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Antti H S Laaksonen 2017-02-21 18:10:08 +02:00
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luku18.tex
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@ -85,9 +85,9 @@ as a sum where each term is a power of two.
\section{Subtrees and paths}
\index{node array}
\index{tree traversal array}
A \key{node array}\footnote{A similar idea is sometimes called the \key{Euler tour technique} \cite{tar84}.} contains the nodes of a rooted tree
A \key{tree traversal array} contains the nodes of a rooted tree
in the order in which a depth-first search
from the root node visits them.
For example, in the tree
@ -155,7 +155,7 @@ a depth-first search proceeds as follows:
\end{tikzpicture}
\end{center}
Hence, the corresponding node array is as follows:
Hence, the corresponding tree traversal array is as follows:
\begin{center}
\begin{tikzpicture}[scale=0.7]
\draw (0,0) grid (9,1);
@ -186,8 +186,8 @@ Hence, the corresponding node array is as follows:
\subsubsection{Subtree queries}
Each subtree of a tree corresponds to a subarray
in the node array,
where the first element is the root node.
in the tree traversal array such that
the first element in the subarray is the root node.
For example, the following subarray contains the
nodes in the subtree of node $4$:
\begin{center}
@ -265,7 +265,7 @@ is $3+4+3+1=11$.
\end{tikzpicture}
\end{center}
The idea is to construct a node array that contains
The idea is to construct a tree traversal array that contains
three values for each node: (1) the identifier of the node,
(2) the size of the subtree, and (3) the value of the node.
For example, the array for the above tree is as follows:
@ -381,7 +381,7 @@ and calculate the sum of values in $O(\log n)$ time.
\subsubsection{Path queries}
Using a node array, we can also efficiently
Using a tree traversal array, we can also efficiently
calculate sums of values on
paths from the root node to any other
node in the tree.
@ -484,7 +484,7 @@ For example, the following array corresponds to the above tree:
When the value of a node increases by $x$,
the sums of all nodes in its subtree increase by $x$.
For example, if the value of node 4 increases by 1,
the node array changes as follows:
the array changes as follows:
\begin{center}
\begin{tikzpicture}[scale=0.7]
@ -650,7 +650,7 @@ performed in $O(\log n)$ time.
\subsubsection{Method 2}
Another way to solve the problem is based on
a node array.
a tree traversal array \cite{ben00}.
Once again, the idea is to traverse the nodes
using a depth-first search:
@ -689,12 +689,13 @@ using a depth-first search:
\end{tikzpicture}
\end{center}
However, in this problem,
we add each node to the node array \emph{always}
However, we use a bit different technique where
we add each node to the tree traversal array \emph{always}
when the depth-first search visits the node,
and not only at the first visit.
Hence, a node that has $k$ children appears $k+1$ times
in the node array, and there are a total of $2n-1$
in the array, and there are a total of $2n-1$
nodes in the array.
We store two values in the array: