My tree nodes have the 3 string fields and 3 node fields which are left, middle and right.
One of the problems is that the method can only take string as a parameter
This is what I have
public TreeNode findNode(String name) {
TreeNode pointer = this.getRoot();
if (pointer.getName().equals(name))
return pointer;
if (pointer.getLeft() != null)
pointer = pointer.getLeft();
findNode(name);
if (pointer.getMiddle() != null)
pointer = pointer.getMiddle();
findNode(name);
if (pointer.getRight() != null)
pointer = pointer.getRight();
findNode(name);
return null;
}
This causes a stack overflow error because I just keep setting the pointer to root. But I have to start somewhere and my only parameters for the method can be name. I can't seem to see how to do this.
You can use a list as a parameter stack.
public TreeNode findNode(String name) {
List<TreeNode> stack = new ArrayList<TreeNode>();
stack.add(this.getRoot());
while (!stack.isEmpty())
{
TreeNode node = stack.remove(0);
if (node.getName().equals(name))
return node;
if (pointer.getLeft() != null)
stack.add(node.getLeft());
if (node.getMiddle() != null)
stack.add(node.getMiddle());
if (node.getRight() != null)
stack.add(node.getRight());
}
return null;
}
You can remove from the end of the list instead of the front of the list if you want to search depth-first.
Im guessing you cannot change the signature of this function. Have a helper function that takes in two parameters, (Node and name) that you call with root and name.
In all three cases (left, middle, right), you are calling findNode(name) but not for those objects, instead it is for this. That's why you get stack overflow.
Use an auxiliary method that takes in a TreeNode parameter in addition to the string:
public TreeNode findNode(String name) {
return auxFindNode(this.getRoot(), name);
}
private TreeNode auxFindNode(TreeNode node, String name) {
//perform your recursive traversal here
}
Your code as it stands will never work because you keep setting pointer to the root of the tree at the beginning of the method. So all your recursive calls start with the root of the tree.
If you prefer not to use another method, you can traverse the tree iteratively by using stack:
public TreeNode findNode(String name) {
Stack<TreeNode> stack = new Stack<TreeNode>();
TreeNode foundNode = null;
while(!stack.empty() && foundNode == null) {
TreeNode node = stack.pop();
if(node.getName().equals(name)) {
foundNode = node;
} else {
if(node.getLeft() != null) {
stack.push(node.getLeft();
}
if(node.getMiddle() != null) {
stack.push(node.getMiddle());
}
if(node.getRight() != null) {
stack.push(node.getRight());
}
}
}
return foundNode;
}
Related
I'm trying to search for a node in a non-binary tree without actually passing a node to the search method.
Each node has a name variable. The findChild() method takes a name, and searches through the tree it was called on to find the node with that name.
To do the recursive search, I call findChild() on the child node rather than passing the child node to the findChild() method. Print statements show me that the method gets down through the tree, but the result variable gets set to null as the stack is unwinded, so the method always returns null. I understand why it's doing this, but I don't understand how to unwind this type of recursion. Any help is appreciated!
My findChild() method:
public FileNode findChild(String name) {
FileNode result = null;
for (FileNode child : this.getChildren()) {
if (child.getName() == name) {
return child;
} else {
child.findChild(name);
}
}
return result;
}
Will the following small change help? Your else condition is never assigning a value.
public FileNode findChild(String name) {
FileNode result = null;
for (FileNode child : this.getChildren()) {
if (child.getName() == name) {
result = child;
break;
} else {
result = child.findChild(name);
if (result != null)
break;
}
}
return result;
}
You're throwing away the result of FileNode#findChild in the else block
Try this
if (child.getName().equals(name)) {
return child;
} else {
FileNode childResult = child.findChild(name);
if (childResult != null) {
return childResult;
}
}
I'd like to implement and test a tree search function. I'm using the DefaultTreeModel of javax.swing. I operato on my own defined Employee objects as the tree data. I tried to get the following to work:
private DefaultMutableTreeNode search(int id, DefaultMutableTreeNode node){
if(node != null){
Employee emp = (Employee) node.getUserObject();
if(emp.getId() == id){
return node;
} else {
DefaultMutableTreeNode foundNode = search(id, node.getPreviousSibling());
if(foundNode == null) {
foundNode = search(id, node.getNextSibling());
}
return foundNode;
}
} else {
return null;
}
}
but it can only find the element which is among one parent's siblings. And like to find it in entire tree, from root to leaves. How do I do it?
