debug result
Attached my code for trying to delete a node in bst.
If I want to delete node 1, when specifying tmp = del in "if (del_node->l_ == NULL)", and remove tmp, then del is removed as well, and the tree data is lost. how can I solve this issue?
Example tree:
3
/ \
1 5
\
2
all data members and functions are declared public for simplicity.
void BST::DeleteNode(int data) {
BinaryTreeNode* &del_node = BST_Search(head_, data);
if (!del_node->l_ && !del_node->r_)
{
delete del_node;
del_node = nullptr;
return;
}
if (del_node->l_ == NULL)
{
BinaryTreeNode* tmp = del_node;
del_node = del_node->r_;
tmp = nullptr;
delete tmp;
return;
}
if (del_node->r_ == NULL)
{
BinaryTreeNode* tmp = del_node;
del_node = del_node->l_;
delete tmp;
return;
}
else
{
del_node->data_ = smallestRightSubTree(del_node->r_);
}
}
int BST::smallestRightSubTree(BinaryTreeNode* rightroot)
{
// if rightroot has no more left childs
if (rightroot && !rightroot->l_)
{
int tmpVal = rightroot->data_;
BinaryTreeNode* tmp = rightroot;
rightroot = rightroot->r_;
delete tmp;
return tmpVal;
}
return smallestRightSubTree(rightroot->l_);
}
int main()
{
BST bst;
bst.BST_Insert(bst.head_, 3);
bst.BST_Insert(bst.head_, 5);
bst.BST_Insert(bst.head_, 1);
bst.BST_Insert(bst.head_, 2);
bst.DeleteNode(1);
return 0;
}
Thanks for help!
EDIT: this is how tmp and del_node look like after the line "del_node = del_node->r_)" in the condition "if(del->l = null)"
void BST::BST_Insert(BinaryTreeNode*& head, int data) {
if (head == nullptr) {
head = new BinaryTreeNode(data, nullptr, nullptr);
return;
}
if (data > head->data_) {
BST_Insert(head->r_, data);
}
else {
BST_Insert(head->l_, data);
}
}
BinaryTreeNode* BST::BST_Search(BinaryTreeNode* root, int key) {
if (root == nullptr || root->data_ == key)
return root;
if (key > root->data_)
return BST_Search(root->r_, key);
return BST_Search(root->l_, key);
}
If your BST_Search returns a BinaryTreeNode* by value, what is the reference in BinaryTreeNode* &del_node = BST_Search(head_, data); actually referencing? It allocates a new temporary and references that. You probably wanted it to reference the variable that is holding the pointer in the tree so that you can modify the tree.
Your BST_Search would have to look like this:
BinaryTreeNode*& BST::BST_Search(BinaryTreeNode*& root, int key) {
if (root == nullptr || root->data_ == key)
return root;
if (key > root->data_)
return BST_Search(root->r_, key);
return BST_Search(root->l_, key);
}
I can't check whether this actually works, because you didn't provide a self-contained compilable example. But something along these lines.
Related
The TreeNode class is defined with only left and right child.
public class TreeNode {
public int val;
public TreeNode left, right;
public TreeNode(int val) {
this.val = val;
}
}
My code finds the next lowest node in O(n). I was wondering if it's possible to find it in lg(N) given that the node doesn't have a pointer to its parent node.
// run time O(n)
public static Integer findNextLowest(TreeNode root, int target) {
Stack<TreeNode> stack = new Stack<>();
TreeNode cur = root;
while (cur != null || stack.size() > 0) {
while (cur != null) {
stack.push(cur);
cur = cur.right;
}
TreeNode node = stack.pop();
if (node.val < target) return node.val; // found the next lowest
cur = node.left;
}
return null;
}
private static TreeNode findNextLowest(TreeNode root, int target){
TreeNode node = root;
TreeNode res = null;
while(node != null){
while(node != null && node.val >= target){
node = node.left;
}
while(node != null && node.val < target){
res = node;
node = node.right;
}
}
return res;
}
No, because you haven't implemented a Binary Search Tree, just a Binary Tree.
