克隆带有随机指针的二叉树

Question

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.

Answer 1

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 (ie, 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).

Answer 2

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 A s 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

Answer 3

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…