You could navigate to the root of the tree first in a different method, then call your current recursive method.
private DefaultMutableTreeNode search(int id, DefaultMutableTreeNode node){
if(node == null){
return null;
}
node = node.getRoot(); //Turns out that this class has a getRoot() method.
return searchInternal(id,node); //Then does a depth-first search from the root node.
}
private DefaultMutableTreeNode searchInternal(int id, DefaultMutableTreeNode node){
if(node == null){ //I also switched this around, for good practice.
return null;
}
Employee emp = (Employee) node.getUserObject();
if(emp.getId() == id){ //Found it!
return node;
}
DefaultMutableTreeNode foundNode = searchInternal(id, node.getPreviousSibling());
if(foundNode == null) {
foundNode = search(id, node.getNextSibling());
}
return foundNode;
}
In general, recursive methods tend to run a bit slower in Java, and they are prone to stack overflowing. It's usually better to use a custom stack for depth-first search, and this doesn't require you to make a different method. Recursive methods might make the code more readable in some cases.
What also works, in this case:
private DefaultMutableTreeNode search(int id, DefaultMutableTreeNode node){
if(node == null){
return null;
}
Enumeration enumeration = node.getRoot().depthFirstEnumeration(); //Enumerate over all nodes in the tree.
while(enumeration.hasMoreElements()){
DefaultMutableTreeNode next = enumeration.next();
if(((Employee)next.getUserObject()).getId() == id){ //Found it!
return next;
}
}
return null;
}
Turns out that depth-first searching is already implemented by Swing. This is a post-order enumeration though. See the javadoc, there are also methods for the other types of traversals.
I made a binary search tree in Java but I'm having troubles whit the deleting nodes part. I managed to erase the node when it has only 1 son, and I have the idea to make the deletion when it has 2 sons, anyways the method I'm using when it has no sons (when it's a leaf) is not working in Java. Normally in C++ I would assign the Node "null" but it doesn't work here.
if (numberOfSons(node) == 0) {
node= null;
return true;
}
That's the portion of the code that takes care of the nulling part. When I debug it, it is referencing the correct node and it's assigning it the null value, but when I return to the Frame where I'm calling the delete method for my tree the node is still there. What's the correct way to "null" an object in Java? I thought everything was a pointer in here and therefore this would work, but I think it doesn't.
When you're nulling something you just make the reference in the scope you're in null. It doesn't affect anything outside.
Let me explain by example. Say you have a method foo:
public void foo(Node node) {
node = null;
if(node == null) {
System.out.println("node is null");
} else {
System.out.println("node is not null");
}
}
Now you call it like this:
public void doSomething() {
Node node = new Node();
foo(node);
if(node == null) {
System.out.println("Original node is null");
} else {
System.out.println("Original node is not null");
}
}
In your console you'll get:
node is null
original node in not null
The reason is that it's not a pointer, it's a reference. When you're nulling a reference, you just say "make this reference synonym to null". It doesn't mean that the object is deleted, it may still exist in other places. There is no way to delete objects in java. All you can do is make sure no other object points to them, and the garbage collector will delete the objects (sometime).
Nothing remains but to reinsert either left or right subtree. For instance:
class BinaryTree<T extends Comparable<T>> {
class Node {
Node left;
Node right;
T value;
}
Node root;
void delete(T soughtValue) {
root = deleteRec(root, soughtValue);
}
Node deleteRec(Node node, T soughtValue) {
if (node == null) {
return null;
}
int comparison = soughtValue.compareTo(node.value);
if (comparison < 0) {
node.left = deleteRec(node.left, soughtValue);
} else if (comparison > 0) {
node.right = deleteRec(node.right, soughtValue);
} else {
if (node.left == null) {
return node.right;
} else if (node.right == null) {
return node.left;
} else {
// Two subtrees remain, do for instance:
// Return left, with its greatest element getting
// the right subtree.
