924 lines
27 KiB
TeX
924 lines
27 KiB
TeX
\chapter{Tree queries}
|
|
|
|
\index{tree query}
|
|
|
|
This chapter discusses techniques for
|
|
efficiently processing queries related
|
|
to subtrees and paths of a rooted tree.
|
|
For example, possible queries are:
|
|
|
|
\begin{itemize}
|
|
\item what is the $k$th ancestor of a node?
|
|
\item what is the sum of values in the subtree of a node?
|
|
\item what is the sum of values in a path between two nodes?
|
|
\item what is the lowest common ancestor of two nodes?
|
|
\end{itemize}
|
|
|
|
\section{Finding ancestors}
|
|
|
|
\index{ancestor}
|
|
|
|
The $k$th \key{ancestor} of a node $x$ in a rooted tree
|
|
is the node that we will reach if we move $k$
|
|
levels up from $x$.
|
|
Let $f(x,k)$ denote the $k$th ancestor of $x$.
|
|
For example, in the following tree, $f(2,1)=1$ and $f(8,2)=4$.
|
|
|
|
\begin{center}
|
|
\begin{tikzpicture}[scale=0.9]
|
|
\node[draw, circle] (1) at (0,3) {$1$};
|
|
\node[draw, circle] (2) at (2,1) {$2$};
|
|
\node[draw, circle] (3) at (-2,1) {$4$};
|
|
\node[draw, circle] (4) at (0,1) {$5$};
|
|
\node[draw, circle] (5) at (2,-1) {$6$};
|
|
\node[draw, circle] (6) at (-3,-1) {$3$};
|
|
\node[draw, circle] (7) at (-1,-1) {$7$};
|
|
\node[draw, circle] (8) at (-1,-3) {$8$};
|
|
\path[draw,thick,-] (1) -- (2);
|
|
\path[draw,thick,-] (1) -- (3);
|
|
\path[draw,thick,-] (1) -- (4);
|
|
\path[draw,thick,-] (2) -- (5);
|
|
\path[draw,thick,-] (3) -- (6);
|
|
\path[draw,thick,-] (3) -- (7);
|
|
\path[draw,thick,-] (7) -- (8);
|
|
|
|
\path[draw=red,thick,->,line width=2pt] (8) edge [bend left] (3);
|
|
\path[draw=red,thick,->,line width=2pt] (2) edge [bend right] (1);
|
|
\end{tikzpicture}
|
|
\end{center}
|
|
|
|
An easy way to calculate the value of $f(x,k)$
|
|
is to perform a sequence of $k$ moves in the tree.
|
|
However, the time complexity of this method
|
|
is $O(n)$, because the tree may contain
|
|
a chain of $O(n)$ nodes.
|
|
|
|
Fortunately, it turns out that
|
|
using a technique similar to that
|
|
used in Chapter 16.3, any value of $f(x,k)$
|
|
can be efficiently calculated in $O(\log k)$ time
|
|
after preprocessing.
|
|
The idea is to precalculate all values $f(x,k)$
|
|
where $k$ is a power of two.
|
|
For example, the values for the tree above
|
|
are as follows:
|
|
|
|
\begin{center}
|
|
\begin{tabular}{r|rrrrrrrrr}
|
|
$x$ & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 \\
|
|
\hline
|
|
$f(x,1)$ & 0 & 1 & 4 & 1 & 1 & 2 & 4 & 7 \\
|
|
$f(x,2)$ & 0 & 0 & 1 & 0 & 0 & 1 & 1 & 4 \\
|
|
$f(x,4)$ & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
|
|
$\cdots$ \\
|
|
\end{tabular}
|
|
\end{center}
|
|
|
|
The value $0$ means that the $k$th ancestor
|
|
of a node does not exist.
|
|
|
|
The preprocessing takes $O(n \log n)$ time,
|
|
because each node can have at most $n$ ancestors.
|
|
After this, any value of $f(x,k)$ can be calculated
|
|
in $O(\log k)$ time by representing $k$
|
|
as a sum where each term is a power of two.
