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Visit the current node for pre-order traversal. For each i from 1 to the current node's number of subtrees − 1, or from the latter to the former for reverse traversal, do: Recursively traverse the current node's i-th subtree. Visit the current node for in-order traversal. Recursively traverse the current node's last subtree.
Level The level of a node is the number of edges along the unique path between it and the root node. [4] This is the same as depth. Width The number of nodes in a level. Breadth The number of leaves. Forest A set of one or more disjoint trees. Ordered tree A rooted tree in which an ordering is specified for the children of each vertex. Size of ...
It is also possible to use depth-first search to linearly order the vertices of a graph or tree. There are four possible ways of doing this: A preordering is a list of the vertices in the order that they were first visited by the depth-first search algorithm. This is a compact and natural way of describing the progress of the search, as was ...
Also called a level-order traversal. In a complete binary tree, a node's breadth-index ( i − (2 d − 1)) can be used as traversal instructions from the root. Reading bitwise from left to right, starting at bit d − 1, where d is the node's distance from the root ( d = ⌊log 2 ( i +1)⌋) and the node in question is not the root itself ( d ...
An alternative algorithm for topological sorting is based on depth-first search.The algorithm loops through each node of the graph, in an arbitrary order, initiating a depth-first search that terminates when it hits any node that has already been visited since the beginning of the topological sort or the node has no outgoing edges (i.e., a leaf node):
6-ary tree represented as a binary tree. Every multi-way or k-ary tree structure studied in computer science admits a representation as a binary tree, which goes by various names including child-sibling representation, [1] left-child, right-sibling binary tree, [2] doubly chained tree or filial-heir chain.
function lookupByPositionIndex(i) node ← head i ← i + 1 # don't count the head as a step for level from top to bottom do while i ≥ node.width[level] do # if next step is not too far i ← i - node.width[level] # subtract the current width node ← node.next[level] # traverse forward at the current level repeat repeat return node.value end ...
In fact in order to answer a level ancestor query, the algorithm needs to jump from a path to another until it reaches the root and there could be Θ(√ n) of such paths on a leaf-to-root path. This leads us to an algorithm that can pre-process the tree in O(n) time and answers queries in O(√ n). In order to reach the optimal query time, we ...