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.
Related
I'm trying to write an algorithm for this problem:
Merge three binary search trees into one sorted array, using O(n) time and O(1) additional space.
I think the straightforward answer is to do an in-order traversal of all three trees at once and compare the elements while traversing. But how can I do such a traversal in all three trees at once? Especially when the trees don't all have the same number of elements.
Your idea seems right.
In each tree, maintain a pointer (iterator).
Initially, the iterator should point to the leftmost node of the tree.
In every iteration, select the minimum of the elements under the three current pointers (it is O(1) time and memory).
Then put that minimum into the resulting array.
After that, advance the corresponding pointer so that it points to the leftmost unvisited element of the tree.
To be able to do that in O(1) memory, the tree should allow some way to go to this next unvisited element: it is sufficient to have a pointer to parent in each node.
Proceed with such iterations until all nodes are visited.
The traversal of a whole tree of n elements takes O(n) time: there are n-1 edges, and the process moves twice along each edge, once up and once down.
So the resulting complexity is 3*O(n) = O(n).
The algorithm to find the next unvisited node is as follows.
Note that, when we are at a node, its left subtree is already fully visited.
The steps are as follows:
While there is no unvisited right child, go up to the parent once.
If, in doing so, we went up and right (we were at the left child), stop right there at the parent.
If we were at the root, terminate the traversal.
Assuming we did not stop yet, there's a right child.
Go there.
Then while there's a left child, go to the left child.
Stop.
The best way to grasp it is perhaps to visualize the steps on some non-trivial picture of a binary search tree. For example, there are explanatory pictures at the Wikipedia article on tree traversal.
I have to create and describe an algorithm for my university course that gets a BST tree T and creates new BST tree T' that satifies properties (and is as fast as possible):
1) T' has the same exact key values as T
2) T' is a Red-Black tree
So far I've had only one idea: randomize 0 or 1. In case of 0, get the max key node from left subtree of T and insert it into T', otherwise get the min key node from right subtree of T and insert it into T'. This is to ensure that Red-Black tree is at least somewhat balanced. The insertion would be any standard RB insertion.
Complexity of getting min/max is O(h), and since this needs to be repeated for each of the nodes in T, this would get quite high. I could also keep a pointer at the max node of left subtree and min node of the right subtree, which would solve the problem of traversing the whole height of the tree every time.
What do you think about this solution? I'm pretty sure it can be done better. Sorry if there is an obvious better solution, but I couldn't find answer to this on the internet, also it's only my 2nd semester at the university and I don't have much experience with programming.
Unless you have some other constraints or information, the fastest way is to forget about the shape of the original BST.
Just put the keys in an ordered list, and build a complete binary tree from it, all in O(N) time.
Then, if there's a partially filled leaf level, then color those nodes red. The rest are black.
I have an one lecture slides says following:
To find middle element in AVL tree, I traverse elements in order until It reaches the moddile element. It takes O(N).
If I know correctly, in tree structure, finding element takes base 2 O(logn) since AVL is binary tree that always divided into 2 childs.
But why it says O(N)?
I am just trying to elaborate 'A. Mashreghi' comment.
Since, the tree under consideration is AVL tree - the guaranteed finding of element in O(log n) holds as log as you have the element(key) to find.
The problem is - you are trying to identify a middle element in the given data structure. As it is AVL tree (self balanced BST) in-order travel gives you elements in ascending order. You want to use this property to find the middle element.
Algorithm goes like - have a counter increment for every node traversed in-order and return # n/2th position. This sums to O(n/2) and hence the overall complexity O(n).
Being divided into 2 children does not guarantee perfect symmetry. For instance, consider the most unbalanced of all balanced binary trees: each right child has a depth one more than its corresponding left child.
In such a tree, the middle element will be somewhere down in the right branch's left branch's ...
You need to determine how many nodes N you have, then locate the N/2th largest node. This is not O(log N) process.
I am having trouble with these questions:
A binary tree with N nodes is at least how deep?
How deep is it at most?
Would the maximum depth just be N?
There are two extremes that you need to consider.
Every node has just a left(or right) child, but not right child. In which case your binary search tree is merely a linkedlist in practice.
Every level in your tree is full, maybe except the last level. This type of trees are called complete.
Third type of tree that I know may not be relevant to your question. But it is called full tree and every node is either a leaf or has n number of childs for an n-ary tree.
So to answer your question. Max depth is N. And at least it has log(N) levels, when it is a complete tree.
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.