Can anyone explain the way of cloning the binary tree with random pointers apart from left to right? every node has following structure.
struct node {
int key;
struct node *left,*right,*random;
}
This is very popular interview question and I am able to figure out the solution based on hashing(which is similar to cloning of linked lists). I tried to understand the solution given in Link (approach 2) but am not able to figure out what does it want to convey by reading code also.
I don't expect solution based on hashing as it is intuitive and pretty straight forward. Please explain solution based on modifying binary tree and cloning it.
The solution presented is based on the idea of interleaving both trees, the original one and its clone.
For every node A in the original tree, its clone cA is created and inserted as A's left child. The original left child of A is shifted one level down in the tree structure and becomes a left child of cA.
For each node B, which is a right child of its parent P (i.e., B == P->right), a pointer to its clone node cB is copied to a clone of its parent.
P P
/ \ / \
/ \ / \
A B cP B
/ \ / \ / \
/ \ / \ / \
X Z A cB Z
/ \ /
cA cZ
/
X
/
cX
Finally we can extract the cloned tree by traversing the interleaved tree and unlinking every other node on each 'left' path (starting from root->left) together with its 'rightmost' descendants path and, recursively, every other 'left' descendant of those and so on.
What's important, each cloned node is a direct left child of its original node. So in the middle part of the algorithm, after inserting the cloned nodes but before extracting them, we can traverse the whole tree walking on original nodes, and whenever we find a random pointer, say A->random == Z, we can copy the binding into clones by setting cA->random = cZ, which resolves to something like
A->left->random = A->random->left;
This allows cloning random pointers directly and does not require additional hash maps (at the cost of interleaving new nodes into the original tree and extracting them later).
The interleaving method can be simplified a little, I think.
1) For every node A in the original tree, create clone cA with the same left and right pointers as A. Then, set As left pointer to cA.
P P
/ \ /
/ \ /
A B cP
/ \ / \
/ \ / \
X Z A B
/ /
cA cB
/ \
X Z
/ /
cX cZ
2) Now given a node and it's clone (which is just node.left), the random pointer for the clone is: node.random.left (if node.random exists).
3) Finally, the binary tree can be un-interleaved.
I find this interleaving makes reasoning about the code much simpler.
Here is the code:
def clone_and_interleave(root):
if not root:
return
clone_and_interleave(root.left)
clone_and_interleave(root.right)
cloned_root = Node(root.data)
cloned_root.left, cloned_root.right = root.left, root.right
root.left = cloned_root
root.right = None # This isn't necessary, but doesn't hurt either.
def set_randoms(root):
if not root:
return
cloned_root = root.left
set_randoms(cloned_root.left)
set_randoms(cloned_root.right)
cloned_root.random = root.random.left if root.random else None
def unterleave(root):
if not root:
return (None, None)
cloned_root = root.left
cloned_root.left, root.left = unterleave(cloned_root.left)
cloned_root.right, root.right = unterleave(cloned_root.right)
return (cloned_root, root)
def cloneTree(root):
clone_and_interleave(root)
set_randoms(root)
cloned_root, root = unterleave(root)
return cloned_root
The terminology used in those interview questions is absurdly bad. It’s the case of one unwitting kuckledgragger somewhere calling that pointer the “random” pointer and everyone just nods and accept this as if it was some CS mantra from an ivory tower. Alas, it’s sheer lunacy.
Either what you have is a tree or it isn’t. A tree is an acyclic directed graph with at most a single edge directed toward any node, and adding extra pointers can’t change it - the things the pointers point to must retain this property.
But when the node has a pointer that can point to any other node, it’s not a tree. You got a proper directed graph with cycles in it, and looking at it as if it were a tree is silly at this point. It’s not a tree. It’s just a generic directed edge graph that you’re cloning. So any relevant directed graph cloning technique will work, but the insistence on using the terms “tree” and “random pointer” obscure this simple fact, and confuse the matters terribly.
This snafu indicates that whoever came up with the question was not qualified to be doing any such interviewing. This stuff is covered in any decent introductory data structure textbook so you’d think it shouldn’t present some astronomical uphill effort to just articulate what you need in a straightforward manner. Let the interviewees deal with users who can’t articulate themselves once they get that job - the data structure interview is neither the place nor time for that. It reeks of stupidity and carelessness, and leaves permanently bad aftertaste. It’s probably yet another stupid thing that ended up in some “interview question bank” because one poor soul got it asked by a careless idiot once and now everyone treats it as gospel. It’s yet again the blind leading the blind and cluelessness abounds.
