For the below tree,
typedef struct SiblingTreeNode{
void *item;
struct SiblingTreeNode *parent;
struct SiblingTreeNode *firstChild;
struct SiblingTreeNode *nextSibling;
}SibTreeNode;
typedef struct Tree{
SibTreeNode *root;
int size; // number of nodes in tree
}Tree;
1)
Assuming the huge depth of a tree, To avoid stack overflow, Can we perform DFS without using recursion?
2)
Generally, a tree node has n child pointers and data pointer(item), From performance aspect of operations on tree, What are the advantages of maintaining sibling(nextSibling) and first child(firstChild) and parent pointer(parent)?
Yes. It's always possible by using an explicit data structure
instead as needed (e.g. your own explicit stack) to keep track of
where you are or what you still need to do. A reason for doing this could be limited call stack space. Some languages support a simple case of recursion called "tail recursion" in a way that avoids stack overhead automatically.
Edit: In this specific case, you don't need to keep track of more than the current node, see code added below.
The advantage of the "left child / left sibling" structure is that you can have any number of children without the need of an additional data structure (e.g. a list or array) to manage the children. A disadvantage is that you can't directly access the nth child -- accessing the nth child is a O(n) operation.
Non-recursive DFS for the given data structure:
void Dfs(Tree* tree) {
SiblingTreeNode* current = tree.root;
while (current != nullptr) {
visit (current);
if (current->firstChild != nullptr) {
current = current->firstChild;
} else if (current->nextSibling != nullptr) {
current = current->nextSibling;
} else {
do {
current = current->parent;
} while (current != nullptr && current->nextSibling == nullptr);
if (current != nullptr) {
current = current->nextSibling;
}
}
} // while
} // Dfs
Related
i have a program with various recursive functions.
I now need to optimize the code to run the program faster: i checked with profiler and, a part from the biggest function with lots of checks, i have two functions that require a lot of time every run.
One (Unmarked_Nodes) is like this:
typedef struct node* tree;
struct node{
char* data;
tree left;
tree right;
int marker;
};
static int remaining = 0;
int main(){
...
}
int Unmarked_Nodes(tree root) {
if (root != NULL) {
Unmarked_Nodes(root->left);
if (root->marker == 0)
remaining++;
Unmarked_Nodes(root->right);
}
return remaining;
}
The other is similar but instead of the if cycle it has a printf of data.
The other, however, is faster than this... why? Or instead: how can i improve the code to make it run faster?
Thanks in advance
Candidate improvements: might help a little although answer remains O(n).
Recurse less often
Loop inside the function for one of the children.
Avoid global
Simply not needed.
Use const
No so much a speed improvement, yet allows for use with constant data.
Avoid hiding pointers
int Unmarked_Nodes(const struct node *root) {
int remaining = 0;
while (root != NULL) {
remaining += Unmarked_Nodes(root->left);
if (root->marker == 0) {
remaining++;
}
root = root->right;
}
return remaining;
}
Perhaps only recurse when both children are non-NULL. Test null-ness at the end of the loop since it is initially false for all recursive entry.
static int Unmarked_Nodes2r(const struct node *root) {
int remaining = 0;
do {
if (root->marker == 0) {
remaining++;
}
if (root->left) {
if (root->right) {
remaining += Unmarked_Nodesr(root->right);
}
root = root->left;
// continue; // Could skip loop test.
} else {
root = root->right;
}
} while (root);
return remaining;
}
int Unmarked_Nodes2(const struct node *root) {
return root ? Unmarked_Nodes2r(root) : 0;
}
In the absence of more information, it would seem that you likely "visit" the tree three times: once for 'marking' nodes (for whatever purpose), once to 'print' marked (or unmarked) nodes, and once more to reset those marks.
Presuming that 'marked nodes' are the interesting ones, consider using a dynamic array of pointers (malloc/realloc in suitable increments) to build a list of only those nodes, print from that list (no 2nd tree traversal), then free() the list (no 3rd tree traversal).
