I'm creating a Doubly Linked List from a Binary Search Tree using recursion and it works perfectly fine when the BST is already populated, i.e. >=2 nodes.
However, I tried running it for a BST that is dynamically getting populated and it gives me a StackOverFlowError as soon as I insert a child to the root node in the BST.
Here is the code (in Java) I've written
public class BSTtoDLL {
/* Binary Search Tree to Doubly Linked List conversion*/
// head --> Pointer to head node of created doubly linked list
static BTNode head;
// Initialize previously visited node as NULL. This is
// static so that the same value is accessible in all recursive
// calls
static BTNode prev = null;
/* BSTtoDLL Construtor */
public BSTtoDLL(){
head = null;
prev = null;
}
// A simple recursive function to convert a given Binary tree
// to Doubly Linked List
// root --> Root of Binary Tree
void BinaryTree2DoubleLinkedList(BTNode root)
{
// Base case
if (root == null)
return;
// Recursively convert left subtree
if(root.left!=null)
BinaryTree2DoubleLinkedList(root.left);
// Now convert this node
if (prev == null){
head = root;
}
else
{
prev.right = root;
root.left = prev;
}
prev = root;
// Finally convert right subtree
BinaryTree2DoubleLinkedList(root.right);
}
And the the console response:
Binary Tree Test
Converting to a DLL
Data--34--Left is Null----Right
is Null--- Binary Tree Test Converting to a DLL Exception in thread
"main" java.lang.StackOverflowError at
com.techwealth.BSTtoDLL.BinaryTree2DoubleLinkedList(BSTtoDLL.java:32)
at
com.techwealth.BSTtoDLL.BinaryTree2DoubleLinkedList(BSTtoDLL.java:32)
at
com.techwealth.BSTtoDLL.BinaryTree2DoubleLinkedList(BSTtoDLL.java:32)
at
com.techwealth.BSTtoDLL.BinaryTree2DoubleLinkedList(BSTtoDLL.java:32)
at
com.techwealth.BSTtoDLL.BinaryTree2DoubleLinkedList(BSTtoDLL.java:32)
Clearly, you are missing the termination condition(the base case is invalidated as you are using static variables). That is the reason why you are seeing stackoverflow error. Go through this link for more info regarding this error: What is a StackOverflowError?
Related
I am converting a binary search tree to a doubly linked list and using recursion for the same. However, the recursion never ends and gives me a StackOverFlow error. I am not able to understand where I've gone wrong.
Here is the code I've written.
/* Binary Search Tree to Doubly Linked List conversion*/
static Node root2;
// head --> Pointer to head node of created doubly linked list
static Node head;
// Initialize previously visited node as NULL. This is
// static so that the same value is accessible in all recursive
// calls
static Node prev = null;
// A simple recursive function to convert a given Binary tree
// to Doubly Linked List
// root --> Root of Binary Tree
void BinaryTree2DoubleLinkedList(Node root2)
{
// Base case
if (root2 == null){
return;
}
// Recursively convert left subtree
BinaryTree2DoubleLinkedList(root2.left);
// Now convert this node
if (prev == null)
head = root2;
else
{
root2.left = prev;
prev.right = root2;
}
prev = root2;
// Finally convert right subtree
BinaryTree2DoubleLinkedList(root2.right);
}
/* Function to print nodes in a given doubly linked list */
void printList(Node node)
{
while (node != null)
{
System.out.print(node.data + " ");
node = node.right;
}
}
However when I'm populating my BST, as soon as I enter two values, it gives me a StackOverlow Exception.
Can anybody please take a look at this?
I am confused by this code:
void in_order_traversal_iterative(BinaryTree *root) {
stack<BinaryTree*> s;
BinaryTree *current = root;
while (!s.empty() || current) {
if (current) {
s.push(current);
current = current->left;
} else {
current = s.top();
s.pop();
cout << current->data << " ";
current = current->right;
}
}
}
We set a pointer to point to root. Then if it exists, then push the current (which is root currently) into the stack. I do not see why we push the whole tree into the stack initially, instead of just the value of the data the node holds. Am I missing something completely or not understanding why it would work this way? I cannot comprehend why we push the whole tree in, rather than a single node...