|
|
|
|
\section{Subtrees and paths}
|
|
|
|
\index{node array}
|
|
|
|
A \key{node 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
|
|
\begin{center}
|
|
\begin{tikzpicture}[scale=0.9]
|
|
\node[draw, circle] (1) at (0,3) {$1$};
|
|
\node[draw, circle] (2) at (-3,1) {$2$};
|
|
\node[draw, circle] (3) at (-1,1) {$3$};
|
|
\node[draw, circle] (4) at (1,1) {$4$};
|
|
\node[draw, circle] (5) at (3,1) {$5$};
|
|
\node[draw, circle] (6) at (-3,-1) {$6$};
|
|
\node[draw, circle] (7) at (-0.5,-1) {$7$};
|
|
\node[draw, circle] (8) at (1,-1) {$8$};
|
|
\node[draw, circle] (9) at (2.5,-1) {$9$};
|
|
|
|
\path[draw,thick,-] (1) -- (2);
|
|
\path[draw,thick,-] (1) -- (3);
|
|
\path[draw,thick,-] (1) -- (4);
|
|
\path[draw,thick,-] (1) -- (5);
|
|
\path[draw,thick,-] (2) -- (6);
|
|
\path[draw,thick,-] (4) -- (7);
|
|
\path[draw,thick,-] (4) -- (8);
|
|
\path[draw,thick,-] (4) -- (9);
|
|
\end{tikzpicture}
|
|
\end{center}
|
|
a depth-first search proceeds as follows:
|
|
\begin{center}
|
|
\begin{tikzpicture}[scale=0.9]
|
|
\node[draw, circle] (1) at (0,3) {$1$};
|
|
\node[draw, circle] (2) at (-3,1) {$2$};
|
|
\node[draw, circle] (3) at (-1,1) {$3$};
|
|
\node[draw, circle] (4) at (1,1) {$4$};
|
|
\node[draw, circle] (5) at (3,1) {$5$};
|
|
\node[draw, circle] (6) at (-3,-1) {$6$};
|
|
\node[draw, circle] (7) at (-0.5,-1) {$7$};
|
|
\node[draw, circle] (8) at (1,-1) {$8$};
|
|
\node[draw, circle] (9) at (2.5,-1) {$9$};
|
|
|
|
\path[draw,thick,-] (1) -- (2);
|
|
\path[draw,thick,-] (1) -- (3);
|
|
\path[draw,thick,-] (1) -- (4);
|
|
\path[draw,thick,-] (1) -- (5);
|
|
\path[draw,thick,-] (2) -- (6);
|
|
\path[draw,thick,-] (4) -- (7);
|
|
\path[draw,thick,-] (4) -- (8);
|
|
\path[draw,thick,-] (4) -- (9);
|
|
|
|
|
|
\path[draw=red,thick,->,line width=2pt] (1) edge [bend right=15] (2);
|
|
\path[draw=red,thick,->,line width=2pt] (2) edge [bend right=15] (6);
|
|
\path[draw=red,thick,->,line width=2pt] (6) edge [bend right=15] (2);
|
|
\path[draw=red,thick,->,line width=2pt] (2) edge [bend right=15] (1);
|
|
\path[draw=red,thick,->,line width=2pt] (1) edge [bend right=15] (3);
|
|
\path[draw=red,thick,->,line width=2pt] (3) edge [bend right=15] (1);
|
|
\path[draw=red,thick,->,line width=2pt] (1) edge [bend right=15] (4);
|
|
\path[draw=red,thick,->,line width=2pt] (4) edge [bend right=15] (7);
|
|
\path[draw=red,thick,->,line width=2pt] (7) edge [bend right=15] (4);
|
|
\path[draw=red,thick,->,line width=2pt] (4) edge [bend right=15] (8);
|
|
\path[draw=red,thick,->,line width=2pt] (8) edge [bend right=15] (4);
|
|
\path[draw=red,thick,->,line width=2pt] (4) edge [bend right=15] (9);
|
|
\path[draw=red,thick,->,line width=2pt] (9) edge [bend right=15] (4);
|
|
\path[draw=red,thick,->,line width=2pt] (4) edge [bend right=15] (1);
|
|
\path[draw=red,thick,->,line width=2pt] (1) edge [bend right=15] (5);
|
|
\path[draw=red,thick,->,line width=2pt] (5) edge [bend right=15] (1);
|
|
|
|
\end{tikzpicture}
|
|
\end{center}
|
|
Hence, the corresponding node array is as follows:
|
|
\begin{center}
|
|
\begin{tikzpicture}[scale=0.7]
|
|
\draw (0,0) grid (9,1);
|
|
|
|
\node at (0.5,0.5) {$1$};
|
|
\node at (1.5,0.5) {$2$};
|
|
\node at (2.5,0.5) {$6$};
|
|
\node at (3.5,0.5) {$3$};
|
|
\node at (4.5,0.5) {$4$};
|
|
\node at (5.5,0.5) {$7$};
|
|
\node at (6.5,0.5) {$8$};
|
|
\node at (7.5,0.5) {$9$};
|
|
\node at (8.5,0.5) {$5$};
|
|
|
|
\footnotesize
|
|
\node at (0.5,1.4) {$1$};
|
|
\node at (1.5,1.4) {$2$};
|
|
\node at (2.5,1.4) {$3$};
|
|
\node at (3.5,1.4) {$4$};
|
|
\node at (4.5,1.4) {$5$};
|
|
\node at (5.5,1.4) {$6$};
|
|
\node at (6.5,1.4) {$7$};
|
|
\node at (7.5,1.4) {$8$};
|
|
\node at (8.5,1.4) {$9$};
|
|
\end{tikzpicture}
|
|
\end{center}
|
|
|
|
\subsubsection{Subtree queries}
|
|
|
|
Each subtree of a tree corresponds to a subarray
|
|
in the node array,
|
|
where the first element is the root node.
