (parent.getId().equals(child.getId())) {
            continue;
        }
        if (hasChildren(child, menuItems)) {
            parent.setChildrenItems(arrangeMegaMenuTree(child, menuItems
                    .subList(menuItems.indexOf(child), menuItems.size())));
        } else {
            List<MegaMenuDTO> tempList = new ArrayList<MegaMenuDTO>();
            tempList.add(child);
            return tempList;
        }
    }
    return null;
}

private boolean hasChildren(MegaMenuDTO parent, List<MegaMenuDTO> children) {
    for (MegaMenuDTO child : children) {

        if (child.getParentId().equals(parent.getId())) {
            return true;
        }
    }
    return false;
}

Answer 1

假设您的Node结构类似于：

Map<Object, Node> temp = new HashMap<>();
for (Pair pair: inputList) {
  Node parent = temp.getOrDefault(pair.parent.id, new Node(pair.parent.id));
  Node child = temp.getOrDefault(pair.child.id, new Node(pair.child.id));
  parent.children.add(child);
  child.parent = parent;
  temp.put(parent.id, parent);
  temp.put(child.id, child);
}

然后你首先开始迭代你的输入列表，并从ids-＆gt;节点创建一个映射（这用于在树仍然是非结构化的情况下获取节点）;

for (Node n: temp.values()) {
  if (n.parent==null) {
    root = n; break;
  }
}

现在，您可以在地图上迭代搜索树的根

awk

此代码假定您的数据“有效”（没有重复的子条目，单根等）。否则您可以轻松地进行调整。

Answer 2

这是基于第一个答案和问题更新的替代解决方案...... :)

主要方法

import java.util.ArrayList;
import java.util.HashMap;
import java.util.List;
import java.util.Map;

public class Main2 {

    public static void main(String[] args) {

        // input
        ArrayList<Pair> pairs = new ArrayList<Pair>();
        pairs.add(new Pair( "H" , "G"));
        pairs.add(new Pair( "F" , "G"));
        pairs.add(new Pair( "G" , "D"));
        // ...


        // Arrange
        // String corresponds to the Id
        Map<String, MegaMenuDTO> hm = new HashMap<>();


        // you are using MegaMenuDTO as Linked list with next and before link 

        // populate a Map
        for(Pair p:pairs){

            //  ----- Child -----
            MegaMenuDTO mmdChild ;
            if(hm.containsKey(p.getChildId())){
                mmdChild = hm.get(p.getChildId());
            }
            else{
                mmdChild = new MegaMenuDTO();
                hm.put(p.getChildId(),mmdChild);
            }           
            mmdChild.setId(p.getChildId());
            mmdChild.setParentId(p.getParentId());
            // no need to set ChildrenItems list because the constructor created a new empty list



            // ------ Parent ----
            MegaMenuDTO mmdParent ;
            if(hm.containsKey(p.getParentId())){
                mmdParent = hm.get(p.getParentId());
            }
            else{
                mmdParent = new MegaMenuDTO();
                hm.put(p.getParentId(),mmdParent);
            }
            mmdParent.setId(p.getParentId());
            mmdParent.setParentId("null");                              
            mmdParent.addChildrenItem(mmdChild);


        }

        // Get the root
        List<MegaMenuDTO> DX = new ArrayList<MegaMenuDTO>(); 
        for(MegaMenuDTO mmd : hm.values()){
            if(mmd.getParentId().equals("null"))
                DX.add(mmd);
        }

        // Print 
        for(MegaMenuDTO mmd: DX){
            System.out.println("DX contains "+DX.size()+" that are : "+ mmd);
        }

    }

}

配对类：

public class Pair {
    private String childId ;
    private String parentId;

    public Pair(String childId, String parentId) {
        this.childId = childId;
        this.parentId = parentId;
    }
    public String getChildId() {
        return childId;
    }
    public void setChildId(String childId) {
        this.childId = childId;
    }
    public String getParentId() {
        return parentId;
    }
    public void setParentId(String parentId) {
        this.parentId = parentId;
    }