You're missing the fact that after a node is popped, its right child must still be traversed:
current = s.top();
s.pop();
cout << current->data << " ";
current = current->right;
If you had only the data on the stack, this would be impossible. The loop invariant is that the stack holds exactly those nodes with un-traversed right children.
Another way to see what's going on is to transform the recursive traversal to the iterative by algebra:
traverse(node) {
if (node) {
traverse(node->left);
visit(node);
traverse(node->right);
}
}
First convert the tail call to iteration. We do this by updating the argument and replacing the recursive call with a goto the start of the function:
traverse(node) {
start:
if (node) {
traverse(node->left);
visit(node);
node = node->right;
goto start;
}
}
The goto and if are the same as a while, so we have so far
traverse(node) {
while (node) {
traverse(node->left);
visit(node);
node = node->right;
}
}
Replacing the other recursive call requires us to simulate the call stack of the compiler's runtime environment. We do that with an explicit stack.
traverse(node) {
start:
while (node) {
stack.push(node); // save the value of the argument.
node = node->left; // redefine it the same way the recursive call would have
goto start; // simulate the recursive call
// recursive call was here; it's gone now!
recursive_return: // branch here to simulate return from recursive call
visit(node);
node = node->right;
}
// simulate the recursive return: if stack has args, restore and go to return site
if (!stack.empty()) {
node = stack.pop(); // restore the saved parameter value
goto recursive_return;
}
}
Though it's ugly, this is a way that always works to implement iterative versions of recursive code. (It's more complicated if there are multiple non-tail recursive calls, but not much.) And I'm sure you can see the similarity to your code.
We can even get rid of the ugliness with more algebra. First, it's not hard to see this code:
start:
while (node) {
stack.push(node); // save the value of the argument.
node = node->left; // redefine it the same way the recursive call would have
goto start; // simulate the recursive call
when executed beginning with start is equivalent to
while (node) {
stack.push(node); // save the value of the argument.
node = node->left; // redefine it the same way the recursive call would have
}
We can also replace
if (!stack.empty()) {
node = stack.pop(); // restore the saved parameter value
goto recursive_return;
}
with the following
if (!stack.empty()) {
node = stack.pop(); // restore the saved parameter value
visit(node);
node = node->right;
goto start;
}
We have merely copied the three instructions after recursive_return: into the if body.
With this, there is no way left to arrive at the recursive_return label, so we can delete it along with the two following statements:
// Dead code! Delete me!
recursive_return:
visit(node);
node = node->right;
We now have:
traverse(node) {
start:
while (node) {
stack.push(node); // save the value of the argument.
node = node->left; // redefine it the same way the recursive call would have
}
if (!stack.empty()) {
node = stack.pop(); // restore the saved parameter value
visit(node);
node = node->right;
goto start;
}
}
We can get rid of the last goto start by replacing it with an endless loop:
traverse(node) {
loop {
while (node) {
stack.push(node); // save the value of the argument
node = node->left; // redefine it the same way the recursive call would have
}
if (stack.empty()) break; // original code returns, so does this!
node = stack.pop(); // restore the saved parameter value
visit(node);
node = node->right;
}
}
Note we are returning under the same conditions as the previous code: the stack is empty!
I will let you prove to yourself that this code does the same as what you presented, only it's a bit more efficient because it avoids some comparisons! We never had to reason at all about pointers and stack elements. It "just happened."
It's not pushing the whole tree into the stack, it pushes the left-most part of the tree. Then it begin to pop the elements and push their right-most counterparts, in ascending order.
Q (tldr;): How do I use the JavaScanner in android-lint to check if a particular function call with a specific string as a parameter has been surrounded by a try/catch block.
