This is task I found from old exam and I try it for solved because they can ask similar question for me on friday.
For solution I have cheap solution but I think it all matter of definition of binary search tree.
I make first tree:
1
\
1
\
1
and here is second tree
1
/
1
/
1
When you make pre order traversal you have same output for both tree.. because same element, and both have degenerated tree. But you don't have same tree! So statement is false.
Only problem is are my tree binary search tree... I think yes because binary search tree the element can have greater / lower equal element?
Please halp when I asked our teacher he say I can ask him when vacation is over but when vacation is over my exam is over.... No good thing for me.
Your answer is perfectly fine given the standard definition of BSTs. By the standard definition, BSTs can have repeated elements and identical elements can go in either subtree.
If the question intends the case where there are no duplicates, or where duplicates must exist in only the left (or only the right) subtree, then a pre-order traversal is sufficient to reconstruct the tree.
If duplicates are not allowed, construct the tree recursively as follows: make the root the first node, and then make the left and right subtrees recursively using as input traversals the portions of the original traversal with nodes less than (for left subtree) and greater than (for right subtree) the root node. If duplicates are allowed but are constrained to either the left or the right subtree, then use the same procedure but adding "or equal" to either the less than or the greater than, but not both.
Related
If we some how transform a Binary Search Tree into a form where no node other than root may have both right and left child and the nodes the right sub-tree of the root may only have right child, and vice versa, such a configuration of BST is inherently sorted with its root being approximately in the middle (in case of nearly complete BST’s). To to this we need to do reverse rotations. Unlike AVL and red black trees, where roatations are done to make the tree balanced, we would do reversed rotations.
I would like to explain the pseudo code and logical implementation of the algorithm through the following images. The algorithm is to first sort the left subtree with respect to the root and then the right subtree. These two subparts will be opposite to each other, that is, left would interchange with right. For simplicity I have taken a BST with right subtree, with respect to root, sorted.
To improve the complexity as compared to tree sort we can augment the above algorithm. We can add a flag to each node where 0 stands for a normal node while 1 is when the node has non null right child, in the original unsorted BST. The nodes with flag 1 have an entry in a hash table with key being their pointers and the values being the right most node. For example node 23's pointer would map to 30.5's pointer. Then we would not have to traverse all the nodes in between for the iteration. If we have 23's pointer and 30.5's pointer we can do the required operation in O(1). This will bring down time complexity , as compared to tree sort.
Please review the algorithm and give suggestion if this algorithm is usefull.
Suppose we have a set of binary trees with their inorder and preorder traversals given,and where no tree is a subtree of another tree in the given set. Now another binary tree Q is given.find whether it can be formed by joining the binary trees from the given set.(while joining each tree in the set should be considered atmost once) joining operation means:
Pick the root of any tree in the set and hook it to any vertex of another tree such that the resulting tree is also a binary tree.
Can we do this using LCA (least common ancestor)?or does it needs any special datastructure to solve?
I think a Binary tree structure should be enough. I don't think you NEED any other special data structure.
And I don't understand how you would use LCA for this. As far as my knowledge goes, LCA is used for knowing the lowest common Ancestor for two NODES in the same tree. It would not help in comparing two trees. (which is what I would do to check if Q can be formed)
My solution in words.
The tree Q that has to be checked if it can be made from the set of trees, So I would take a top-down approach. Basically comparing Q with the possible trees formed from the set.
Logic:
if Q.root does not match with any of the roots of the trees in the set (A,B,C....Z...), No solution possible.
if Q.root matches a Tree root (say A) check corresponding children and mark A as used/visited. (Which is what I understand from the question: a tree can be used only once)
We should continue with A in our solution only if all of Q's children match the corresponding children of A. (I would do Depth First traversal, Breadth First would work as well).
We can add append a new tree from the set (i.e. append a new root (tree B) only at leaf nodes of A as we have to maintain binary tree). Keep track of where the B was appended.
Repeat same check with corresponding children comparison as done for A. If no match, remove B and try to add C tree at the place where B was Added.
We continue to do this till we run out of nodes in Q. (unless we want IDENTICAL MATCH, in which case we would try other tree combinations other than the ones that we have, which match Q but have more nodes).
Apologies for the lengthy verbose answer. (I feel my pseudo code would be difficult to write and be riddled with comments to explain).