|
|
For example, the following subarray contains the
|
|
nodes in the subtree of node $4$:
|
|
\begin{center}
|
|
\begin{tikzpicture}[scale=0.7]
|
|
\fill[color=lightgray] (4,0) rectangle (8,1);
|
|
\draw (0,0) grid (9,1);
|
|
|
|
\node at (0.5,0.5) {$1$};
|
|
\node at (1.5,0.5) {$2$};
|
|
\node at (2.5,0.5) {$6$};
|
|
\node at (3.5,0.5) {$3$};
|
|
\node at (4.5,0.5) {$4$};
|
|
\node at (5.5,0.5) {$7$};
|
|
\node at (6.5,0.5) {$8$};
|
|
\node at (7.5,0.5) {$9$};
|
|
\node at (8.5,0.5) {$5$};
|
|
|
|
\footnotesize
|
|
\node at (0.5,1.4) {$1$};
|
|
\node at (1.5,1.4) {$2$};
|
|
\node at (2.5,1.4) {$3$};
|
|
\node at (3.5,1.4) {$4$};
|
|
\node at (4.5,1.4) {$5$};
|
|
\node at (5.5,1.4) {$6$};
|
|
\node at (6.5,1.4) {$7$};
|
|
\node at (7.5,1.4) {$8$};
|
|
\node at (8.5,1.4) {$9$};
|
|
\end{tikzpicture}
|
|
\end{center}
|
|
Using this fact, we can efficiently process queries
|
|
that are related to subtrees of a tree.
|
|
As an example, consider a problem where each node
|
|
is assigned a value, and our task is to support
|
|
the following queries:
|
|
\begin{itemize}
|
|
\item update the value of a node
|
|
\item calculate the sum of values in the subtree of a node
|
|
\end{itemize}
|
|
|
|
Consider the following tree where the blue numbers
|
|
are the values of the nodes.
|
|
For example, the sum of the subtree of node $4$
|
|
is $3+4+3+1=11$.
|
|
|
|
\begin{center}
|
|
\begin{tikzpicture}[scale=0.9]
|
|
\node[draw, circle] (1) at (0,3) {$1$};
|
|
\node[draw, circle] (2) at (-3,1) {$2$};
|
|
\node[draw, circle] (3) at (-1,1) {$3$};
|
|
\node[draw, circle] (4) at (1,1) {$4$};
|
|
\node[draw, circle] (5) at (3,1) {$5$};
|
|
\node[draw, circle] (6) at (-3,-1) {$6$};
|
|
\node[draw, circle] (7) at (-0.5,-1) {$7$};
|
|
\node[draw, circle] (8) at (1,-1) {$8$};
|
|
\node[draw, circle] (9) at (2.5,-1) {$9$};
|
|
|
|
\path[draw,thick,-] (1) -- (2);
|
|
\path[draw,thick,-] (1) -- (3);
|
|
\path[draw,thick,-] (1) -- (4);
|
|
\path[draw,thick,-] (1) -- (5);
|
|
\path[draw,thick,-] (2) -- (6);
|
|
\path[draw,thick,-] (4) -- (7);
|
|
\path[draw,thick,-] (4) -- (8);
|
|
\path[draw,thick,-] (4) -- (9);
|
|
|
|
\node[color=blue] at (0,3+0.65) {2};
|
|
\node[color=blue] at (-3-0.65,1) {3};
|
|
\node[color=blue] at (-1-0.65,1) {5};
|
|
\node[color=blue] at (1+0.65,1) {3};
|
|
\node[color=blue] at (3+0.65,1) {1};
|
|
\node[color=blue] at (-3,-1-0.65) {4};
|
|
\node[color=blue] at (-0.5,-1-0.65) {4};
|
|
\node[color=blue] at (1,-1-0.65) {3};
|
|
\node[color=blue] at (2.5,-1-0.65) {1};
|
|
\end{tikzpicture}
|
|
\end{center}
|
|
|
|
The idea is to construct a node 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:
|
|
|
|
\begin{center}
|
|
\begin{tikzpicture}[scale=0.7]
|
|
\draw (0,1) grid (9,-2);
|
|
|
|
\node at (0.5,0.5) {$1$};
|
|
\node at (1.5,0.5) {$2$};
|
|
\node at (2.5,0.5) {$6$};
|
|
\node at (3.5,0.5) {$3$};
|
|
\node at (4.5,0.5) {$4$};
|
|
\node at (5.5,0.5) {$7$};
|
|
\node at (6.5,0.5) {$8$};
|
|
\node at (7.5,0.5) {$9$};
|
|
\node at (8.5,0.5) {$5$};
|
|
|
|
\node at (0.5,-0.5) {$9$};
|
|
\node at (1.5,-0.5) {$2$};
|
|
\node at (2.5,-0.5) {$1$};
|
|
\node at (3.5,-0.5) {$1$};
|
|
\node at (4.5,-0.5) {$4$};
|
|
\node at (5.5,-0.5) {$1$};
|
|
\node at (6.5,-0.5) {$1$};
|
|
\node at (7.5,-0.5) {$1$};
|
|
\node at (8.5,-0.5) {$1$};
|
|
|
|
\node at (0.5,-1.5) {$2$};
|
|
\node at (1.5,-1.5) {$3$};
|
|
\node at (2.5,-1.5) {$4$};
|
|
\node at (3.5,-1.5) {$5$};
|
|
\node at (4.5,-1.5) {$3$};
|
|
\node at (5.5,-1.5) {$4$};
|
|
\node at (6.5,-1.5) {$3$};
|
|
\node at (7.5,-1.5) {$1$};
|
|
\node at (8.5,-1.5) {$1$};
|
|
|
|
\footnotesize
|
|
\node at (0.5,1.4) {$1$};
|
|
\node at (1.5,1.4) {$2$};
|
|
\node at (2.5,1.4) {$3$};
|
|
\node at (3.5,1.4) {$4$};
|
|
\node at (4.5,1.4) {$5$};
|
|
\node at (5.5,1.4) {$6$};
|
|
\node at (6.5,1.4) {$7$};
|
|
\node at (7.5,1.4) {$8$};
|
|
\node at (8.5,1.4) {$9$};
|
|
\end{tikzpicture}
|
|
\end{center}
|
|
|
|
Using this array, we can calculate the sum of values
|
|
in any subtree by first finding out the size of the subtree
|
|
and then the values of the corresponding nodes.
