*Tree child->parent relationships

Posted 2020-06-07

Given a list of child->parent relationships, build a binary tree out of it. All the element Ids inside the tree are unique. 

Example: 

Given the following relationships: 

Child Parent IsLeft 
15 20 true 
19 80 true 
17 20 false 
16 80 false 
80 50 false 
50 null false 
20 50 true 


You should return the following tree: 
50 
/ \ 
20 80 
/ \ / \ 
15 17 19 16 


Function Signature 


/** 
* Represents a pair relation between one parent node and one child node inside a binary tree 
* If the _parent is null, it represents the ROOT node 
*/ 
public class Relation { 
public Integer _parent; 
public Integer _child; 
public boolean _isLeft; 
} 


/** 
* Represents a single Node inside a binary tree 
*/ 
public class Node { 
public Integer _id; 
public Node _left; 
public Node _right; 
} 

/** 
* Implement a method to build a tree from a list of parent-child relationships 
* And return the root Node of the tree 
*/ 
public Node buildTree (List<Relation> data) 
{ 
}

 

解法一：One simple approach to this problem can be: we know the ROOT of the tree is the one where the parent is null in the list. Once we figure out the parent, we can iteratively figure out its children and their children- by looping over the complete list and finding out the ones that point the current node as its parent. To build tree efficiently, we can use queue to keep track of the tree built till then. The running time would be O(n^2), with constant space (not really, we are keeping a queue as well)

public static Node buildTree(List<Relation> data){
        if(data==null) return new Node();
        Node root=new Node();
        int children=0;
        for(int i=0;i<data.size();i++){
            Relation x=data.get(i);
            if(x.parent==null){
                root=new Node(x.child,null,null);
                break;
            }
        }
        if(root==null) return root;
        Queue<Node> q=new LinkedList<Node>();
        q.add(root);
        while(!q.isEmpty()){
            Node x=q.poll();
            for(int i=0;i<data.size();i++){
                Relation y=data.get(i);
                if(y.parent==x.id){
                    Node n=new Node(y.child,null,null);
                    if(y.isLeft)
                        x.left=n;
                    else x.right=n;
                    q.add(n);
                    children++;
                    if(children==2){
                        children=0;
                        break;
                    }
                }
            }
        }
        return root;
    }

解法二：

Another way to approach this problem for a better running time could be by using a HashMap. We can hash the list with key as the parent and value as a list of its children. And then iteratively generating the tree. This solution would be O(n) time complexity and O(n) space complexity.

public static Node buildTree(List<Relation> data){
        if(data==null) return new Node();
        Node root=new Node();
        HashMap<Integer,ArrayList<Relation>> tree = new HashMap<Integer,ArrayList<Relation>>();
        for(Relation d:data){
            if(d.parent==null)
                root=new Node(d.child,null,null);
            else{
                if(tree.containsKey(d.parent)){
                    ArrayList<Relation> value=tree.get(d.parent);
                    value.add(d);
                } else {
                    ArrayList<Relation> value = new ArrayList<Relation>();
                    value.add(d);
                    tree.put(d.parent,value);
                }
            }
        }
        if(root==null) return root;
        Queue<Node> q = new LinkedList<Node>();
        q.add(root);
        while(!q.isEmpty()){
            Node x = q.poll();
            if(tree.containsKey(x.id)){
                ArrayList<Relation> value=tree.get(x.id);
                for(Relation v:value){
                    Node child = new Node(v.child,null,null);
                    q.add(child);
                    if(v.isLeft)    
                        x.left=child;
                    else x.right=child;
                }
            }
        }
        return root;
    }

reference：

http://www.careercup.com/question?id=5668114807128064