Details: I have completed the android-lint tutorials on the official site and have gone through the source of the existing lint-checks. However, I can't seem to grasp the workflow for this AST-based parsing of JavaScanner. What I'm trying to achieve is to catch a function that sets a specific property and surround it with a try/catch block. For example:
MyPropertySettings.set("SOME_PROPERTY", "SOME_VAL");
should not trigger the lint rule but:
MyPropertySettings.set("SOME_SENSITIVE_PROPERTY", "SOME_VAL");
should because it's not surrounded by a try/catch block with SetPropertyException. I don't want to introduce the try/catch to the function itself because it only throws the exception in extremely rare cases (and the internals of the function are based on some reflection mojo).
For this question, even a workflow/hint would be fine. If I can get the first few steps, I might be able to grasp it better.
Update:
After some more study, I have found that I need to set the set function above in getApplicableMethodNames() and then, somehow read the property of that function to decide if the check applies. That part should be easy.
Surrounding try/catch would be more difficult and I gather I would need to do some "flow analysis". How is the question now.
Well, along with the getApplicableMethodNames() method, you need to override the visitMethod() function. You will get the MethodInvocationNode. Just fetch the arguments passed in the invocation using the node.astArguments() function. This returns a list of arguments that you can iterate through using a StrictListAccessor. Check the arguments passed and if it matches your criterion, run a loop and keep calculating the parent node of the invocation node till a try node is found. If it is a try node, then you can get a list of catches using node.astCatches(). Scan the list and find the appropriate exception. If not found, then report.
You can code like this:
check if it is surrounded by try/catch:
#Override
public void visitMethod(JavaContext context, AstVisitor visitor, MethodInvocation node) {
// check the specified class that invoke the method
JavaParser.ResolvedMethod method = (JavaParser.ResolvedMethod) context.resolve(node);
JavaParser.ResolvedClass clzz = method.getContainingClass();
boolean isSubClass = false;
// sSupportSuperType = {"class name"};
for (int i = 0; i < sSupportSuperType.length; i++) {
if (clzz.isSubclassOf(sSupportSuperType[i], false)) {
isSubClass = true;
break;
}
}
if (!isSubClass) return;
// check if surrounded by try/catch
Node parent = node;
while (true) {
Try tryCatch = context.getParentOfType(parent, Try.class);
if (tryCatch == null) {
break;
} else {
for (Catch aCatch : tryCatch.astCatches()) {
TypeReference catchType = aCatch.astExceptionDeclaration().astTypeReference();
}
parent = tryCatch;
}
}
// get the arguments string
String str = node.astArguments().first().toString();
if (!str.startsWith("\"SOME_PROPERTY\"")) {
context.report(ISSUE, node, context.getLocation(node), "message");
}
}
before this you have to define the specific method by override:
#Override
public List<String> getApplicableMethodNames() {
return Collections.singletonList("set");
}
I am learning algorithms and data structures and to train I am trying to design and implement a binary tree using objective-c.
So far I have the following Classes:
main - for testing
Node - node of tree
BinaryTree - for all methods related to the tree
One of the first methods in BinaryTree class I implemented is insertNode:forRoot:.
- (void)insertNodeByRef:(Node **)node forRoot:(Node **)root{
if (head == NULL) {
head = *node;
}
// Case 2 root is null so can assign the value of the node to it
if (root == NULL) {
root = node;
} else {
if (node.data > root.data) { // to the right
[self insertNode:node forRoot:root.right];
} else if (node.data < root.data) { //or to the left
[self insertNode:node forRoot:root.left];
}
}
}
Where the interface of Node class looks like:
#interface Node : NSObject
#property(nonatomic, assign) int data;
#property(nonatomic, strong) Node * right;
#property(nonatomic, strong) Node * left;
#end
My problem is that I don't know how to access the Node class member variables if I am passing Node as a reference. Whenever I try to access the node properties (like data, left or right) I am getting the following error message:
Member reference base type 'Node *__autoreleasing *' is not a structure or union
So my questions is:
how can I access those properties (data, left or right) and use them to store either int data or reference to another Node?
