I have a question about a method that should find a node in a binary tree, that contains a given value. The method provided below doesn't work, and the question is why.
public Node search(Node node, int value) {
if(node.value == value) return node;
if(node.left != null) search(node.left, value);
if(node.right != null) search(node.right, value);
return null;
}
The problem is that this method sometimes returns null when there actually is a node with a given value in the tree. Why is that?
You shouldn't ignore the value returned by the recursive call, and your method should not use root, but the passed node :
public Node search(Node node, int value) {
if(node.value == value) return node;
Node found = null;
if(node.left != null)
found = search(node.left, value);
if(found == null && node.right != null)
found = search(node.right, value);
return found;
}
Like Eran said, your calls to search via the if statements aren't returning a node, as your search function is supposed to. The only situation in which your current code would work is if the first node n satisfies n.value == value.
Related
I'm doing a BST project which is search a value in the BST and return if it was found. I used a test method to check the codes, and it work fine. Problem came from the return type I guessed.
public BSTNode contains(BSTNode root, String needle) {
BSTNode current = root;
while (current != null) {
if (current.getData().compareTo(needle) > 0)
current=current.getLeft();
else if (current.getData().compareTo(needle) < 0)
current=current.getRight();
else if (current.getData().compareTo(needle) == 0)
return current;
else current=null;
}
return current;
}
result:
BSTNode node = bst.contains(root, "delta");
Assertions.assertTrue(node instanceof BSTNode);
false;
BSTNode node = bst.contains(root, "delta");
Assertions.assertTrue(true);
true;
As my understanding, I believe the codes work fine and the value return was right. I just don't understand the "node instance of BSTNode" why it was false and how can I fix it?
Thank you in advanced
Your method, as written, can only ever return null. You only exit the while loop when current is null, and then you return current. null instanceof Anything is always false.
The line current=current; will also cause an infinite loop if the value is found.
Both of these can be addressed at once: you should return current when the comparison is 0.
As you already got the point where you were actually wrong, I am adding one more constraint, by modifying your existing code.
In case of BST, there would be 3 points:
current node is equal to key node
current node is greater than the key node
current node is lesser than the key node
Therefore you need iterate until unless you found the key you were looking for or you traversed all possible keys. In either case you need to stop your iteration.
public BSTNode contains(BSTNode root, String needle) {
BSTNode current = root;
/* the variable you will return at the end */
BSTNode result = null;
/* iterate the loop until current becomes null or result becomes not null */
while (current != null && result == null) {
/* when you found what you were looking for */
if (current.getData().compareTo(needle) == 0){
result = current;
}else if (current.getData().compareTo(needle) > 0){
current=current.getLeft();
}else{
current=current.getRight();
}
}
/* return the result at the end, it will be either null or the value you were looking */
return result;
}
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;
}
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 understand the algorithms but I am not sure how to put it into actual codes. Please help! And also please explain in details. I really want to understand this besides just copying down the answer. ;)
Here are my codes:
public boolean getLeftChild(){
Node insertNode = root;
while(insertNode!=null){
insertNode = insertNode.left;
}
return true;
}
public Boolean removeMin(){
Node insertNode = root;
Node parentNode =root;
if (insertNode.left ==null){
insertNode.right = parentNode;
insertNode = null;
}else if (getLeftChild() ==true && insertNode.right != null){
insertNode.left = null;
}else{
parentNode.left = insertNode.right;
}
return true;
}
First things first: For trees I highly recommend recursion.
Just one example:
getSmallestNode(Node node){
if(node.left != null){
return getSmallestNode(node.left)
}
return node;
}
For the deletion, there can be two cases if you want do delete the smallest (and therefore the "most left leaf" child) of a binary tree.
Case 1: The leaf has no child nodes, in that case just set the according entry in the parent to null (mostLeftChild.getParent().left = null)
Case 2: The leaf has a right child node (there can't be a left child node because that means there would be a smaller node and your currently selected node isn't the smallest) in that case you replace the current left node with the smallest node of the right subtree mostLeftChild.getParent().left = getSmallestFromSubtree(mostLeftChild.right)
So now to make that into code, it could look something like this (No guarantee that it really works)
public Node deleteSmallest(Node node){
// haven't reached leaf yet
if(node.left != null{
return deleteSmallest(node.left)
}
// case 1, no child nodes
if(node.right == null){
node.getParent().left = null;
} else { // case 2, right child node
node.getParent().left = deleteSmallest(node.right)
}
return node;
}
And you would call it with deleteSmallest(root)