You wouldn't need to 'mark/unmark' anything. Interesting nodes added to the suggested list mean that those nodes are 'marked', and 'unmarked' when the list is erased.
You may need to consider if 'marking' may encounter unwanted duplicates.
Another suggestion is to consider transforming the tree into a list once it is filled. Then, use conventional binary search of that list to mark 'nodes', and a sweep through to erase marks (presuming the same list is to be reused multiple times.
Another suggestion relates to whether you are marking to include or to exclude from the print traversal. If marked nodes are included, then simply 'unmark' them as you print them. If marked nodes are excluded, then mark all those other unmarked nodes being printed that haven't previously been 'excluded' and remember whether '0' means 'marked' or if '1' means marked for the next time it comes to searching/marking.
I am having a very difficult time understanding what my book is wanting me to learn. I am currently re-vising some algorithms and data structure material I studied quite some years ago. One chapter is about Binary Search Tree and their application, etc. The book talks about a BST (binary search tree) as a tree with nodes that has the following qualities: int data, node left, node right, node parent. I understand this and I have implemented a search function.
Now to the issue: How do I create a delete function when the arguments I have are N* delete(Tree* tree, N* delete_node);?
There are lots of recursive functions but they rely on the fact that we recursively pass a pointer of a node and a key value. This is not what the exercise in the book is looking for. An example of a method that I have tried and works is: source code for C Program for Insertion and Deletion in Binary Search Tree. Two other great sources that fails to meet these criteria are codezclub and geeksforgeeks. They solve the "delete" function but their function have other parameters in their function. My book specifies
A modifying operation that, given a pointer delete_node to an
element in the BST, tree, deletes delete_node from the BST tree and returns its pointer.
Function declaration: N* delete(Tree* tree, N* delete_node);
Anybody got any ideas? Also minimally reproducible examples are given in the sources above with the caviat that the structs are missing "parent" node. Below are my definitions for the tree and the node:
typedef struct N {
int data;
struct N* left;
struct N* right;
struct N* parent;
} N;
typedef struct Tree {
N* root;
} Tree;
You can use CLRS as a reference for understanding Binary search tree data structures and operations on it. Let's implement delete procedure first and then explain necessary parts.
We need 2 procedures, SUCCESSOR and transplant other than delete itself to make it complete. required procedures are implement before delete.
In order to move subtrees around within the binary search tree, we define a
subroutine transplant, which replaces one subtree as a child of its parent with
another subtree. When transplant replaces the subtree rooted at node u with
the subtree rooted at node v, node u’s parent becomes node v’s parent, and u’s
parent ends up having v as its appropriate child.
void transplant(Tree** root, Tree* u, Tree* v) {
if(u == NULL)
return;
else if(u->parent == NULL)
*root = v;
else if(u->parent->left == u)
u->parent->left = v;
else
u->parent->right = v;
if(v != NULL)
v->parent = u->parent;
}
We need to implement Tree* SUCCESSOR(Tree* T) procedure which returns the node where node->key is the next larger element greater than T->key, if it exists. otherwise, returns NULL. We need SUCCESSOR when delete_node has 2 children. It'll be replaced with the SUCCESSOR(delete_node).
Tree* SUCCESSOR(Tree* T) {
if(T == NULL)
return T; // Since T==NULL, it's equivalent to return NULL
else if(T->left != NULL) {
while(T->left != NULL)
T=T->left;
return T;
}
else {
while(T->parent != NULL && T->parent->right == T)
T = T->parent;
return T->parent;
}
}
We can all needed to implement delete procedure.