Hope it makes sense. Thanks!
Your code is mixing two common approaches to the task, hence the problem. You are also using an abstract data type (ADT) type approach, rather than an object-oriented one, so there are three approaches to consider.
In both ADT approaches your tree is represented by a reference to its root, in Objective-C this is probably stored in an instance variable:
Node *TreeRoot;
Note also that both of these algorithms use field references, a->b, rather than property references, a.b - this is because the former references a variable and the second algorithm requires passing a reference to a variable.
Functional ADT: Pass-by-value and assign result
In this approach a node is inserted into a tree and a modified tree is returned which is assigned back, e.g. the top-level call to insert a Node nodeToInsert would be:
TreeRoot = insertNode(nodeToInsert, TreeRoot);
and the insertNode function looks like:
Node *insertNode(Node *node, Node *root)
{
if(root == nil)
{ // empty tree - return the insert node
return node;
}
else
{ // non-empty tree, insert into left or right subtree
if(node->data > root->data) // to the right
{
root->right = insertNode(node, root->right);
}
else if(node->data < root->data)//or to the left
{
root->left = insertNode(node, root->left);
}
// tree modified if needed, return the root
return root;
}
}
Note that in this approach in the case of a non-empty (sub)tree the algorithm performs a redundant assignment into a variable - the assigned value is what is already in the variable... Because of this some people prefer:
Procedural ADT: Pass-by-reference
In this approach the variable holding the root of the (sub)tree is passed-by-reference, rather than its value being passed, and is modified by the called procedure as needed. E.g. the top-level call would be:
insertNode(nodeToInsert, &TreeRoot); // & -> pass the variable, not its value
and the insertNode procedure looks like:
void insertNode(Node *node, Node **root)
{
if(*root == nil)
{ // empty tree - insert node
*root = node;
}
else
{ // non-empty tree, insert into left or right subtree
Node *rootNode = *root;
if(node->data > rootNode->data) // to the right
{
insertNode(node, &rootNode->right);
}
else if(node->data < rootNode->data)//or to the left
{
insertNode(node, &root->left);
}
}
}
You can now see that your method is a mixture of the above two approaches. Both are valid, but as you are using Objective-C it might be better to take the third approach:
Object-Oriented ADT
This is a variation of the procedural ADT - rather than pass a variable to a procedure the variable, now called an object, owns a method which updates itself. Doing it this way means you must test for an empty (sub)tree before you make a call to insert a node, while the previous two approaches test in the call. So now we have the method in Node:
- (void) insert:(Node *)node
{
if(node.data > self.data) // using properties, could also use fields ->
{
if(self.right != nil)
[self.right insert:node];
else
self.right = node;
}
else if(node.data < rootNode.data)
{
if(self.left != nil)
[self.left insert:node];
else
self.left = node;
}
}
You also need to change the top level call to do the same test for an empty tree:
if(TreeRoot != nil)
[TreeRoot insert:nodeToInsert];
else
TreeRoot = nodeToInsert;
And a final note - if you are using MRC, rather than ARC or GC, for memory management you'll need to insert the appropriate retain/release calls.
Hope that helps you sort things out.
First of all, don't write your methods to take Node **. It's just confusing.
Second, think about how it should work. Describe to yourself how it should work at a pretty abstract level. Translate that description directly into code, inventing new (not-yet-written!) messages where necessary. If there are steps you don't know how to do yet, just punt those off to new messages you'll write later. I'll walk you through it.