Copying arbitrary graphs is a well solved problem and in all cases you need to retain the state of your traversal somehow. Whether it’s done by inserting nodes into the original graph to mark the progress - one could call it intrusive marking - or by adding data to the copy in progress and removing it when done, or by using an auxiliary structure such as a hash, or by doing repeat traversal to check it you made a copy of that node elsewhere - is of secondary importance, since the purpose Is always the same: to retain the same state information, just encoding it in various ways, trading off speed and memory use (as always).
When thinking of this problem, you need to tell yourself what sort of state you need to finish the copy, and abstract it away, and implement the copy using this abstract interface. Then you can implement it in a few ways, but at that point the copy itself doesn’t obscure things since you look at this simple abstract state-preserving interface and not at the copy process.
In real life the choice of any particular implementation highly depends on the amount and structure of data being copied, and the extent you have control over it all. If you’re the one controlling the structure of the nodes, then you’ll usually find that they have some padding that you could use to store a bit of state information. Or you’ll find that the memory block allocated for the nodes is actually larger than requested: malloc will often end up providing a block larger than asked for, and all reasonable platforms have APIs that let you retrieve the actual size of the block and thus check if there’s maybe some leftover space just begging to be used. These APIs are not always fast so be careful there of course. But you see where this is going: such optimization requires benchmarks and a clear need driven by demands of the application. Otherwise, use whatever is least likely to be buggy - ideally a C library that provides data structures that you could use right away. If you need a cyclic graph there are libraries that do just that - use them first.
But boy, do I hate that idiotic “random” name of the pointer. Who comes up with this nonsense and why do they pollute so many minds? There’s nothing random about it. And a tree that’s not a tree is not a tree. I’d fail that interviewer in a split second…
Related
I'm reading a book about Data structures and I'm getting into trouble with the implementation example of binary tree in that book. The problem is I need to calculate and implement this parse tree below:
This is the source code of the example I mentioned above:
I've known about trees but I cannot understand what the source code really means because the book I'm reading does not explain each step. I really need the deeper explanation for the source code.
EDIT : You can focus on the loop step, it is the most difficult one for me to understand
This code seems to implement the Reverse polish notation, i.e. a notation where operators follow their operands. It reads an expression recorded in the RPN and builds the corresponding binary tree. For example, for the tree above the RPN form will be:
ABC+DE**F+*
The logic is pretty straightforward and is based on a stack that contains nodes of the tree:
Every time you encounter an operand (i.e. a letter), you create a new leaf node with an operand and push it to the stack.
Every time you encounter an operator, you create a new operator node, that replaces the two top-most nodes from the stack. The replaced nodes become the new node's children.
In the end, you get the expression tree on the top of the stack.
Update: As for the specific lines you mentioned: z is a special kind of tree node, a sentinel, that is depicted as a tiny rectangle on the picture. That's a no-value node, which allows you to know when you reach the tree bottom. Another way is just to use a null pointer (the link above compares the approaches).
z->l = z;
z->r = z;
is what makes the node it's own child. A sentinel node can also represent an empty tree.
Now in the loop:
x->info = c;
x->l = z;
x->r = x;
creates a new leaf node (operand nodes don't have children). If we then find than the node is actually an operator, the children are immediately replaced with the operands from the stack.
I have a game I am working on that uses a linked list for the entities in the game. I have found what I think to be some sort of bug. Note, I'm coding in C. But after this trouble with C pointers I'm thinking about trying C++ techniques.
In my debug testing two projectiles were colliding which blows both of them up. Basically the situation is this:
Starting in Entity's move function:
1) Projectile entity moves
2) Loop through all entities checking collision at this new location
3) If collision, in this case between projectiles, remove both
I pass a double pointer of the entity to the function that does collision. That entity may be removed but I still need to use it for advancing the entity to the next one in the list (in the while loop). If that didn't make sense it is seen as this:
ENTITY *node;
while (node)
{
...
entity_do_collision (&node); // <-- node may be removed in this function
//Debug
if (node == global_node)
{
}
else
node = global_node;
node = node->next; // <-- pass a double pointer above so this works here
}
So, I've ran through the code so many times and don't see any illegal operations. The part that gets me is sometimes the double pointer will work and sometimes it won't. I tested using a global entity pointer (that always works) to compare back in my entity move function to test if the node being removed matches what it was set to in the entity remove function.
This description is a little abstract, so let me know if I need to explain more.
There are zillion solutions which would or would not work for your exact problem.
Here are some ideas to start with:
Do not delete objects from the container on collision, but mark them "dead". Clean up "dead" bodies in separate pass after collision detection finished.
