I am currently faced with the following problem:
Given is a tree with an unchangeable root node and n children.
I need to optimize this tree so that:
The children count of any node is minimized (only talking about the direct children of a node here, not their children or the like)
As a result of this, the tree height is maximized
The tree is descending in order, so that always node > child
All nodes are < root node.
However, sometimes a node is only < root node and neither < or > than another node.
Any ideas, hints or the like would be greatly appreciated.
Thank you.
From your description, it sounds as if you just want to: (1) sort the nodes into descending order, then (2) make each node a child of its predecessor if its value is strictly smaller than the predecessor's, and a sibling of its predecessor otherwise. This way, the height of the tree is simply the number of distinct values, which is the biggest it can possibly be given your third condition.
I can't help suspecting that you're wanting something more complicated. Am I missing the point somehow?
I agree with Alexei, I think what you want to do is a linked list with a custom insertion function that inserts the elements in a specific order.
This was based on your question.
Now, I don't now what exactly you are trying to do here, but if the goal is to keep an efficient sorted collection of objects, my guess would be to use a Binary Search Tree.
Traversing this tree in order will be very easy and return a sorted list of items. Insertions and deletions are pretty easy too and only have an average complexity of O(log n).
Related
I'm aware of ways to keep binary search trees balanced/self-balancing using rotations.
I am not sure if my case needs to be that complicated. I don't need to maintain any sorted order property like with self-balancing BSTs. I just have an ordinary binary tree that I may need to delete nodes or insert nodes. I need try to maintain balance in the tree. For simplicity, my binary tree is similar to a segment tree, and every time a node is deleted, all the nodes along the path from the root to this node will be affected (in my case, it's just some subtraction of the nodal values). Similarly, every time a node is inserted, all the nodes from the root to the inserted node's final location will be affected (an addition to nodal values this time).
What would be the most straightforward way to keep a tree such as this balanced? It doesn't need to be strictly as height balanced as AVL trees, but something like RB trees or maybe slightly less balanced is acceptable as well.
If a new node does not have to be inserted at a particular spot -- possibly determined by its own value and the values in the tree -- but you are completely free to choose its location, then you could maintain the shape of the tree as a complete tree:
In a complete binary tree every level, except possibly the last, is completely filled, and all nodes in the last level are as far left as possible.
An array is a very efficient data structure for a complete tree, as you can store the nodes in their order in a breadth-first traversal. Because the tree is given to be complete, the array has no gaps. This structure is commonly used for heaps:
Heaps are usually implemented with an array, as follows:
Each element in the array represents a node of the heap, and
The parent / child relationship is defined implicitly by the elements' indices in the array.
Example of a complete binary max-heap with node keys being integers from 1 to 100 and how it would be stored in an array.
In the array, the first index contains the root element. The next two indices of the array contain the root's children. The next four indices contain the four children of the root's two child nodes, and so on. Therefore, given a node at index i, its children are at indices 2i + 1 and 2i + 2, and its parent is at index floor((i-1)/2). This simple indexing scheme makes it efficient to move "up" or "down" the tree.
Operations
In your case, you would define the insert/delete operations as follows:
Insert: append the node to the end of the array. Then perform the mutation needed to its ancestors (as you described in your question)
Delete: replace the node to be deleted with the node that currently sits at the very end of the array, and shorten the array by 1. Make the updates needed that follow from the change at these two locations -- so two paths from root-to-node are impacted.
When balancing non-BSTs, the big question to ask is
Can your tree efficiently support rotations?
Some types of binary trees, like k-d trees, have a specific layer-by-layer structure that makes rotations infeasible. Others, like range trees, have auxiliary metadata in each node that's expensive to update after a rotation. But if you can handle rotations, then you can use just about any of the balancing strategies out there. The simplest option might be to model your tree on a treap: put a randomly-chosen weight field into each node, and then, during insertions, rotate your newly-added leaf up until its weight is less than its parent. To delete, repeatedly rotate the node with its lighter child until it's a leaf, then delete it.