|
|
For example, the values in the subtree of node $4$
|
|
can be found as follows:
|
|
|
|
\begin{center}
|
|
\begin{tikzpicture}[scale=0.7]
|
|
\fill[color=lightgray] (4,1) rectangle (5,0);
|
|
\fill[color=lightgray] (4,0) rectangle (5,-1);
|
|
\fill[color=lightgray] (4,-1) rectangle (8,-2);
|
|
\draw (0,1) grid (9,-2);
|
|
|
|
\node at (0.5,0.5) {$1$};
|
|
\node at (1.5,0.5) {$2$};
|
|
\node at (2.5,0.5) {$6$};
|
|
\node at (3.5,0.5) {$3$};
|
|
\node at (4.5,0.5) {$4$};
|
|
\node at (5.5,0.5) {$7$};
|
|
\node at (6.5,0.5) {$8$};
|
|
\node at (7.5,0.5) {$9$};
|
|
\node at (8.5,0.5) {$5$};
|
|
|
|
\node at (0.5,-0.5) {$9$};
|
|
\node at (1.5,-0.5) {$2$};
|
|
\node at (2.5,-0.5) {$1$};
|
|
\node at (3.5,-0.5) {$1$};
|
|
\node at (4.5,-0.5) {$4$};
|
|
\node at (5.5,-0.5) {$1$};
|
|
\node at (6.5,-0.5) {$1$};
|
|
\node at (7.5,-0.5) {$1$};
|
|
\node at (8.5,-0.5) {$1$};
|
|
|
|
\node at (0.5,-1.5) {$2$};
|
|
\node at (1.5,-1.5) {$3$};
|
|
\node at (2.5,-1.5) {$4$};
|
|
\node at (3.5,-1.5) {$5$};
|
|
\node at (4.5,-1.5) {$3$};
|
|
\node at (5.5,-1.5) {$4$};
|
|
\node at (6.5,-1.5) {$3$};
|
|
\node at (7.5,-1.5) {$1$};
|
|
\node at (8.5,-1.5) {$1$};
|
|
|
|
\footnotesize
|
|
\node at (0.5,1.4) {$1$};
|
|
\node at (1.5,1.4) {$2$};
|
|
\node at (2.5,1.4) {$3$};
|
|
\node at (3.5,1.4) {$4$};
|
|
\node at (4.5,1.4) {$5$};
|
|
\node at (5.5,1.4) {$6$};
|
|
\node at (6.5,1.4) {$7$};
|
|
\node at (7.5,1.4) {$8$};
|
|
\node at (8.5,1.4) {$9$};
|
|
\end{tikzpicture}
|
|
\end{center}
|
|
|
|
To support the queries efficiently,
|
|
it suffices to store the values of the
|
|
nodes in a binary indexed tree or segment tree.
|
|
After this, we can both update a value
|
|
and calculate the sum of values in $O(\log n)$ time.
|
|
|
|
\subsubsection{Path queries}
|
|
|
|
Using a node array, we can also efficiently
|
|
calculate sums of values on
|
|
paths between the root node and any other
|
|
node in the tree.
|
|
Let us next consider a problem where our task
|
|
is to support the following queries:
|
|
\begin{itemize}
|
|
\item change the value of a node
|
|
\item calculate the sum of values on a path between
|
|
the root node and a node
|
|
\end{itemize}
|
|
|
|
For example, in the following tree, the sum of
|
|
values from the root to node 8 is $4+5+3=12$.