Tree* delete(Tree* root, Tree* delete_node) {
if(delete_node == NULL)
return root;
if(delete_node->left== NULL)
transplant(&root, delete_node, delete_node->right);
else if(delete_node->right == NULL)
transplant(&root, delete_node, delete_node->left);
else {
Tree* succ = SUCCESSOR(delete_node);
if(delete_node->right != succ) {
transplant(&root, succ , succ ->right);
succ->right = delete_node->right;
succ->right->parent = succ;
}
transplant(&root, delete_node, succ);
succ->left = delete_node->left;
succ->left->parent = succ;
}
return root;
}
Here's how delete procedure works:
The procedure for deleting a given node delete_node from a binary search tree root takes as arguments pointers to root and delete_node. It organizes its cases a bit differently from the three mentioned above.
If delete_node has no left child (first if statement do this part), then we replace delete_node by its right child, which may or may not be NULL. When delete_node's right child is NULL, this case deals with the situation in which delete_node has no children. When delete_node's right child is non-NULL, this case handles the situation in which delete_node has just one child, which is its right child.
If delete_node has just one child, which is its left child (else if part handles this case), then we replace delete_node by its left child.
Otherwise, delete_node has both a left and a right child. We find delete_node’s successor succ, which lies in delete_node’s right subtree and has no left child (If it has a left, child, then it's not the smallest number greater than delete_node->key. We get contradiction). We want to splice succ out of its current location and have it replace delete_node in the tree.
If succ is delete_node's right child, then we replace delete_node by succ.
Otherwise, succ lies within delete_node’s right subtree but is not delete_node's right child. In this case, we first replace succ by its own right child, and then we replace delete_node by succ.
Since both quotes are taken fromCLRS, I encourage you to review it in case of having problem in understanding why it works.
Of course you can do deletion without recursive function.
typedef struct N {
int data;
struct N* left;
struct N* right;
struct N* parent;
} N;
typedef struct Tree {
N* root;
} Tree;
N* delete(Tree* tree, N* delete_node);
Lets start from this definition. Well, have to say the delete function is uncommon, given the fact that in most cases we are using:
void delete(Tree* tree, int element);
Because we don't know whether the element exists or not (it need to be determined by contains() or find() function when we execute the deletion), and we don't need any return value most likely.
Anyway lets put it your way.
Deleting from BST is complicated and you should proceed with three basic situation:
The node you want to delete is leaf, i.e. it has no child;
The node has one child, either left or right;
The node have both right and left child.
These situations are different in processing, and for simplicity we will omit some implementation details and safety checks, but don't forget them in production code.
In situation 1, you simply free the memory, and change the pointer in its parent to NULL:
if (delete_node->parent->data < delete_node->data)
delete_node->parent->right = NULL; // WARNING: corner case
else
delete_node->parent->left= NULL; // WARNING: corner case
free(delete_node);
Situation 2 is pretty like Situation 1; instead of NULL, we need to change the pointer in parent node to its only child, then we free the node. The logic is similar and we will skip the code.
Situation 3 is a little more complicated. Generally we have two ways and both of them works; either we replace the node's data with largest (in BST we can say "rightest") data of its left subtree, or we replace the node's data with smallest/leftest data of its right subtree. After the replacing, we remove that node instead. Let's examine the second approach, i.e. replace the node's data with smallest data of its right subtree. The first approach has very similar logic.
For this approach, we need this function implemented:
N* findMin(N* root_of_subtree);
It will take root_of_subtree as input (search will start from this node), and returns the pointer to the desired, smallest node.
Assume we have found the desired smallest node in the right subtree of delete_node:
N* smallestNodeOfRightSubtree = findMin(delete_node->right);
delete_node->data = smallestNodeOfRightSubtree->data;
smallestNodeOfRightSubtree->parent->left = smallestNodeOfRightSubtree->right; // WARNING, corner case
free(smallestNodeOfRightSubtree);
We directly changed parent->left here, since in general case (corner case 2 explains what is not general), that node has two charateristics:
It has no left child.
It must be on the left subtree of its parents.