A BST will constrain its values such that left.val < val < right.val, so you can do
// run time O(log(n)) if cur is balanced
public static Integer findNextLowest(TreeNode cur, int target) {
if (target < cur.val) { return cur.left != null ? findNextLowest(cur.left, target) : null; }
if (curr.right != null)
{
Integer result = findNextLowest(cur.right, target);
if (result != null) { return result; }
}
return cur.val;
}
You should use something like a R-B tree to ensure it is balanced
package LinkList2;
//import java.util.*;
public class Duplicates {
public static void removeDuplicates(LinkedListNode head)
{
LinkedListNode current = head;
while(current!= null && current.next!= null)
{
LinkedListNode curr = current;
while(curr!=null)
{
if(curr.next.data==current.data) //Getting error at this line
curr.next = curr.next.next;
else
curr = curr.next;
}
current = current.next;
}
}
public static void main(String args[])
{
LinkedListNode first = new LinkedListNode(0,null,null);
LinkedListNode head = first;
LinkedListNode second = first;
for(int i=1; i< 8; i++)
{
second = new LinkedListNode(i%2, null, null);
first.setNext(second);
second.setPrevious(first);
}
System.out.println(head.printForward());
removeDuplicates(head);// Getting error at this line
}
}
Getting null pointer exception in the above code. When I try to run the above code, it gives null pointer exception.
Please help me with my mistake.
Below is the implementation of LinkList where all the methods are defined
class LinkedListNode {
public LinkedListNode next;
public LinkedListNode prev;
public LinkedListNode last;
public int data;
public LinkedListNode(int d, LinkedListNode n, LinkedListNode p) {
data = d;
setNext(n);
setPrevious(p);
}
public void setNext(LinkedListNode n) {
next = n;
if (this == last) {
last = n;
}
if (n != null && n.prev != this) {
n.setPrevious(this);
}
}
public void setPrevious(LinkedListNode p) {
prev = p;
if (p != null && p.next != this) {
p.setNext(this);
}
}
public String printForward() {
if (next != null) {
return data + "->" + next.printForward();
} else {
return ((Integer) data).toString();
}
}
public LinkedListNode clone() {
LinkedListNode next2 = null;
if (next != null) {
next2 = next.clone();
}
LinkedListNode head2 = new LinkedListNode(data, next2, null);
return head2;
}
}
You are getting an exception just because of the following condition:
while(curr != null)
Replace it with while(curr != null && curr.next != null) this way you can check if you have the next element.
Hope this helps.
The problem is that here:
while(curr != null)
{
if(curr.next.data==current.data) //Getting error at this line
curr.next = curr.next.next;
else
curr = curr.next;
}
You are accessing the curr.next.data where you are not checking if that node is null or not. This through your NullPointerException.
To fix your problem is to check on the while loop, if the .next is also not null.
while(curr != null && curr.next != null)
{
if(curr.next.data==current.data) //Getting error at this line
curr.next = curr.next.next;
else
curr = curr.next;
}
In other words, you are not checking if your next node is actually the end of the linked list (i.e null). If you need in your program logic to handle this separately, then you should remove the check from the while loop, and implement this check differently.
This is not an homework. I am just totally blocked on this. I know what to do but I am having difficulty manipulating the tree. please help.
I am trying to delete and node from an BST. I am able to lookup and find the parent and store it an tree.
package com.test.binarytree;
public class BinaryTreeDelete {
private Node root;
//create null binary tree
public BinaryTreeDelete(){
root = null;
}
//delete
public void delete(int target){
root = delete(root, target);
}
public Node delete(Node node, int target){
NodeWithParent temp = lookupFindParent(root, null, target);
if( node == null){
return null;
}
else{
if( node.left == null || node.right == null) //leaf node
{
//WHAT DO I DO HERE
//temp.parent.left = null;
//temp.parent.right = null;
//return null;
}
if( node.left != null && node.right == null ) //one child only on left
{
//WHAT DO I DO HERE
}
if( node.right != null && node.left == null ) //one child only on right
{
//WHAT DO I DO HERE
}
if( node.left != null && node.right != null ) //two children
{
//WHAT DO I DO HERE
}
}
return null;
}
private NodeWithParent lookupFindParent(Node node, Node parentNode, int target){
if( node == null ){
return null;
}
if( node.data == target){
return new NodeWithParent(node, parentNode);
}
else if( node.data > target ){
parentNode = node;
return lookupFindParent(node.left, parentNode, target);
}
else{
parentNode = node;
return lookupFindParent(node.right, parentNode, target);
}
}
//insert
public void insert(int data){
root = insert(root, data);
}
public Node insert (Node node, int data){
if(node == null){
node = new Node(data);
}
else{
if( data <= node.data ){
node.left = insert(node.left, data);
}
else{
node.right = insert(node.right, data);
}
}
return node;
}
//print tree
public void printTree(){
printTree(root);
System.out.println();
}
//print tree
private void printTree(Node node) {
if (node == null) return;
// left, node itself, right
printTree(node.left);
System.out.print(node.data + " ");
printTree(node.right);
}
//node class
public static class Node{
Node left;
Node right;
int data;
Node(int newNode){
data = newNode;
left = null;
right = null;
}
}
//node class
public static class NodeWithParent{
Node current;
Node parent;
NodeWithParent(Node current, Node parent){
this.current = current;
this.parent = parent;
}
}
public static void main(String[] args) {
BinaryTreeDelete bt = new BinaryTreeDelete();
//insert with inserts - tree increases on right if inserted in order
bt = new BinaryTreeDelete();
bt.insert(5);
bt.insert(3);
bt.insert(7);
bt.insert(1);
bt.insert(4);
bt.insert(6);
bt.insert(9);
bt.printTree();
//bt.delete(3);
//bt.delete(4);
//bt.delete(6);
bt.delete(9);
//bt.delete(5);
bt.printTree();
}
}
I'm going to provide you the logic (that means you have to write the code yourself) of how to delete a node in a BST.
