[英]Finding Cousins in binary search tree
In a binary tree two nodes are cousins if they are on same level and they have different parent. 在二叉树中,如果两个节点处于同一级别并且具有不同的父级,则它们是表兄弟。
For this in a binary search tree, I associated with every key a level using a tree map and also associated with every key a parent using a tree map. 为此,在二叉搜索树中,我使用树形图将每个键关联到一个级别,还使用树形图将每个键与父键关联。 Then i invoke BFS on root which sets the levels of various keys.
然后,我在根目录上调用BFS来设置各种密钥的级别。
But my isCousins function is giving false even for nodes that are cousins.For example in the binary tree i have created in my code, 12 and 50 are cousins but still it is printing false. 但是我的isCousins函数甚至为表亲的节点都给出了false,例如在我在代码中创建的二叉树中,12和50是表亲,但它仍然打印出false。
Here is my source code. 这是我的源代码。
import java.util.*;
class BST
{
Node root;
LinkedList<Node> q=new LinkedList<Node>();
TreeMap<Integer,Integer> level=new TreeMap<Integer,Integer>();
TreeMap<Integer,Node> parent=new TreeMap<Integer,Node>();
Node insert(Node x,int key)
{
if(x==null)
{
parent.put(key,null);
return new Node(key,null,null,null);
}
else if(x.key<key)
{
x.right=insert(x.right,key);
x.right.parent=x;
parent.put(key,x.right.parent);
return x;
}
else if(x.key>key)
{
x.left=insert(x.left,key);
x.left.parent=x;
parent.put(key,x.left.parent);
return x;
}
else { x.key=key; return x;}
}
public void BFS(Node r)
{
if(r==null)
return;
if(r.parent==null)
level.put(r.key,0);
else level.put(r.key,level.get(r.parent.key)+1);
q.add(r);
while(q.size()!=0)
{
Node n=q.poll();
BFS(n.left);
BFS(n.right);
}
}
public boolean isCousins(int a,int b)
{
BFS(root);
if(level.get(a)==level.get(b)&&parent.get(a)!=parent.get(b))
return true;
else return false;
}
public static void main(String []args)
{
BST tree1=new BST();
tree1.root=null;
tree1.root=tree1.insert(tree1.root,15);
tree1.root=tree1.insert(tree1.root,66);
tree1.root=tree1.insert(tree1.root,5);
tree1.root=tree1.insert(tree1.root,3);
tree1.root=tree1.insert(tree1.root,12);
tree1.root=tree1.insert(tree1.root,75);
tree1.root=tree1.insert(tree1.root,50);
System.out.println(tree1.isCousins(12,50));
}
}
class Node
{
Node left,right,parent;
int key;
int level;
Node(int k,Node l,Node r,Node p)
{
key=k;
left=l;
right=r;
parent=p;
}
}
Courtesy of Patrick Murphy, wrong output was coming due to small mistake which i easily corrected using following contion: 帕特里克·墨菲(Patrick Murphy)提供的错误输出是由于小错误而引起的,我可以使用以下条件轻松纠正该错误:
if(parent.get(key)==null) 如果(parent.get(键)== NULL)
The correct code is: 正确的代码是:
import java.util.*;
class BST
{
Node root;
LinkedList<Node> q=new LinkedList<Node>();
TreeMap<Integer,Integer> level=new TreeMap<Integer,Integer>();
TreeMap<Integer,Node> parent=new TreeMap<Integer,Node>();
Node insert(Node x,int key)
{
if(x==null)
{
parent.put(key,null);
return new Node(key,null,null,null);
}
else if(x.key<key)
{
x.right=insert(x.right,key);
x.right.parent=x;
if(parent.get(key)==null)
parent.put(key,x);
return x;
}
else if(x.key>key)
{
x.left=insert(x.left,key);
x.left.parent=x;
if(parent.get(key)==null)
parent.put(key,x);
return x;
}
else { x.key=key; return x;}
}
public void BFS(Node r)
{
if(r==null)
return;
if(r.parent==null)
level.put(r.key,0);
else level.put(r.key,level.get(r.parent.key)+1);
q.add(r);
while(q.size()!=0)
{
Node n=q.poll();
BFS(n.left);
BFS(n.right);
}
}
public static void main(String []args)
{
BST tree1=new BST();
tree1.root=null;
tree1.root=tree1.insert(tree1.root,15);
tree1.root=tree1.insert(tree1.root,66);
tree1.root=tree1.insert(tree1.root,5);
tree1.root=tree1.insert(tree1.root,3);
tree1.root=tree1.insert(tree1.root,12);
tree1.root=tree1.insert(tree1.root,75);
tree1.root=tree1.insert(tree1.root,50);
System.out.println(tree1.isCousins(66,75));
}
public boolean isCousins(int a,int b)
{
BFS(root);
if(level.get(a)==level.get(b)&&parent.get(a)!=parent.get(b))
return true;
else return false;
}
}
class Node
{
Node left,right,parent;
int key;
int level;
Node(int k,Node l,Node r,Node p)
{
key=k;
left=l;
right=r;
parent=p;
}
}
Node is not a primitive type, so you can't have code that reads: Node不是原始类型,因此您不能拥有以下代码:
parent.get(a)!=parent.get(b);
Using primitive operators to compare non-primitive types just compares the object references, not the actual contents of the objects themselves. 使用原始运算符比较非原始类型只会比较对象引用,而不是对象本身的实际内容。 Instead, write:
相反,写:
!parent.get(a).equals(parent.get(b));
Since the value in your TreeMap level is an Integer and not an int, you shouldn't compare them with ==. 由于TreeMap级别中的值是Integer而不是int,因此不应将它们与==进行比较。 Instead, use the .equals() method again.
而是,再次使用.equals()方法。
level.get(a).equals(level.get(b));
Also, 也,
level.get(a).equals(level.get(b))&&!parent.get(a).equals(parent.get(b));
is a boolean statement, so you could just put 是布尔型语句,因此您可以将
return level.get(a).equals(level.get(b))&&!parent.get(a).equals(parent.get(b));
instead of 代替
if(level.get(a).equals(level.get(b))&&!parent.get(a).equals(parent.get(b)))
return true;
return false;
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