But, it is not always general, we still have some corner cases need to cover:
What if delete_node is root node? It has no parent, and code in Situation 1 will be undefined behavior.
What if the smallest node of the right subtree of delete_node is delete_node->right, i.e. the right subtree only have 1 node. In this case, we cannot change smallestNodeOfRightSubtree->parent->left; instead, we should change parent->right.
THAT'S IT. Now you know the logics, and some corner cases. GL & HF.
Recursion is soo confusing for me...
below are the structs that are used to create the binary-like tree:
struct parent {
char *name;
Child *children;
};
typedef struct parent Parent;
struct child {
struct Parent *Pptr;
struct child *next;
};
typedef struct child Child;
I want to iterate through the entire tree and on each child/parent (basically every node) call a function called birthdaygift().
Below is what I've tried so far, but I don't think it works.
tree(Parent *parent) {
Child* child = parent->children;
birthdaygift(parent);
if (child == NULL) {
return;
}
while (child->next != NULL) {
Parent * recurseparent = child->Pptr;
tree(recurseparent);
}
birthdaygift(parent);
}
It'd be really helpful if someone can give me some pointers?
Your data structure looks very, very odd, it looks like you have the child nodes pointing back to the parents and are somehow maintaining the list of children in the child node as a linked list.... I'll give my answer assuming you are intending to work with a binary tree.
Now, assuming that is the case, wouldn't going up to the parent from one child and then going up from the next child take you to the same node? Typically, binary trees are parsed by starting from the top and going down.
A traditional binary tree has pointers from the parent (for this example, *left and *right), and you would traverse the entire tree by using either a Depth First Search algorithm (basically, you keep going left recursively until you run out of nodes, and then you go right), in pseudocode
function gift_node(*node) {
birthday_gift(node);
if (node->left != NULL) gift_node(node->left);
if (node->right != NULL) gift_node(node->right);
}
Now, if you were to watch the process parsing this binary tree, you would see it start at the top and keep following the left node. It would then backtrack, process the right node and it's subordinate nodes and keep going until it has visited every node.
The struct you declared is not a tree, it is similar to a doubly-linked list.
The struct for a binary tree node would be:
struct BST_Node {
Value value;
BST_Node *right;
BST_Node *left;
};
For traversing a tree, you want to visit every single node. One way to do this is to visit all the nodes to the left recursively, then the node itself, then all the nodes to the right recursively (this is known as in-order traversal, but this isn't important for this situation).
If you want to call birthdaygift() on each node:
void traverse(Node *n) {
if (n->left)
traverse(n->left);
birthdaygift(n);
if (n->right)
traverse(n->right);
}
A discussion of other tree traversal algorithms can be found here.
How can I free this tree effectively? That algorithm should work for any given node in such tree. So I'll have pointer to node, and that node will be "root" node. And I want to free everything below that node.
Every node in tree is this struct:
typedef struct tag
{
struct tag* parent;
struct tag* nextSibling;
struct tag* previousSibling;
struct tag* firstChild;
struct tag* lastChild;
char* name;
char* text;
}node;
I imagine this would work. But in reality, Dariusz is correct. You just use a valid tree traversal, and perform your operation on each node.
The question changed: And since you want this to operate on any node in the tree, just find the root first. It's much easier to write a tree traversal that progresses in one direction, than up and down the tree.
You've changed the question from deleting a tree, to deleting a subset of a tree. So, instead, let's do this. Remove the element from the tree first (remove_node). and then perform the same free that we would have done before.
void remove_node(node *self) {
if (self->previousSibling)
self->previousSibling->nextSibling = self->nextSibling;
if (self->nextSibling)
self->nextSibling->previousSibling = self->previousSibling;
if (self->parent && self->parent->firstChild == self)
self->parent->firstChild = self->nextSibling;
if (self->parent && self->parent->lastChild == self)
self->parent->lastChild = self->previousSibling;
}
void free_node(node *self) {
// Free one node. Perhaps this is:
free(self->name);
free(self->text);
free(self);
}
void iterate_nodes(node *root, void op(node *self) ) {
if (root == NULL)
return;
iterate_nodes(root->nextSibling, op);
iterate_nodes(root->firstChild, op);
op(root);
}
int main() {
node *node = NULL; // Some node in the tree...