There are three cases.
Node to be deleted has both left and right child as null: Delete the node and make the parent point to null.
Node to be deleted has either left or right child (but not both) as null: Delete the node but make sure that the parent points to the valid child of the to-be-deleted node.
Node to be deleted has nether left child nor right child as null: In this case, you have to find the next greater element of the to-be-deleted node. This next greater element is the least element of the right subtree of the to-be-deleted node. Since this is the least element, it has at least one of its child as null. So swap the values of the to-be-deleted node with the next greater node. After you swap, delete this next greater node using points 1 and 2 (whichever is fitting to the situation). Now, why the next greater node and not any node. Because if you replace a node with its next greater node, the BST remains a BST. Try it out in an example and it will be clear.
I am using poppler, and I want to access topic or headings of a particular page number using poppler, so please tell me how to do this using poppler.
Using the glib API. Don't know which API you want.
I'm pretty sure there is no topic/heading stored with a particular page.
You have to walk the index, if there is one.
Walk the index with backtracking. If you are lucky, each index node contains a PopplerActionGotoDest (check type!).
You can grab the title from the PopplerAction object (gchar *title) and get the page number from the included PopplerDest (int page_num).
page_num should be the first page of the section.
Assuming your PDF has an index containing PopplerActionGotoDest objects.
Then you simply walk it, checking for the page_num.
If page_num > searched_num, go back one step.
When you are at the correct parent, walk the childs. This should give you the best match.
I just made some code for it:
gchar* getTitle(PopplerIndexIter *iter, int num, PopplerIndexIter *last,PopplerDocument *doc)
{
int cur_num = 0;
int next;
PopplerAction * action;
PopplerDest * dest;
gchar * title = NULL;
PopplerIndexIter * last_tmp;
do
{
action = poppler_index_iter_get_action(iter);
if (action->type != POPPLER_ACTION_GOTO_DEST) {
printf("No GOTO_DEST!\n");
return NULL;
}
//get page number of current node
if (action->goto_dest.dest->type == POPPLER_DEST_NAMED) {
dest = poppler_document_find_dest (doc, action->goto_dest.dest->named_dest);
cur_num = dest->page_num;
poppler_dest_free(dest);
} else {
cur_num = action->goto_dest.dest->page_num;
}
//printf("cur_num: %d, %d\n",cur_num,num);
//free action, as we don't need it anymore
poppler_action_free(action);
//are there nodes following this one?