Hope this helps.
An alternate solution: Will be much less efficient (try only if there are relatively less number of trees) : forming all possible set of trees ( first in 2s then 3s ....N) and and Checking the formed trees if they are identical to Q.
the comparing part can be referred here:
http://www.geeksforgeeks.org/write-c-code-to-determine-if-two-trees-are-identical/
Given a set of values, it's possible for there to be many different possible binary search trees that can be formed from those values. For example, for the values 1, 2, and 3, there are five BSTs we can make from those values:
1 1 2 3 3
\ \ / \ / /
2 3 1 3 1 2
\ / \ /
3 2 2 1
Many data structures that are based on balanced binary search trees use tree rotations as a primitive for reshaping a BST without breaking the required binary search tree invariants. Tree rotations can be used to pull a node up above its parent, as shown here:
rotate
u right v
/ \ -----> / \
v C A u
/ \ <----- / \
A B rotate B C
left
Given a BST containing a set of values, is it always possible to convert that BST into any arbitrary other BST for the same set of values? For example, could we convert between any of the five BSTs above into any of the other BSTs just by using tree rotations?
The answer to your question depends on whether you are allowed to have equal values in the BST that can appear different from one another. For example, if your BST stores key/value pairs, then it is not always possible to turn one BST for those key/value pairs into a different BST for the same key/value pairs.
The reason for this is that the inorder traversal of the nodes in a BST remains the same regardless of how many tree rotations are performed. As a result, it's not possible to convert from one BST to another if the inorder traversal of the nodes would come out differently. As a very simple case, suppose you have a BST holding two copies of the number 1, each of which is annotated with a different value (say, A or B). In that case, there is no way to turn these two trees into one another using tree rotations:
1:a 1:b
\ \
1:b 1:a
You can check this by brute-forcing the (very small!) set of possible trees you can make with the rotations. However, it suffices to note that an inorder traversal of the first tree gives 1:a, 1:b and an inorder traversal of the second tree gives 1:b, 1:a. Consequently, no number of rotations will suffice to convert between the trees.
On the other hand, if all the values are different, then it is always possible to convert between two BSTs by applying the right number of tree rotations. I'll prove this using an inductive argument on the number of nodes.
As a simple base case, if there are no nodes in the tree, there is only one possible BST holding those nodes: the empty tree. Therefore, it's always possible to convert between two trees with zero nodes in them, since the start and end tree must always be the same.
For the inductive step, let's assume that for any two BSTs of 0, 1, 2, .., n nodes with the same values, that it's always possible to convert from one BST to another using rotations. We'll prove that given any two BSTs made from the same n + 1 values, it's always possible to convert the first tree to the second.
To do this, we'll start off by making a key observation. Given any node in a BST, it is always possible to apply tree rotations to pull that node up to the root of the tree. To do this, we can apply this algorithm:
while (target node is not the root) {
if (node is a left child) {
apply a right rotation to the node and its parent;
} else {
apply a left rotation to the node and its parent;
}
}
The reason that this works is that every time a node is rotated with its parent, its height increases by one. As a result, after applying sufficiently many rotations of the above forms, we can get the root up to the top of the tree.
This now gives us a very straightforward recursive algorithm we can use to reshape any one BST into another BST using rotations. The idea is as follows. First, look at the root node of the second tree. Find that node in the first tree (this is pretty easy, since it's a BST!), then use the above algorithm to pull it up to the root of the tree. At this point, we have turned the first tree into a tree with the following properties:
The first tree's root node is the root node of the second tree.
The first tree's right subtree contains the same nodes as the second tree's right subtree, but possibly with a different shape.
The first tree's left subtree contains the same nodes as the second tree's left subtree, but possibly with a different shape.
Consequently, we could then recursively apply this same algorithm to make the left subtree have the same shape as the left subtree of the second tree and to make the right subtree have the same shape as the right subtree of the second tree. Since these left and right subtrees must have strictly no more than n nodes each, by our inductive hypothesis we know that it's always possible to do this, and so the algorithm will work as intended.
To summarize, the algorithm works as follows:
If the two trees are empty, we are done.
Find the root node of the second tree in the first tree.
Apply rotations to bring that node up to the root.
Recursively reshape the left subtree of the first tree to have the same shape as the left subtree of the second tree.