Mark them dead on collision, but not delete at all. Just reuse marked nodes for future entities.
Improvement of (2): sort your container, so "dead" entities went to the tail and mark size of the container as it would contain only "living" ones
Improvement of (1), (2), (3): implement some kind of "garbage collection", so "dead" entities would be cleaned up let's say once a second or once a frame or when memory threshold reached.
etc.
Sidenote: You should never use linked lists in 21th's century (an era of hierarchies of caches, prefetching, out-of-order execution and mutithreading), unless you really, really have no other choice and you understand what you are doing. Use arrays by default, swith to something else only if you find it reasonable.
More info:
What is “cache-friendly” code?
Stop Using Linked-Lists
Bjarne Stroustrup: Why you should avoid Linked Lists (video)
Original code:
ENTITY *node;
while (node)
{
...
entity_do_collision (&node); // <-- node may be removed in this function
node = node->next;
/* the function can have changed node's value
** but on the next iteration ( on the **value** of node->next)
** the original node->next will not be affected!
*/
}
Sample code using pointer tot pointer:
ENTITY **pp;
for (pp = &global_node; *pp; pp = &(*pp)->next)
{
...
entity_do_collision (pp); // <-- *pp (= node) may be removed in this function
...
}
I recently implemented binary search tree, linked lists etc as a learning exercise. II implemented several API's like Insert,delete etc.
For example the Insert node API looks like
void insertNode(Node** root, Node* node)
The application would allocate memory for the node to be inserted. i.e node and assign the value/key and pass to this function.
1) My question is whether this is right approach in general? Or does the application only need to pass the value/key to insertNode function and the this function allocates memory?
i.e void insertNode(Node** root, int key)
{
malloc for node here
}
2) What is a good design practice- the application handles allocating memory and free or this library of APi's ?
Thanks
General principles:
Whoever allocates memory should also have responsibility to delete it.
Make the client's job as easy as possible
Think about thread safety
So, don't have client allocate the memory being managed by the tree. However we then need to think carefully about what a find or search should return.
You might look at example code to see what policies other folk have taken for collection APIs. For example.
In passing you don't need the double pointer for your root.
If someone has root pointing to an object { left -> ..., right -> ...} then if you pass root as Node* your code can write
root->left = newValue;
you're modifying what root points to not root itself. By passing the double pointer you're allowing the structure to be completely relocated because someone can write:
*root = completelyNewTree
which I doubt you intend.
Now for inserting I'd suggest passing in the new value, but have the collection allocate space and copy it. The collection now has complete control over the data it holds, and anything it returns should be copied. We don't want clients messing with the tree directly (thread safety).
This implies that for result of a find either the caller must pre-allocate a buffer or we must clearly document that the collection is returning an allocated copy that the caller must delete.
And yes, working in any Garbage Collecting language is much easier!
I'm trying to resolve this for fun but I'm having a little bit of trouble on the implementation, the problem goes like this:
Having n stacks of blocks containing m blocks each, design a program in c that controlls a robotic arm that moves the blocks form an inicial configuration to a final one using the minimum amount of movements posible, your arm can only move one block at a time and can only take the block at the top of the stack, your solution should use either pointers or recursive methods
In other words the blocks should go from this(suposing there are 3 stacks and 3 blocks):
| || || |
|3|| || |
|2||1|| |
to this:
| ||1|| |
| ||2|| |
| ||3|| |
using the shortest amount of movements printing each move
I was thinking that maybe I could use a tree of some sorts to solve it (n-ary tree maybe?) since that is the perfect use of pointers and recursive methods but so far it has proved unsuccesfull, I'm having lots of trouble defining the estructure that will store all the movements since I would have to check every time I want to add a new move to the tree if that move has not been done before, I want each leaf to be unique so when I find the solution it will give me the shortest path.
This is the data structure I was thinking of:
typedef struct tree(
char[MAX_BLOCK][MAX_COL] value;
struct tree *kids
struct tree *brothers;
)Tree;
(I'm really new at C so sorry beforehand if this is all wrong, I'm more used to Java)
How would you guys do it? Do you have any good ideas?
You have the basic idea - though I am not sure why you have elected to choose brothers over the parent.
You can do this problem with a simple BFS search, but it is a slightly less interesting solution, and not the one you for which seemed to have set yourself up.
I think it will help if we concisely and clearly state our approach to the problem as a formulation of either Dijkstra's, A*, or some other search algorithm.
If you are unfamiliar with Dijkstra's, it is imperative that you read up on the algorithm before attempting any further. It is one of the foundational works in shortest path exploration.