Node leftsRightmost = node.left;
while (leftsRightmost.right != null) {
leftsRightmost = leftsRightmost.right;
}
leftsRightmost.right = node.right;
return node.left;
}
}
return node;
}
}
As Java does not have aliases parameters as in C++ Node*& - a kind of in-out parameter, I use the result of deleteRec here. In java any function argument that is an object variable will never change the variable with another object instance. That was one of the language design decisions like single inheritance.
I'm trying to implement an Iterator in my own TreeSet class.
However my attempt at creating it only works until the current node is the root.
The Iterator looks like this:
Constructor:
public TreeWordSetIterator()
{
next = root;
if(next == null)
return;
while(next.left != null)
next = next.left;
}
hasNext:
public boolean hasNext()
{
return next != null;
}
Next:
public TreeNode next()
{
if(!hasNext()) throw new NoSuchElementException();
TreeNode current = next;
next = findNext(next); // find next node
return current;
}
findNext:
private TreeNode findNext(TreeNode node)
{
if(node.right != null)
{
node = node.right;
while(node.left != null)
node = node.left;
return node;
}
else
{
if(node.parent == null)
return null;
while(node.parent != null && node.parent.left != node)
node = node.parent;
return node;
}
}
This works fine up until I get to my root node. So I can only iterate through the left child of root, not the right. Can anyone give me a few tips on what I'm doing wrong? I don't expect a solution, just a few tips.
Question: How can I find the next node in a TreeSet given each node points to its parent, left-child and right-child.
Thanks in advance
It helps to consider the rules of a Binary Search Tree. Let's suppose the previously returned node is n:
If n has a right subtree, then the node with the next value will be the leftmost node of the right subtree.
If n does not have a right subtree, then the node with the next value will be the first ancestor of n that contains n in its left subtree.
Your code is correctly handling the first case, but not the second. Consider the case where node is the leftmost leaf of the tree (the starting case). node has no right child, so we go straight to the else. node has a parent, so the if-clause is skipped. node.parent.left == node, so the while clause is skipped without executing at all. The end result is that node gets returned. I'd expect your iterator to continue returning the same node forever.
There are 3 main ways you can iterate a binarry tree
private void inOrder(TreeNode node) {
if(isEmpty())return;
if(node.getLeftNode()!=null)inOrder(node.getLeftNode());
System.out.print(node.getNodeData()+" ");
if(node.getRightNode()!=null)inOrder(node.getRightNode());
}
private void preOrder(TreeNode node) {
if(isEmpty())return;
System.out.print(node.getNodeData()+" ");
if(node.getLeftNode()!=null)preOrder(node.getLeftNode());
if(node.getRightNode()!=null)preOrder(node.getRightNode());
}
private void postOrder(TreeNode node) {
if(isEmpty())return;
if(node.getLeftNode()!=null)postOrder(node.getLeftNode());
if(node.getRightNode()!=null)postOrder(node.getRightNode());
System.out.print(node.getNodeData()+" ");
}
//use
inOrder(root);
preOrder(root);
postOrder(root);
Its simple as that ,your code doesn't really makes sense to me, is there something else you are trying to do besides iterating in one of this ways?
I think you need to save previous point in your iterator so you know where you've been before
Here some code but be aware that it is not complete you should do it by yourself and it's just to show you the idea. it also doesn't handle the root node.
findNext(TreeNode node, TreeNode previousNode) {
if(node.left != null && node.left != previousNode && node.right != previousNode){ //go left if not been there yet
return node.left;
}
if(node.right != null && node.right != previousNode){ //go right if not been there yet
return node.right;
}
return findNext(node.parent, node); //go up and pass current node to avoid going down
}
A good approach is to use a stack to manage sequencing, which is sort of done for you if you use a recursive traversal (instead of trying to build an Iterator at all) as described in SteveL's answer.
As you want to start from the left, you first load onto the stack the root node and its leftmost children in the proper order (push while going down to the left from the root).
Always pop the next from the top of the stack, and push its right child (if any) and all its leftmost children before returning the one you just popped, so that they're next in line.
By this approach, the top of the stack will always be the next to return, and when the stack is empty, there's no more...