Recursively reshape the right subtree of the first tree to have the same shape as the right subtree of the second tree.
To analyze the runtime of this algorithm, note that applying steps 1 - 3 requires at most O(h) steps, where h is the height of the first tree. Every node will be brought up to the root of some subtree exactly once, so we do this a total of O(n) times. Since the height of an n-node tree is never greater than O(n), this means that the algorithm takes at most O(n2) time to complete. It's possible that it will do a lot better (for example, if the two trees already have the same shape, then this runs in time O(n)), but this gives a nice worst-case bound.
Hope this helps!
For binary search trees this can actually be done in O(n).
Any tree can be "straightened out", ie put into a form in which all nodes are either the root or a left child.
This form is unique (reading down from root gives the ordering of the elements)
A tree is straightened out as follows:
For any right child, perform a left rotation about itself. This decreases the number of right children by 1, so the tree is straightened out in O(n) rotations.
If A can be straightened out into S in O(n) rotations, and B into S in O(n) rotations, then since rotations are reversible one can turn A -> S -> B in O(n) rotations.
I have some questions on binary trees:
Wikipedia states that a binary tree is complete when "A complete binary tree is a binary tree in which every level, except possibly the last, is completely filled, and all nodes are as far left as possible." What does the last "as far left as possible" passage mean?
A well-formed binary tree is said to be "height-balanced" if (1) it is empty, or (2) its left and right children are height-balanced and the height of the left tree is within 1 of the height of the right tree, taken from How to determine if binary tree is balanced?, is this correct or there's "jitter" on the 1-value? I read on the answer I linked that there could be also a difference factor of 4 between the height of the right and the left tree
Do the complete and height-balanced definitions just apply to binary tree or just any other tree?
Following the reference of the definition in wikipedia, I got to
this page. The definition was taken from there but modified:
Definition: A binary tree in which every level, except possibly the deepest, is completely filled. At depth n, the height of the
tree, all nodes must be as far left as possible.
It continues with a note below though,
A complete binary tree has 2k nodes at every depth k < n and between 2n and 2^(n+1) - 1 nodes altogether.
Sometimes, definitions vary according to convenience (be useful for something). That passage might be a variation which, as I understand, requires leaf nodes to fill first the left side of the deepest level (that is, fill from left to right). The definition that I usually found is exactly as described above but without that
passage.
Usually the definition taken for height-balanced tree is the one you
described. In other words:
A tree is balanced if and only if for every node the heights of its two subtrees differ by at most 1.
That definition was taken from here. Again, sometimes definitions are made more flexible to serve specific purposes. For example, the definition of an AVL tree says that
In an AVL tree, the heights of the two child subtrees of any node
differ by at most one
Still, I remember once I had to rewrite an algorithm so that the tree
would be considered height-balanced if the two child subtrees of any
node differed by at most 2. Note that the definition you gave is recursive, this is very common for binary trees.
In a tree whose number of children is variable, you wouldn't be able to say that it is complete (any parent could have the number of children that you want). Still, it can apply to n-ary trees (with a fixed amount of n children).
Do the complete and height-balanced definitions just apply to binary
tree or just any other tree?
Short answer: Yes, it can be extended to any n-ary tree.
I need to write function, which receives some key x and split 2-3 tree into 2 2-3 trees. In first tree there are all nodes which are bigger than x, and in second which are less. I need to make it with complexity O(logn). thanks in advance for any idea.
edited
I thought about finding key x in the tree. And after split its two sub-trees(bigger or lesser if they exist) into 2 trees, and after begin to go up and every time to check sub-trees which I've not checked yet and to join to one of the trees. My problem is that all leaves must be at the same level.
If you move from the root to your key and split each node so one points at the nodes larger than the key and the other at the rest and then make the larger node be a part of your larger tree, say by having the leftmost node at one level higher point at it, (don't fix the tree yet, do it at the end) until you reach the key you will get your trees. Then you just need to fix both trees on the path you used (note that the same path exists on both trees).
Assuming you have covered 2-3-4 trees in the lecture already, here is a hint: see whether you can apply the same insertion algorithm for 2-3 trees also. In particular, make insertions always start in the leaf, and then restructure the tree appropriately. When done, determine the complexity of the algorithm you got.