remove_node(node);
iterate_nodes(node, free_node);
}
Use any of the standard tree-traversal mechanisms and delete all elements.
http://en.wikipedia.org/wiki/Tree_traversal
You can use a standard post-order tree traversal algorithms to free the whole tree. To make it work from any given node, just start by traversing all of the parent links to the root. For example:
void free_tree(node *n)
{
// Find the root node of the tree
while(n->parent)
n = n->parent;
free_tree_helper(n);
}
void free_tree_helper(node *n)
{
// Free all children of this node in post-order traversal
node *child = n->firstChild;
while(child)
{
// Save the next sibling pointer to avoid dangling pointers
node *next = child->nextSibling;
free_tree_helper(child);
child = next;
}
// All the children have been freed, now free the parent
free(n->name);
free(n->text);
free(n);
}
Alternatively, if you use a memory pool to allocate your tree nodes, all of your nodes come from the same pool, and the pool contains no other tree's nodes, you can instead just free the entire memory pool at once, saving you from having to traverse the entire tree. This will be much more efficient, since it avoids a lot of cache misses in a potentially large tree, and it also avoids certain memory fragmentation problems.
I have an AVL Tree. Each node looks like this:
typedef struct {
Node *parent; // the parent node
Node *left; // the node left of this node
Node *right; // the node right of this node
int height; // the height of this node
void *value; // payload
} Node;
Is it possible to iterate over an AVL tree with these nodes in O(1) space, without recursion, if yes, how?
If not, a solution with sub-O(n) space or iff really necessary O(N) space is appreciated as well.
With iterating over the tree I mean I want to visit each node once, and if possible in order (from the mostleft to the mostright node).
If you store the last node you have visited, you can derive the next node to visit in an iterator.
If the last node was your parent, go down the left subtree.
If the last node was your left subtree, go down the right subtree.
If the last node was your right subtree, go to your parent.
This algorithm gives you a traversal in O(1) for the tree. You need to flesh it out a little for the leaves and decide what kind of iterator (pre/in/post-order) you want to decide where the iterator should and wait for incrementation.
It is possible to get the next in-order node given a pointer to some node, as long as you keep parent pointers. This can be used to iterate the tree, starting with the leftmost node. From my implementation of AVL tree:
BAVLNode * BAVL_GetNext (const BAVL *o, BAVLNode *n)
{
if (n->link[1]) {
n = n->link[1];
while (n->link[0]) {
n = n->link[0];
}
} else {
while (n->parent && n == n->parent->link[1]) {
n = n->parent;
}
n = n->parent;
}
return n;
}
To get the leftmost node:
BAVLNode * BAVL_GetFirst (const BAVL *o)
{
if (!o->root) {
return NULL;
}
BAVLNode *n = o->root;
while (n->link[0]) {
n = n->link[0];
}
return n;
}
Here, node->link[0] and node->link[1] are the left and right child of the node, respectively, and node->parent is the pointer to the parent node (or NULL for root).
A single GetNext() operation has O(logn) time complexity. However, when used to iterate the entire tree, you get O(n) amortized time complexity.
"Datastructures and their algorithms" by Harry Lewis and Larry Denenberg describe link inversion traversal for constant space traversal of a binary tree. For this you do not need parent pointer at each node. The traversal uses the existing pointers in the tree to store path for back tracking. 2-3 additional node references are needed. Plus a bit on each node to keep track of traversal direction (up or down) as we move down. In my implementation of this algorithms from the book, profiling shows that this traversal has far less memory / processor time. An implementation in java (c would be faster i guess) is here.