In code:
import java.util.Iterator;
import java.util.NoSuchElementException;
import java.util.Stack;
public class TreeNodeInOrderIterator<T> implements Iterator<T> {
private final Stack<TreeNode<T>> stack;
public TreeNodeInOrderIterator(TreeNode<T> root) {
this.stack = new Stack<TreeNode<T>>();
pushLeftChildren(root);
}
#Override
public boolean hasNext() {
return !stack.isEmpty();
}
#Override
public T next() {
if (!hasNext())
throw new NoSuchElementException();
TreeNode<T> top = stack.pop();
pushLeftChildren(top.right);
return top.val;
}
#Override
public void remove() {
throw new UnsupportedOperationException();
}
private void pushLeftChildren(TreeNode<T> cur) {
while (cur != null) {
stack.push(cur);
cur = cur.left;
}
}
}
In this code, TreeNode is defined by
public class TreeNode<T> {
T val;
TreeNode<T> left;
TreeNode<T> right;
TreeNode(T x) { val = x; }
}
If you want to have each node also know its parent, that's ok, but all the traversal is by using what's on the stack and adding to the stack using the left and right child links.
I would like using my own Node class to implement tree structure in Java. But I'm confused how to do a deep copy to copy a tree.
My Node class would be like this:
public class Node{
private String value;
private Node leftChild;
private Node rightChild;
....
I'm new to recursion, so is there any code I can study? Thank you!
try
class Node {
private String value;
private Node left;
private Node right;
public Node(String value, Node left, Node right) {
this.value = value;
...
}
Node copy() {
Node left = null;
Node right = null;
if (this.left != null) {
left = this.left.copy();
}
if (this.right != null) {
right = this.right.copy();
}
return new Node(value, left, right);
}
}
Doing it recursively using pre-order traversal.
public static Node CopyTheTree(Node root)
{
if (root == null)
{
return null;
}
Node newNode = new Node(null, null, root.Value);
newNode.Left= CopyTheTree(root.Left);
newNode.Right= CopyTheTree(root.Right);
return newNode;
}
You can use something like this. It will go though the old tree depth first wise and create a copy of it.
private Tree getCopyOfTree(oldTree) {
Tree newTree = new Tree();
newTree.setRootNode(new Node());
copy(oldTree.getRootNode(), newTree.getRootNode())
return newTree;
}
private void copy(Node oldNode, Node newNode) {
if (oldNode.getLeftChild != null) {
newNode.setLeftChild(new Node(oldNode.getLeftChild()));
copy(oldNode.getLeftChild, newNode.getLeftChild());
}
if (oldNode.getRightChild != null) {
newNode.setRightChild(new Node(oldNode.getRightChild()));
copy(oldNode.getRightChild, newNode.getRightChild());
}
}
I like Evgeniy Dorofeev's answer above, but sometimes you might not be able to add a method to the type Node as you might not own it. In that case(this is in c#):
public static TreeNode CopyTree(TreeNode originalTreeNode)
{
if (originalTreeNode == null)
{
return null;
}
// copy current node's data
var copiedNode = new TreeNode(originalTreeNode.Data);
// copy current node's children
foreach (var childNode in originalTreeNode.Children)
{
copiedNode.Children.Add(CopyTree(childNode));
}
return copiedNode;
}
Not sure but try something with post order traversal of your tree and creating a new node for each node you traverse. You might require stack for storing the nodes you created to make left and right child links.
public static TreeNode copy( TreeNode source )
{
if( source == null )
return null;
else
return new TreeNode( source.getInfo( ), copy( source.getLeft( ) ), copy( source.getRight( ) ) );
}
/Sure. Sorry for the delay. Anyway... any recursive method has a base case, and one or more recursive cases. In this instance, the first line is obvious... if the argument to the parameter 'source' is null (as it eventually evaluates to in order to end the method's operation), it will return null; if not, the recursive case is initiated. In this case, you're returning the entire copied tree once the recursion is complete.
The 'new' operator is used, indicating the instantiation of a TreeNode with each visit to the various nodes of the tree during the traversal, occurring through recursive calls to 'copy', whose arguments become references to the left and right TreeNodes (if there are any). Once source becomes null in each argument, the base case is initiated, releasing the recursion stack back to the original call to 'copy', which is a copy of the root of the original tree./
Node copy(Node node)
{
if(node==null) return node;
Node node1 =new Node(node.data);
node1.left=copy(node.left);
node1.right=copy(node.right);
return node1;
}