Presumably you want the public API of BinaryTree to include this message:
#interface BinaryTree
- (void)insertValue:(int)value;
So how do you implement insertValue:? Pretend you're the BinaryTree object. What's your high-level description of what you need to do to insert a value? You want to create a new Node. Then you want to insert that new Node into yourself. Translate that description directly into code:
#implementation BinaryTree {
Node *root_; // root node, or nil for an empty tree
}
- (void)insertValue:(int)value {
Node *node = [[Node alloc] initWithData:value];
[self insertNode:node];
}
Now think about how you do the inserting. Well, if you are an empty tree, your root_ is nil and you can just set it to the new node. Otherwise, you can just ask your root node to insert the new node under himself. Translate that description directly into code:
- (void)insertNode:(Node *)node {
if (root_ == nil) {
root_ = node;
} else {
[root_ insertNode:node];
}
}
Now pretend you're a Node. You've been asked to insert a new Node under yourself. How do you do it? You have to compare the new node's value to your value. If the new node's value is less than your value, you want to insert the new node on your left side. Otherwise, you want to insert it on your right side. Translate that description directly into code:
#implementation Node
- (void)insertNode:(Node *)node {
if (node.data < self.data) {
[self insertNodeOnLeftSide:node];
} else {
[self insertNodeOnRightSide:node];
}
}
Now you're still a Node, and you've been asked to insert a new node on your left side. How do you do it? Well, if you don't have a child on your left side yet, just use the new node as your left child. Otherwise, you ask your left child to insert the new node under himself. Translate that description directly into code:
- (void)insertNodeOnLeftSide:(Node *)node {
if (self.left == nil) {
self.left = node;
} else {
[self.left insertNode:node];
}
}
I'll leave the implementation of insertNodeOnRightSide: as an exercise for the reader. ;^)
Your code, in my opinion, has a lot of logic errors. Maybe consider reviewing what a pointer-to-pointer is to insure you're designing the desired effect. Likewise, you need to dereference node/root to access them in normal state. Otherwise, the error is valid, Node** is not type of structure or union.
(Node **)node is a pointer to an object pointer so node.something is invalid because you are a reference to far away from the object.
But (*node).something will work.
Addition for comments :
When you originally call this method : -(void)insertNodeByRef:(Node **)node forRoot:(Node **)root how do you call it?
From the error you've post in your comment it look to me that you are doing :
Node *n = [[Node alloc] init];
[aNode insertNodeByRef:n forRoot:aRoot];
when your method signature state that you need to call it like this :
[aNode insertNodeByRef:&n forRoot:&aRoot];
To pass the address of the pointer to the object.
I'm saying this because your error is now stating that your are sending Node * instead of Node ** which are 2 different thing. (( Incompatible pointer types sending 'Node *' to parameter of type 'Node **' ) I've remove the __autoreleasing between the 2 *, it was obscuring the error message.)
So in other word you are passing a pointer to an object when your method is asking for a pointer TO A pointer to an object.
Is there an algorithm to optimize a highly recursive function into an iteration over a data structure? For example, given the following function...
// template <typename T> class BSTNode is defined somewhere else
// It's a generic binary search tree.
template <typename T, typename R>
void in_order(BSTNode<T>* root, R (*routine)(T))
{
if (root)
{
in_order(root->leftChild);
routine(root->data);
in_order(root->rightChild);
}
}
... is it possible to optimize it into...
// template <typename> class Stack is defined somewhere else
// It's a generic LIFO array (aka stack).
template <typename T, typename R>
void in_order(BSTNode<T>* root, R (*routine)(T))
{
if (!root) return;
Stack<BSTNode*> stack;
stack.push(NULL);
line1:
while (root->leftChild)
{
stack.push(root);
root = root->leftChild;
}
line2:
routine(root->data);
if (root->rightChild)
{
root = root->rightChild;
goto line1;
}
else if (root = stack.pop())
goto line2;
}
(Of course, the difference is that, instead of filling the call stack, the latter fills another stack in the heap.)
EDIT: I meant an algorithm that can be executed by the compiler, so I don't have to optimize it by hand.
Yes, you can do this.
However, aside from possibly exhausting stack space with very deep trees, there is no "optimization" here. If you need speed gains, consider threading your trees instead.
The only general recursive optimisation I've come across is that of optimising tail recursion. This is frequently done in functional languages and basically involves the compiler/interpreter changing a function where the final operation is the recursive call into a fully iterative function (so no problems with stack, etc).