}

MegaMenuDTO课程已更新

import java.util.ArrayList;
import java.util.List;

public class MegaMenuDTO {

    private String Id;
    private String name;
    private String parentId;
    private List<MegaMenuDTO> childrenItems; 

    public MegaMenuDTO() {
        this.Id = "";
        this.name = "";     
        this.parentId = "";
        this.childrenItems = new ArrayList<MegaMenuDTO>();
    }

    public String getId() {
        return Id;
    }
    public void setId(String id) {
        Id = id;
    }
    public String getName() {
        return name;
    }
    public void setName(String name) {
        this.name = name;
    }
    public String getParentId() {
        return parentId;
    }
    public void setParentId(String parentId) {
        this.parentId = parentId;
    }
    public List<MegaMenuDTO> getChildrenItems() {
        return childrenItems;
    }
    public void setChildrenItems(List<MegaMenuDTO> childrenItems) {
        this.childrenItems = childrenItems;
    }
    public void addChildrenItem(MegaMenuDTO childrenItem){
        if(!this.childrenItems.contains(childrenItem))
            this.childrenItems.add(childrenItem);
    }

    @Override
    public String toString() {
        return "MegaMenuDTO [Id=" + Id + ", name=" + name + ", parentId="
                + parentId + ", childrenItems=" + childrenItems + "]";
    }

}

Answer 3

尝试一下

只需根据您的ID对其进行哈希处理即可建立关系。

完成后找到树的根。

    private MegaMenuDTO getRoot(Map<String, MegaMenuDTO> tree) {
        return tree.values().stream().filter(node -> Objects.isNull(node.getParentId())).findFirst().orElse(null);
    }

    public MegaMenuDTO createTree(List<MegaMenuDTO> list) {
        Map<String, MegaMenuDTO> tree = new HashMap<>();
        list.forEach(current -> tree.put(current.getId(), current));
        list.forEach(current -> {
            String parentId = current.getParentId();
            if (tree.containsKey(parentId)) {
                MegaMenuDTO parent = tree.get(parentId);
                current.setParentId(parentId);
                parent.addChildrenItem(current);
                tree.put(parentId, parent);
                tree.put(current.getId(), current);
            }
        });
        return getRoot(tree);
    }

Answer 4

取决于@Diego Martinoia解决方案和@OSryx解决方案，我问题的最终解决方案

private List<MegaMenuDTO> dooo(List<MegaMenuDTO> input) {
    Map<String, MegaMenuDTO> hm = new HashMap<String, MegaMenuDTO>();
    MegaMenuDTO child = null;
    MegaMenuDTO mmdParent = null;

    for (MegaMenuDTO item : input) {

        // ------ Process child ----
        if (!hm.containsKey(item.getId())) {
            hm.put(item.getId(), item);
        } 
        child = hm.get(item.getId());

        // ------ Process Parent ----
        if (item.getParentId() != null && !item.getParentId().equals("")
                && !item.getParentId().equals("0")) {
            if (hm.containsKey(item.getParentId())) {
                mmdParent = hm.get(item.getParentId());
                mmdParent.getChildrenItems().add(child);
            }
        }

    }

    List<MegaMenuDTO> DX = new ArrayList<MegaMenuDTO>();
    for (MegaMenuDTO mmd : hm.values()) {
        if (mmd.getParentId() == null || mmd.getParentId().equals("")
                || mmd.getParentId().equals("0"))
            DX.add(mmd);
    }
    return DX;
}

Answer 5

这是我的解决方法：

将MegaMenuDTO列表转换为地图：

this.$('.add-resource').click(function () {
     var testId = $(this).attr('id');
     //alert(testId);
   $("#layoutCanvas").find(`[data-id='${testId}']`).addClass('hidden')
     
});

然后将地图转换为树：

<script src="https://cdnjs.cloudflare.com/ajax/libs/jquery/3.3.1/jquery.min.js"></script>
<!-- Latest compiled and minified CSS -->
<link rel="stylesheet" href="https://maxcdn.bootstrapcdn.com/bootstrap/3.4.0/css/bootstrap.min.css">

<a class="btn btn-primary add-resource" id="567">Button</a>
<div id="layoutCanvas">
    <div data-id="567">
        Test 1
    </div>
    
    <div data-id="235">
        Test 2
    </div>
</div>

<style>
.hidden{
  //attribute you want to add
 
}
</style>

将父/子关系的java arrayList转换为树？

5 个答案:

主要方法

配对类：

MegaMenuDTO课程已更新