|
|
|
|
\begin{center}
|
|
\begin{tikzpicture}[scale=0.9]
|
|
\node[draw, circle] (1) at (0,3) {$1$};
|
|
\node[draw, circle] (2) at (-3,1) {$2$};
|
|
\node[draw, circle] (3) at (-1,1) {$3$};
|
|
\node[draw, circle] (4) at (1,1) {$4$};
|
|
\node[draw, circle] (5) at (3,1) {$5$};
|
|
\node[draw, circle] (6) at (-3,-1) {$6$};
|
|
\node[draw, circle] (7) at (-0.5,-1) {$7$};
|
|
\node[draw, circle] (8) at (1,-1) {$8$};
|
|
\node[draw, circle] (9) at (2.5,-1) {$9$};
|
|
|
|
\path[draw,thick,-] (1) -- (2);
|
|
\path[draw,thick,-] (1) -- (3);
|
|
\path[draw,thick,-] (1) -- (4);
|
|
\path[draw,thick,-] (1) -- (5);
|
|
\path[draw,thick,-] (2) -- (6);
|
|
\path[draw,thick,-] (4) -- (7);
|
|
\path[draw,thick,-] (4) -- (8);
|
|
\path[draw,thick,-] (4) -- (9);
|
|
|
|
\node[color=blue] at (0,3+0.65) {4};
|
|
\node[color=blue] at (-3-0.65,1) {5};
|
|
\node[color=blue] at (-1-0.65,1) {3};
|
|
\node[color=blue] at (1+0.65,1) {5};
|
|
\node[color=blue] at (3+0.65,1) {2};
|
|
\node[color=blue] at (-3,-1-0.65) {3};
|
|
\node[color=blue] at (-0.5,-1-0.65) {5};
|
|
\node[color=blue] at (1,-1-0.65) {3};
|
|
\node[color=blue] at (2.5,-1-0.65) {1};
|
|
\end{tikzpicture}
|
|
\end{center}
|
|
|
|
To solve this problem, we can use a similar
|
|
technique as we used for subtree queries,
|
|
but the values of the nodes are stored
|
|
in a special way:
|
|
if the value of a node at position $k$
|
|
increases by $a$,
|
|
the value at position $k$ increases by $a$
|
|
and the value at position $k+c$ decreases by $a$,
|
|
where $c$ is the size of the subtree.
|
|
|
|
\begin{samepage}
|
|
For example, the following array corresponds to the above tree:
|
|
\begin{center}
|
|
\begin{tikzpicture}[scale=0.7]
|
|
\draw (0,1) grid (10,-2);
|
|
|
|
\node at (0.5,0.5) {$1$};
|
|
\node at (1.5,0.5) {$2$};
|
|
\node at (2.5,0.5) {$6$};
|
|
\node at (3.5,0.5) {$3$};
|
|
\node at (4.5,0.5) {$4$};
|
|
\node at (5.5,0.5) {$7$};
|
|
\node at (6.5,0.5) {$8$};
|
|
\node at (7.5,0.5) {$9$};
|
|
\node at (8.5,0.5) {$5$};
|
|
\node at (9.5,0.5) {--};
|
|
|
|
\node at (0.5,-0.5) {$9$};
|
|
\node at (1.5,-0.5) {$2$};
|
|
\node at (2.5,-0.5) {$1$};
|
|
\node at (3.5,-0.5) {$1$};
|
|
\node at (4.5,-0.5) {$4$};
|
|
\node at (5.5,-0.5) {$1$};
|
|
\node at (6.5,-0.5) {$1$};
|
|
\node at (7.5,-0.5) {$1$};
|
|
\node at (8.5,-0.5) {$1$};
|
|
\node at (9.5,-0.5) {--};
|
|
|
|
\node at (0.5,-1.5) {$4$};
|
|
\node at (1.5,-1.5) {$5$};
|
|
\node at (2.5,-1.5) {$3$};
|
|
\node at (3.5,-1.5) {$-5$};
|
|
\node at (4.5,-1.5) {$2$};
|
|
\node at (5.5,-1.5) {$5$};
|
|
\node at (6.5,-1.5) {$-2$};
|
|
\node at (7.5,-1.5) {$-2$};
|
|
\node at (8.5,-1.5) {$-4$};
|
|
\node at (9.5,-1.5) {$-4$};
|
|
|
|
\footnotesize
|
|
\node at (0.5,1.4) {$1$};
|
|
\node at (1.5,1.4) {$2$};
|
|
\node at (2.5,1.4) {$3$};
|
|
\node at (3.5,1.4) {$4$};
|
|
\node at (4.5,1.4) {$5$};
|
|
\node at (5.5,1.4) {$6$};
|
|
\node at (6.5,1.4) {$7$};
|
|
\node at (7.5,1.4) {$8$};
|
|
\node at (8.5,1.4) {$9$};
|
|
\node at (9.5,1.4) {$10$};
|
|
\end{tikzpicture}
|
|
\end{center}
|
|
\end{samepage}
|
|
|
|
For example, the value of node $3$ is $-5$,
|
|
because it is the next node after the subtrees
|
|
of nodes $2$ and $6$ and its own value is $3$.
|
|
So the value decreases by $5+3$ and increases by $3$.
|
|
Note that the array contains an extra index 10
|
|
that only has the opposite number of the value
|
|
of the root node.
|
|
|
|
Using this array, the sum of values in a path
|
|
from the root to node $x$ equals the sum
|
|
of values in the array from the beginning to node $x$.