If you cannot support rotations, you'll need a rebalancing strategy that does not require them. Perhaps the easiest option there is to model your tree after a scapegoat tree, which works by lazily detecting a node that's too deep for the tree to be balanced, then rebuilding the smallest imbalanced subtree possible into a perfectly-balanced tree to get everything back into order. Deletions are handled by rebuilding the whole tree once the number of nodes drops by some constant factor.
Assume we have a tree where every node has pre-decided set of outgoing nodes. Is it possible to come up with a fast way/optimizations to count the number of leaf nodes given a level value? Would be great if someone could suggest any ideas/links/resources to do the same.
No. you'd still have to traverse the entire tree. There's no way of predicting the precise structure - or approximating it - from only the number of childnodes of each node of the tree.
Apart from that: just keep a counter and update it on each insertion. Far simpler and wouldn't change time-complexity of any operation, except for counting leaves, which would be reduced to O(1).
This can actually get pretty tough thing. As it varies of what is the programming language, what is the input data structure, is the tree binary or general tree (arbitrary number of children), size of the tree.
The most general idea is to run a DFS or BFS, starting from the root, to get every node level and then make a list of sets where each set contains the nodes of a single level. The set can be any structure, standard list is fine.
Let's say you are working in C++ which is good, if not the best practical choice if you need performance (even better than C).
Let's say we have a general tree and the input structure is adjacency list as you mentioned.
//nodes are numbered from zero to N-1
vector<vector<int>> adjList;
Then you run either a BFS or DFS, either will do for a tree, keeping a level for each node. The level for a next node is the level of it's parent plus one.
Once you discover the level, you put the node in like this.
vector<vector<int>> nodesPartitionedByLevels(nodeCount);
//run bfs here
//inside it you call
nodesPartitionedByLevels[level].push_back(node)
That's about it.
Then when you have the levels, you iterate all the nodes on that level and you check the adjaceny list if it contans any nodes.
basically you call adjList[node].empty(). If true than that is a leaf node.
We are given a tree with n nodes in form of a pointer to its root node, where each node contains a pointer to its parent, left child and right child, and also a key which is an integer. For each node v I want to add additional field v.bigger which should contain number of nodes with key bigger than v.key, that are in a subtree rooted at v. Adding such a field to all nodes of a tree should take O(n log n) time in total.
I'm looking for any hints that would allow me to solve this problem. I tried several heuristics - for example when thinking about doing this problem in bottom-up manner, for a fixed node v, v.left and v.right could provide v with some kind of set (balanced BST?) with operation bigger(x), which for a given x returns a number of elements bigger than x in that set in logarihmic time. The problem is, we would need to merge such sets in O(log n), so this seems as a no-go, as I don't know any ordered set like data structure which supports quick merging.
I also thought about top-down approach - a node v adds one to some u.bigger for some node u if and only if u lies on a simple path to the root and u<v. So v could update all such u's somehow, but I couldn't come up with any reasonable way of doing that...
So, what is the right way of thinking about this problem?
Perform depth-first search in given tree (starting from root node).
When any node is visited for the first time (coming from parent node), add its key to some order-statistics data structure (OSDS). At the same time query OSDS for number of keys larger than current key and initialize v.bigger with negated result of this query.
When any node is visited for the last time (coming from right child), query OSDS for number of keys larger than current key and add the result to v.bigger.
You could apply this algorithm to any rooted trees (not necessarily binary trees). And it does not necessarily need parent pointers (you could use DFS stack instead).
For OSDS you could use either augmented BST or Fenwick tree. In case of Fenwick tree you need to preprocess given tree so that values of the keys are compressed: just copy all the keys to an array, sort it, remove duplicates, then substitute keys by their indexes in this array.
Basic idea:
Using the bottom-up approach, each node will get two ordered lists of the values in the subtree from both sons and then find how many of them are bigger. When finished, pass the combined ordered list upwards.
Details:
Leaves:
Leaves obviously have v.bigger=0. The node above them creates a two item list of the values, updates itself and adds its own value to the list.