last_tmp = poppler_index_iter_copy(iter);
next = poppler_index_iter_next (iter);
//descend
if (!next || cur_num > num) {
if ((!next && cur_num < num) || cur_num == num) {
//descend current node
if (last) {
poppler_index_iter_free(last);
}
last = last_tmp;
}
//descend last node (backtracking)
if (last) {
/* Get the the action and do something with it */
PopplerIndexIter *child = poppler_index_iter_get_child (last);
gchar * tmp = NULL;
if (child) {
tmp = getTitle(child,num,last,doc);
poppler_index_iter_free (child);
} else {
action = poppler_index_iter_get_action(last);
if (action->type != POPPLER_ACTION_GOTO_DEST) {
tmp = NULL;
} else {
tmp = g_strdup (action->any.title);
}
poppler_action_free(action);
poppler_index_iter_free (last);
}
return tmp;
} else {
return NULL;
}
}
if (cur_num > num || (next && cur_num != 0)) {
// free last index_iter
if (last) {
poppler_index_iter_free(last);
}
last = last_tmp;
}
}
while (next);
return NULL;
}
getTitle gets called by:
for (i = 0; i < num_pages; i++) {
iter = poppler_index_iter_new (document);
title = getTitle(iter,i,NULL,document);
poppler_index_iter_free (iter);
if (title) {
printf("title of %d: %s\n",i, title);
g_free(title);
} else {
printf("%d: no title\n",i);
}
}
I am trying to follow Trees tutorial at: http://cslibrary.stanford.edu/110/BinaryTrees.html
Here is the code I have written so far:
package trees.bst;
import java.util.ArrayList;
import java.util.List;
import java.util.StringTokenizer;
/**
*
* #author sachin
*/
public class BinarySearchTree {
Node root = null;
class Node {
Node left = null;
Node right = null;
int data = 0;
public Node(int data) {
this.left = null;
this.right = null;
this.data = data;
}
}
public void insert(int data) {
root = insert(data, root);
}
public boolean lookup(int data) {
return lookup(data, root);
}
public void buildTree(int numNodes) {
for (int i = 0; i < numNodes; i++) {
int num = (int) (Math.random() * 10);
System.out.println("Inserting number:" + num);
insert(num);
}
}
public int size() {
return size(root);
}
public int maxDepth() {
return maxDepth(root);
}
public int minValue() {
return minValue(root);
}
public int maxValue() {
return maxValue(root);
}
public void printTree() { //inorder traversal
System.out.println("inorder traversal:");
printTree(root);
System.out.println("\n--------------");
}
public void printPostorder() { //inorder traversal
System.out.println("printPostorder traversal:");
printPostorder(root);
System.out.println("\n--------------");
}
public int buildTreeFromOutputString(String op) {
root = null;
int i = 0;
StringTokenizer st = new StringTokenizer(op);
while (st.hasMoreTokens()) {
String stNum = st.nextToken();
int num = Integer.parseInt(stNum);
System.out.println("buildTreeFromOutputString: Inserting number:" + num);
insert(num);
i++;
}
return i;
}
public boolean hasPathSum(int pathsum) {
return hasPathSum(pathsum, root);
}
public void mirror() {
mirror(root);
}
public void doubleTree() {
doubleTree(root);
}
public boolean sameTree(BinarySearchTree bst) { //is this tree same as another given tree?
return sameTree(this.root, bst.getRoot());
}
public void printPaths() {
if (root == null) {
System.out.println("print path sum: tree is empty");
}
List pathSoFar = new ArrayList();
printPaths(root, pathSoFar);
}
///-------------------------------------------Public helper functions
public Node getRoot() {
return root;
}
//Exporting a non public Type through public API
///-------------------------------------------Helper Functions
private boolean isLeaf(Node node) {
if (node == null) {
return false;
}
if (node.left == null && node.right == null) {
return true;
}
return false;
}
///-----------------------------------------------------------
private boolean sameTree(Node n1, Node n2) {
if ((n1 == null && n2 == null)) {
return true;
} else {
if ((n1 == null || n2 == null)) {
return false;
} else {
if ((n1.data == n2.data)) {
return (sameTree(n1.left, n2.left) && sameTree(n1.right, n2.right));
}
}
}
return false;
}
private void doubleTree(Node node) {
//create a copy
//bypass the copy to continue looping
if (node == null) {
return;
}
Node copyNode = new Node(node.data);
Node temp = node.left;
node.left = copyNode;
copyNode.left = temp;
doubleTree(copyNode.left);
doubleTree(node.right);
}
private void mirror(Node node) {
if (node == null) {
return;
}
Node temp = node.left;
node.left = node.right;
node.right = temp;
mirror(node.left);
mirror(node.right);
}
private void printPaths(Node node, List pathSoFar) {
if (node == null) {
return;
}
pathSoFar.add(node.data);
if (isLeaf(node)) {
System.out.println("path in tree:" + pathSoFar);
pathSoFar.remove(pathSoFar.lastIndexOf(node.data)); //only the current node, a node.data may be duplicated
return;
} else {
printPaths(node.left, pathSoFar);
printPaths(node.right, pathSoFar);