|
|
For example, the sum from the root to node $8$
|
|
can be calculated as follows:
|
|
|
|
\begin{center}
|
|
\begin{tikzpicture}[scale=0.7]
|
|
\fill[color=lightgray] (6,1) rectangle (7,0);
|
|
\fill[color=lightgray] (0,-1) rectangle (7,-2);
|
|
\draw (0,1) grid (10,-2);
|
|
|
|
\node at (0.5,0.5) {$1$};
|
|
\node at (1.5,0.5) {$2$};
|
|
\node at (2.5,0.5) {$6$};
|
|
\node at (3.5,0.5) {$3$};
|
|
\node at (4.5,0.5) {$4$};
|
|
\node at (5.5,0.5) {$7$};
|
|
\node at (6.5,0.5) {$8$};
|
|
\node at (7.5,0.5) {$9$};
|
|
\node at (8.5,0.5) {$5$};
|
|
\node at (9.5,0.5) {--};
|
|
|
|
\node at (0.5,-0.5) {$9$};
|
|
\node at (1.5,-0.5) {$2$};
|
|
\node at (2.5,-0.5) {$1$};
|
|
\node at (3.5,-0.5) {$1$};
|
|
\node at (4.5,-0.5) {$4$};
|
|
\node at (5.5,-0.5) {$1$};
|
|
\node at (6.5,-0.5) {$1$};
|
|
\node at (7.5,-0.5) {$1$};
|
|
\node at (8.5,-0.5) {$1$};
|
|
\node at (9.5,-0.5) {--};
|
|
|
|
\node at (0.5,-1.5) {$4$};
|
|
\node at (1.5,-1.5) {$5$};
|
|
\node at (2.5,-1.5) {$3$};
|
|
\node at (3.5,-1.5) {$-5$};
|
|
\node at (4.5,-1.5) {$2$};
|
|
\node at (5.5,-1.5) {$5$};
|
|
\node at (6.5,-1.5) {$-2$};
|
|
\node at (7.5,-1.5) {$-2$};
|
|
\node at (8.5,-1.5) {$-4$};
|
|
\node at (9.5,-1.5) {$-4$};
|
|
|
|
\footnotesize
|
|
\node at (0.5,1.4) {$1$};
|
|
\node at (1.5,1.4) {$2$};
|
|
\node at (2.5,1.4) {$3$};
|
|
\node at (3.5,1.4) {$4$};
|
|
\node at (4.5,1.4) {$5$};
|
|
\node at (5.5,1.4) {$6$};
|
|
\node at (6.5,1.4) {$7$};
|
|
\node at (7.5,1.4) {$8$};
|
|
\node at (8.5,1.4) {$9$};
|
|
\node at (9.5,1.4) {$10$};
|
|
\end{tikzpicture}
|
|
\end{center}
|
|
The sum is
|
|
\[4+5+3-5+2+5-2=12\]
|
|
that equals the sum $4+5+3=12$.
|
|
This method works, because the value of each node
|
|
is added to the sum when the depth-first search
|
|
visits the node for the first time, and the value
|
|
of the node is removed from the sum when the subtree of the
|
|
node has been processed.
|
|
|
|
Once again, we can store the values of the nodes
|
|
in a binary indexed tree or a segment tree,
|
|
so it is possible to both update a value and
|
|
calculate the sum of values efficiently in $O(\log n)$ time.
|
|
|
|
\section{Lowest common ancestor}
|
|
|
|
\index{lowest common ancestor}
|
|
|
|
The \key{lowest common ancestor}
|
|
of two nodes in the tree is the lowest node
|
|
whose subtree contains both the nodes.
|
|
A typical problem is to efficiently process
|
|
queries that ask to find the lowest
|
|
common ancestor of given two nodes.
|
|
For example, in the tree
|
|
\begin{center}
|
|
\begin{tikzpicture}[scale=0.9]
|
|
\node[draw, circle] (1) at (0,3) {$1$};
|
|
\node[draw, circle] (2) at (2,1) {$4$};
|
|
\node[draw, circle] (3) at (-2,1) {$2$};
|
|
\node[draw, circle] (4) at (0,1) {$3$};
|
|
\node[draw, circle] (5) at (2,-1) {$7$};
|
|
\node[draw, circle] (6) at (-3,-1) {$5$};
|
|
\node[draw, circle] (7) at (-1,-1) {$6$};
|
|
\node[draw, circle] (8) at (-1,-3) {$8$};
|
|
\path[draw,thick,-] (1) -- (2);
|
|
\path[draw,thick,-] (1) -- (3);
|
|
\path[draw,thick,-] (1) -- (4);
|
|
\path[draw,thick,-] (2) -- (5);
|
|
\path[draw,thick,-] (3) -- (6);
|
|
\path[draw,thick,-] (3) -- (7);
|
|
\path[draw,thick,-] (7) -- (8);
|
|
\end{tikzpicture}
|
|
\end{center}
|
|
the lowest common ancestor of nodes 5 and 8 is node 2,
|
|
and the lowest common ancestor of nodes 3 and 4 is node 1.
|
|
|
|
Next we will discuss two efficient techniques for
|
|
finding the lowest common ancestor of two nodes.
|
|
|
|
\subsubsection{Method 1}
|
|
|
|
One way to solve the problem is to use the fact
|
|
that we can efficiently find the $k$th
|
|
ancestor of any node in the tree.
|
|
Thus, we can first make sure that
|
|
both nodes are at the same level in the tree,
|
|
and then find the smallest value of $k$
|
|
such that the $k$th ancestor of both nodes is the same.