For other forms of recursion, I'm not aware that any general purpose algorithm for optimising them into iterative functions has been found/created. You can certainly always express such functions in an iterative manner, but the transformation is not a general purpose one.
You're asking for the continuation-passing-style transformation with defunctionalized continuations; it's covered in chapter 6 of Essentials of Programming Languages, with code in Scheme. But it'd be a huge pain to implement for C++. Maybe if you have a compiler frontend that converts C++ into reasonably accessible datastructures, though, and you need to do this to a lot of code. The book also explains how to do this transformation systematically by hand, which is more likely to be practical in your situation.
Technically the answer to this is "yes": any algorithm which can be expressed recursively (i.e. with an implicit stack) can also be reformulated to use iteration (i.e. with an explicit stack or other tracking structure).
However, I suspect that most compilers can't or won't attempt to do this for you. Coming up with a general-purpose automatic algorithm for performing the transformation is likely to be pretty difficult, although I've never tried it, so I shouldn't say that it's intractable.
It is possible to traverse a tree depth-first, without using recursion.
A good example is the following: http://code.activestate.com/recipes/461776/.
The compiler won't do this optimization for you, though. However, the concept is not so hard to grasp. What you're doing is creating a call stack and nodelist yourself, instead of using a function call to go deeper into the tree.
It is possible to traverse a mutable ordered tree iteratively, by recording the node's parent in the branches you take, and knowing the direction you approach a node from ( down, or up from the left or right, which the ordered property of the tree lets you test ):
template <typename T, typename R>
void in_order ( BSTNode<T>* root, R (*routine)(T) ) {
typedef BSTNode<T>* Node;
Node current = root;
Node parent = 0;
bool going_down = true;
while ( current ) {
Node next = 0;
if ( going_down ) {
if ( current -> leftChild ) {
// navigate down the left, swapping prev with the path taken
Node next_child = current -> leftChild;
current -> leftChild = parent;
parent = current;
current = next_child;
} else if ( current -> rightChild ) {
// navigate down the right, swapping prev with the path taken
Node next_child = current -> rightChild;
current -> rightChild = parent;
parent = current;
current = next_child;
} else {
// leaf
routine ( current -> data );
going_down = false;
}
} else {
// moving up to parent
if ( parent ) {
Node next_parent = 0;
// came from the left branch
if ( parent -> data > current -> data ) {
// visit parent after left branch
routine ( parent -> data );
// repair parent
next_parent = parent -> leftChild;
parent -> leftChild = current;
// traverse right if possible
if ( parent -> rightChild ) {
going_down = true;
// navigate down the right, swapping prev with the path taken
Node next_child = parent -> rightChild;
parent -> rightChild = next_parent;
//parent = current;
current = next_child;
continue;
}
} else {
// came from the right branch
next_parent = parent -> rightChild;
parent -> rightChild = current;
}
current = parent;
parent = next_parent;
} else {
break;
}
}
}
}
If rather than storing the children, you store the XOR of the parent and child, then you can get the next node in whatever direction you approach from without having to mutate the tree.
I don't know of anything with automatically transforms non-tail recursive functions into such code. I do know of environments where the call stack is allocated on the heap, which transparently avoid a stack overflow in cases where you can't perform such mutations and have a fixed small stack size. Usually recording the state on a stack takes less space that the call stack, as you're selecting only the essential local state to record, and are not recording return addresses or any caller-save registers ( if you used a functor rather than a function pointer, then it's more likely that the compiler might be able to inline routine and so not save the caller save registers in the simple recursive case, so reducing the amount of stack required for each recursion. YMMV ).
Since you're using templates you should only need to do the traversal code once, and combine it with strategy templates to switch between pre, post and inorder, or whatever other iteration modes you want.
Of course, the difference is that,
instead of filling the call stack, the
latter fills another stack in the
heap.
You answered yourself. What is cheaper, stack or heap? (unless you are running out of stack space)
Sure you can make your own stack.
You want speed? Unless routine(root->data) does almost nothing, you'll never notice the difference.