}
}
private boolean hasPathSum(int pathsum, Node node) {
if (node == null) {
return false;
}
int val = pathsum - node.data;
boolean ret = false;
if (val == 0 && isLeaf(node)) {
ret = true;
} else if (val == 0 && !isLeaf(node)) {
ret = false;
} else if (val != 0 && isLeaf(node)) {
ret = false;
} else if (val != 0 && !isLeaf(node)) {
//recurse further
ret = hasPathSum(val, node.left) || hasPathSum(val, node.right);
}
return ret;
}
private void printPostorder(Node node) { //inorder traversal
if (node == null) {
return;
}
printPostorder(node.left);
printPostorder(node.right);
System.out.print(" " + node.data);
}
private void printTree(Node node) { //inorder traversal
if (node == null) {
return;
}
printTree(node.left);
System.out.print(" " + node.data);
printTree(node.right);
}
private int minValue(Node node) {
if (node == null) {
//error case: this is not supported
return -1;
}
if (node.left == null) {
return node.data;
} else {
return minValue(node.left);
}
}
private int maxValue(Node node) {
if (node == null) {
//error case: this is not supported
return -1;
}
if (node.right == null) {
return node.data;
} else {
return maxValue(node.right);
}
}
private int maxDepth(Node node) {
if (node == null || (node.left == null && node.right == null)) {
return 0;
}
int ldepth = 1 + maxDepth(node.left);
int rdepth = 1 + maxDepth(node.right);
if (ldepth > rdepth) {
return ldepth;
} else {
return rdepth;
}
}
private int size(Node node) {
if (node == null) {
return 0;
}
return 1 + size(node.left) + size(node.right);
}
private Node insert(int data, Node node) {
if (node == null) {
node = new Node(data);
} else if (data <= node.data) {
node.left = insert(data, node.left);
} else {
node.right = insert(data, node.right);
}
//control should never reach here;
return node;
}
private boolean lookup(int data, Node node) {
if (node == null) {
return false;
}
if (node.data == data) {
return true;
}
if (data < node.data) {
return lookup(data, node.left);
} else {
return lookup(data, node.right);
}
}
public static void main(String[] args) {
BinarySearchTree bst = new BinarySearchTree();
int treesize = 5;
bst.buildTree(treesize);
//treesize = bst.buildTreeFromOutputString("4 4 4 6 7");
treesize = bst.buildTreeFromOutputString("3 4 6 3 6");
//treesize = bst.buildTreeFromOutputString("10");
for (int i = 0; i < treesize; i++) {
System.out.println("Searching:" + i + " found:" + bst.lookup(i));
}
System.out.println("tree size:" + bst.size());
System.out.println("maxDepth :" + bst.maxDepth());
System.out.println("minvalue :" + bst.minValue());
System.out.println("maxvalue :" + bst.maxValue());
bst.printTree();
bst.printPostorder();
int pathSum = 10;
System.out.println("hasPathSum " + pathSum + ":" + bst.hasPathSum(pathSum));
pathSum = 6;
System.out.println("hasPathSum " + pathSum + ":" + bst.hasPathSum(pathSum));
pathSum = 19;
System.out.println("hasPathSum " + pathSum + ":" + bst.hasPathSum(pathSum));
bst.printPaths();
bst.printTree();
//bst.mirror();
System.out.println("Tree after mirror function:");
bst.printTree();
//bst.doubleTree();
System.out.println("Tree after double function:");
bst.printTree();
System.out.println("tree size:" + bst.size());
System.out.println("Same tree:" + bst.sameTree(bst));
BinarySearchTree bst2 = new BinarySearchTree();
bst2.buildTree(treesize);
treesize = bst2.buildTreeFromOutputString("3 4 6 3 6");
bst2.printTree();
System.out.println("Same tree:" + bst.sameTree(bst2));
System.out.println("---");
}
}
Now the problem is that netbeans shows Warning: Exporting a non public Type through public API for function getRoot().
I write this function to get root of tree to be used in sameTree() function, to help comparison of "this" with given tree.
Perhaps this is a OOP design issue... How should I restructure the above code that I do not get this warning and what is the concept I am missing here?
The issue here is that some code might call getRoot() but won't be able to use it's return value since you only gave package access to the Node class.
Make Node a top level class with its own file, or at least make it public
I don't really understand why you even created the getRoot() method. As long as you are inside your class you can even access private fields from any other object of the same class.
So you can change
public boolean sameTree(BinarySearchTree bst) { //is this tree same as another given tree?
return sameTree(this.root, bst.getRoot());
}
to
public boolean sameTree(BinarySearchTree bst) { //is this tree same as another given tree?
return sameTree(this.root, bst.root);
}
I would also add a "short path" for the case you pass the same instance to this method:
public boolean sameTree(BinarySearchTree bst) { //is this tree same as another given tree?
if (this == bst) {
return true;
}
return sameTree(this.root, bst.root);
}