|
|
|
|
As an example, let us find the lowest common
|
|
ancestor of nodes $5$ and $8$ in the following tree:
|
|
|
|
\begin{center}
|
|
\begin{tikzpicture}[scale=0.9]
|
|
\node[draw, circle] (1) at (0,3) {$1$};
|
|
\node[draw, circle] (2) at (2,1) {$4$};
|
|
\node[draw, circle] (3) at (-2,1) {$2$};
|
|
\node[draw, circle] (4) at (0,1) {$3$};
|
|
\node[draw, circle] (5) at (2,-1) {$7$};
|
|
\node[draw, circle,fill=lightgray] (6) at (-3,-1) {$5$};
|
|
\node[draw, circle] (7) at (-1,-1) {$6$};
|
|
\node[draw, circle,fill=lightgray] (8) at (-1,-3) {$8$};
|
|
\path[draw,thick,-] (1) -- (2);
|
|
\path[draw,thick,-] (1) -- (3);
|
|
\path[draw,thick,-] (1) -- (4);
|
|
\path[draw,thick,-] (2) -- (5);
|
|
\path[draw,thick,-] (3) -- (6);
|
|
\path[draw,thick,-] (3) -- (7);
|
|
\path[draw,thick,-] (7) -- (8);
|
|
\end{tikzpicture}
|
|
\end{center}
|
|
|
|
Node $5$ is at level $3$, while node $8$ is at level $4$.
|
|
Thus, we first move one step upwards from node $8$ to node $6$.
|
|
After this, it turns out that the parent of both nodes $5$
|
|
and $6$ is node $2$, so we have found the lowest common ancestor.
|
|
|
|
\begin{center}
|
|
\begin{tikzpicture}[scale=0.9]
|
|
\node[draw, circle] (1) at (0,3) {$1$};
|
|
\node[draw, circle] (2) at (2,1) {$4$};
|
|
\node[draw, circle] (3) at (-2,1) {$2$};
|
|
\node[draw, circle] (4) at (0,1) {$3$};
|
|
\node[draw, circle] (5) at (2,-1) {$7$};
|
|
\node[draw, circle,fill=lightgray] (6) at (-3,-1) {$5$};
|
|
\node[draw, circle] (7) at (-1,-1) {$6$};
|
|
\node[draw, circle,fill=lightgray] (8) at (-1,-3) {$8$};
|
|
\path[draw,thick,-] (1) -- (2);
|
|
\path[draw,thick,-] (1) -- (3);
|
|
\path[draw,thick,-] (1) -- (4);
|
|
\path[draw,thick,-] (2) -- (5);
|
|
\path[draw,thick,-] (3) -- (6);
|
|
\path[draw,thick,-] (3) -- (7);
|
|
\path[draw,thick,-] (7) -- (8);
|
|
|
|
\path[draw=red,thick,->,line width=2pt] (6) edge [bend left] (3);
|
|
\path[draw=red,thick,->,line width=2pt] (8) edge [bend right] (7);
|
|
\path[draw=red,thick,->,line width=2pt] (7) edge [bend right] (3);
|
|
\end{tikzpicture}
|
|
\end{center}
|
|
|
|
Using this method, we can find the lowest common ancestor
|
|
of any two nodes in $O(\log n)$ time after an $O(n \log n)$ time
|
|
preprocessing, because both steps can be
|
|
performed in $O(\log n)$ time.
|
|
|
|
\subsubsection{Method 2}
|
|
|
|
Another way to solve the problem is based on
|
|
a node array.
|
|
Once again, the idea is to traverse the nodes
|
|
using a depth-first search:
|
|
|
|
\begin{center}
|
|
\begin{tikzpicture}[scale=0.9]
|
|
\node[draw, circle] (1) at (0,3) {$1$};
|
|
\node[draw, circle] (2) at (2,1) {$4$};
|
|
\node[draw, circle] (3) at (-2,1) {$2$};
|
|
\node[draw, circle] (4) at (0,1) {$3$};
|
|
\node[draw, circle] (5) at (2,-1) {$7$};
|
|
\node[draw, circle] (6) at (-3,-1) {$5$};
|
|
\node[draw, circle] (7) at (-1,-1) {$6$};
|
|
\node[draw, circle] (8) at (-1,-3) {$8$};
|
|
\path[draw,thick,-] (1) -- (2);
|
|
\path[draw,thick,-] (1) -- (3);
|
|
\path[draw,thick,-] (1) -- (4);
|
|
\path[draw,thick,-] (2) -- (5);
|
|
\path[draw,thick,-] (3) -- (6);
|
|
\path[draw,thick,-] (3) -- (7);
|
|
\path[draw,thick,-] (7) -- (8);
|
|
|
|
\path[draw=red,thick,->,line width=2pt] (1) edge [bend right=15] (3);
|
|
\path[draw=red,thick,->,line width=2pt] (3) edge [bend right=15] (6);
|
|
\path[draw=red,thick,->,line width=2pt] (6) edge [bend right=15] (3);
|
|
\path[draw=red,thick,->,line width=2pt] (3) edge [bend right=15] (7);
|
|
\path[draw=red,thick,->,line width=2pt] (7) edge [bend right=15] (8);
|
|
\path[draw=red,thick,->,line width=2pt] (8) edge [bend right=15] (7);
|
|
\path[draw=red,thick,->,line width=2pt] (7) edge [bend right=15] (3);
|
|
\path[draw=red,thick,->,line width=2pt] (3) edge [bend right=15] (1);
|
|
\path[draw=red,thick,->,line width=2pt] (1) edge [bend right=15] (4);
|
|
\path[draw=red,thick,->,line width=2pt] (4) edge [bend right=15] (1);
|
|
\path[draw=red,thick,->,line width=2pt] (1) edge [bend right=15] (2);
|
|
\path[draw=red,thick,->,line width=2pt] (2) edge [bend right=15] (5);
|
|
\path[draw=red,thick,->,line width=2pt] (5) edge [bend right=15] (2);
|
|
\path[draw=red,thick,->,line width=2pt] (2) edge [bend right=15] (1);
|
|
\end{tikzpicture}
|
|
\end{center}
|
|
|
|
However, we add each node to the node 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 $2n-1$
|
|
nodes in the array.