With a familiarity of Dijkstra's, A* can readily be described as
Dijsktra's minimizes distance from the start. A* adds a heuristic which minimizes the (expected) distance to the end.
With this algorithm in mind, lets state the specific inputs to an A* search algorithm.
Given a start configuration S-start, and an ending configuration S-end, can we find the shortest path from S-start to S-end given a set of rules R governed by a reward function T
Now, we can envision our data structure not as a tree, but as a graph. Nodes will be board states, and we can transition from state to state using our rules, R. We will pick which edge to follow using the reward function T, the heuristic to A*.
What is missing from your data-structure is the cost. At each node, you will want to store the current shortest path, and whether it is finalized.
Let's make a modification to your data-structure which will allow us to readily traverse a graph and store the shortest path information.
typedef struct node {
char** boardState;
struct node *children;
struct node *parent;
int distance;
char status; //pseudo boolean
} node;
You may want to stop here if you were interested in discovering the algorithm for yourself.
We now consider the rules of our system: one block at a time, from the top of a stack. Each move will constitue an edge in our graph, whose weight is governed by the shortest number of moves from S-begin plus our added heuristic.
We can then sketch a draft of the algorithm as follows:
node * curr = S-begin;
while (curr != S-end) {
curr->status == 'T'; //T for True
for(Node child : children) {
// Only do this update if it is cheaper than the
int updated = setMin(child->distance, curr->distance + 1 + heuristic(child->board));
if(updated == 1) child->parent = curr;
}
//set curr to the node with global minimum distance who has not been explored
}
You can then find the shortest path by tracing the parents backwards from S-end to S-begin.
If you are interested in these types of problems, you should consider taking a uppergraduate level AI course, where they approach these types of problems :-)
I have a very simple binary tree structure, something like:
struct nmbintree_s {
unsigned int size;
int (*cmp)(const void *e1, const void *e2);
void (*destructor)(void *data);
nmbintree_node *root;
};
struct nmbintree_node_s {
void *data;
struct nmbintree_node_s *right;
struct nmbintree_node_s *left;
};
Sometimes i need to extract a 'tree' from another and i need to get the size to the 'extracted tree' in order to update the size of the initial 'tree' .
I was thinking on two approaches:
1) Using a recursive function, something like:
unsigned int nmbintree_size(struct nmbintree_node* node) {
if (node==NULL) {
return(0);
}
return( nmbintree_size(node->left) + nmbintree_size(node->right) + 1 );
}
2) A preorder / inorder / postorder traversal done in an iterative way (using stack / queue) + counting the nodes.
What approach do you think is more 'memory failure proof' / performant ?
Any other suggestions / tips ?
NOTE: I am probably going to use this implementation in the future for small projects of mine. So I don't want to unexpectedly fail :).
Just use a recursive function. It's simple to implement this way and there is no need to make it more compilcated.
If you were doing it "manually" you'd basically end up implementing the same thing, just that you wouldn't use the system call stack for temporary variables but your own stack. Usually this won't have any advantages outweighing the more complicated code.
If you later find out that a substantial time in your program is spend calculating the sizes of trees (which probably won't happen) you can still start to profile things and try how a manual implementation performs. But then it might also better to do algorithmic improvements like already keeping track of the changes in size during the extraction process.
If your "very simple" binary tree isn't balanced, then the recursive option is scary, because of the unconstrained recursion depth. The iterative traversals have the same time problem, but at least the stack/queue is under your control, so you needn't crash. In fact, with flags and an extra pointer in each node and exclusive access, you can iterate over your tree without any stack/queue at all.
Another option is for each node to store the size of the sub-tree below it. This means that whenever you add or remove something, you have to track all the way up to the root updating all the sizes. So again if the tree isn't balanced that's a hefty operation.
If the tree is balanced, though, then it isn't very deep. All options are failure-proof, and performance is estimated by measurement :-) But based on your tree node struct, either it's not balanced or else you're playing silly games with flags in the least significant bits of pointers...
There might not be much point being very clever with this. For many practical uses of a binary tree (in particular if it's a binary search tree), you realise sooner rather than later that you want it to be balanced. So save your energy for when you reach that point :-)
How big is this tree, and how often do you need to know its size? As sth said, the recursive function is the simplest and probably the fastest.
If the tree is like 10^3 nodes, and you change it 10^3 times per second, then you could just keep an external count, which you decrement when you remove a node, and increment when you add one. Other than that, simple is best.
Personally, I don't like any solution that requires decorating the nodes with extra information like counts and "up" pointers (although sometimes I do it). Any extra data like that makes the structure denormalized, so changing it involves extra code and extra chances for errors.