Search results
Results From The WOW.Com Content Network
Depending on the problem at hand, pre-order, post-order, and especially one of the number of subtrees − 1 in-order operations may be optional. Also, in practice more than one of pre-order, post-order, and in-order operations may be required. For example, when inserting into a ternary tree, a pre-order operation is performed by comparing items.
One problem with this algorithm is that, because of its recursion, it uses stack space proportional to the height of a tree. If the tree is fairly balanced, this amounts to O(log n) space for a tree containing n elements. In the worst case, when the tree takes the form of a chain, the height of the tree is n so the algorithm takes O(n) space. A ...
This unsorted tree has non-unique values (e.g., the value 2 existing in different nodes, not in a single node only) and is non-binary (only up to two children nodes per parent node in a binary tree). The root node at the top (with the value 2 here), has no parent as it is the highest in the tree hierarchy.
The converse is not true: most directed graphs are neither reflexive nor transitive. A preorder that is antisymmetric no longer has cycles; it is a partial order, and corresponds to a directed acyclic graph. A preorder that is symmetric is an equivalence relation; it can be thought of as having lost the direction markers on the edges of the graph.
All of the following problems can be solved in O(Prefix sum(n)) (the time it takes to solve the prefix sum problem in parallel for a list of n items): Classifying advance and retreat edges: Do list ranking on the ETR and save the result in a two-dimensional array A. Then (u,v) is an advance edge iff A(u,v) < A(v,u), and a retreat edge otherwise.
The necessary distinction can be made by first partitioning the edges; i.e., defining the binary tree as triplet (V, E 1, E 2), where (V, E 1 ∪ E 2) is a rooted tree (equivalently arborescence) and E 1 ∩ E 2 is empty, and also requiring that for all j ∈ { 1, 2 }, every node has at most one E j child. [14]
Each node of a ternary search tree stores a single character, an object (or a pointer to an object depending on implementation), and pointers to its three children conventionally named equal kid, lo kid and hi kid, which can also be referred respectively as middle (child), lower (child) and higher (child). [1]
A problem related to the order-maintenance problem is the list-labeling problem in which instead of the order(X, Y) operation the solution must maintain an assignment of labels from a universe of integers {,, …,} to the elements of the set such that X precedes Y in the total order if and only if X is assigned a lesser label than Y.