All other nodes:
Get both lists from sons and merge them in an ordered way. Since they are already sorted, this is O(number of nodes in subtree). During the merge you can also find how many nodes qualify the condition and get the value of v.bigger for the node.
Why is this O(n logn)?
Every node in the tree counts through the number of nodes in its subtree. This means the root counts all the nodes in the tree, the sons of the root each count (combined) the number of nodes in the tree (yes, yes, -1 for the root) and so on all nodes in the same height count together the number of nodes that are lower. This gives us that the number of nodes counted is number of nodes * height of the tree - which is O(n logn)
What if for each node we keep a separate binary search tree (BST) which consists of nodes of the subtree rooted at that node.
For a node v at level k, merging the two subtrees v.left and v.right which both have O(n/2^(k+1)) elements is O(n/2^k). After forming the BST for this node, we can find v.bigger in O(n/2^(k+1)) time by just counting the elements in the right (traditionally) subtree of the BST. Summing up, we have O(3*n/2^(k+1)) operations for a single node at level k. There are a total of 2^k many level k nodes, therefore we have O(2^k*3*n/2^(k+1)) which is simplified as O(n) (dropping the 3/2 constant). operations at level k. There are log(n) levels, hence we have O(n*log(n)) operations in total.
How to find a loop in a binary tree? I am looking for a solution other than marking the visited nodes as visited or doing a address hashing. Any ideas?
Suppose you have a binary tree but you don't trust it and you think it might be a graph, the general case will dictate to remember the visited nodes. It is, somewhat, the same algorithm to construct a minimum spanning tree from a graph and this means the space and time complexity will be an issue.
Another approach would be to consider the data you save in the tree. Consider you have numbers of hashes so you can compare.
A pseudocode would test for this conditions:
Every node would have to have a maximum of 2 children and 1 parent (max 3 connections). More then 3 connections => not a binary tree.
The parent must not be a child.
If a node has two children, then the left child has a smaller value than the parent and the right child has a bigger value. So considering this, if a leaf, or inner node has as a child some node on a higher level (like parent's parent) you can determine a loop based on the values. If a child is a right node then it's value must be bigger then it's parent but if that child forms a loop, it means he is from the left part or the right part of the parent.
3.a. So if it is from the left part then it's value is smaller than it's sibling. So => not a binary tree. The idea is somewhat the same for the other part.
Testing aside, in what form is the tree that you want to test? Remeber that every node has a pointer to it's parent. An this pointer points to a single parent. So depending of the format you tree is in, you can take advantage from this.
As mentioned already: A tree does not (by definition) contain cycles (loops).
To test if your directed graph contains cycles (references to nodes already added to the tree) you can iterate trough the tree and add each node to a visited-list (or the hash of it if you rather prefer) and check each new node if it is in the list.
Plenty of algorithms for cycle-detection in graphs are just a google-search away.
I want to sum all the values in the leaves of a BST. Apparently, I can't get to the leaves without traversing the whole tree. Is this true? Can I get to the leaves without taking O(N) time?
You realize that the leaves themselves will be at least 1/2 of O(n) anyway?
There is no way to get the leaves of a tree without traversing the whole tree (especially if you want every single leaf), which will unfortunately operate in O(n) time. Are you sure that a tree is the best way to store your data if you want to access all of these leaves? There are other data structures which will allow more efficient access to your data.
To access all leaf nodes of a BST, you will have to traverse all the nodes of BST and that would be of order O(n).
One alternative is to use B+ tree where you can traverse to a leaf node in O(log n) time and after that all leaf nodes can be accessed sequentially to compute the sum. So, in your case it would be O(log n + k), where k is the number of leaf nodes and n is the total number of nodes in the B+ tree.
cheers
You will either have to traverse the tree searching for nodes without children, or modify the structure you are using to represent the tree to include a list of the leaf nodes. This will also necessitate modifying your insert and delete methods to maintain the list (for instance, if you remove the last child from a node, it becomes a leaf node). Unless the tree is very large, it's probably nice enough to just go ahead and traverse the tree.