|
|
|
|
We store two values in the array:
|
|
(1) the identifier of the node, and (2) the level of the
|
|
node in the tree.
|
|
The following array corresponds to the above tree:
|
|
|
|
\begin{center}
|
|
\begin{tikzpicture}[scale=0.7]
|
|
|
|
\draw (0,1) grid (15,2);
|
|
%\node at (-1.1,1.5) {\texttt{node}};
|
|
\node at (0.5,1.5) {$1$};
|
|
\node at (1.5,1.5) {$2$};
|
|
\node at (2.5,1.5) {$5$};
|
|
\node at (3.5,1.5) {$2$};
|
|
\node at (4.5,1.5) {$6$};
|
|
\node at (5.5,1.5) {$8$};
|
|
\node at (6.5,1.5) {$6$};
|
|
\node at (7.5,1.5) {$2$};
|
|
\node at (8.5,1.5) {$1$};
|
|
\node at (9.5,1.5) {$3$};
|
|
\node at (10.5,1.5) {$1$};
|
|
\node at (11.5,1.5) {$4$};
|
|
\node at (12.5,1.5) {$7$};
|
|
\node at (13.5,1.5) {$4$};
|
|
\node at (14.5,1.5) {$1$};
|
|
|
|
|
|
\draw (0,0) grid (15,1);
|
|
%\node at (-1.1,0.5) {\texttt{depth}};
|
|
\node at (0.5,0.5) {$1$};
|
|
\node at (1.5,0.5) {$2$};
|
|
\node at (2.5,0.5) {$3$};
|
|
\node at (3.5,0.5) {$2$};
|
|
\node at (4.5,0.5) {$3$};
|
|
\node at (5.5,0.5) {$4$};
|
|
\node at (6.5,0.5) {$3$};
|
|
\node at (7.5,0.5) {$2$};
|
|
\node at (8.5,0.5) {$1$};
|
|
\node at (9.5,0.5) {$2$};
|
|
\node at (10.5,0.5) {$1$};
|
|
\node at (11.5,0.5) {$2$};
|
|
\node at (12.5,0.5) {$3$};
|
|
\node at (13.5,0.5) {$2$};
|
|
\node at (14.5,0.5) {$1$};
|
|
|
|
\footnotesize
|
|
\node at (0.5,2.5) {$1$};
|
|
\node at (1.5,2.5) {$2$};
|
|
\node at (2.5,2.5) {$3$};
|
|
\node at (3.5,2.5) {$4$};
|
|
\node at (4.5,2.5) {$5$};
|
|
\node at (5.5,2.5) {$6$};
|
|
\node at (6.5,2.5) {$7$};
|
|
\node at (7.5,2.5) {$8$};
|
|
\node at (8.5,2.5) {$9$};
|
|
\node at (9.5,2.5) {$10$};
|
|
\node at (10.5,2.5) {$11$};
|
|
\node at (11.5,2.5) {$12$};
|
|
\node at (12.5,2.5) {$13$};
|
|
\node at (13.5,2.5) {$14$};
|
|
\node at (14.5,2.5) {$15$};
|
|
\end{tikzpicture}
|
|
\end{center}
|
|
|
|
Using this array, we can find the lowest common ancestor
|
|
of nodes $a$ and $b$ by finding the node with lowest level
|
|
between nodes $a$ and $b$ in the array.
|
|
For example, the lowest common ancestor of nodes $5$ and $8$
|
|
can be found as follows:
|
|
|
|
\begin{center}
|
|
\begin{tikzpicture}[scale=0.7]
|
|
\fill[color=lightgray] (2,1) rectangle (3,2);
|
|
\fill[color=lightgray] (5,1) rectangle (6,2);
|
|
\fill[color=lightgray] (2,0) rectangle (6,1);
|
|
|
|
\node at (3.5,-0.5) {$\uparrow$};
|
|
|
|
\draw (0,1) grid (15,2);
|
|
\node at (0.5,1.5) {$1$};
|
|
\node at (1.5,1.5) {$2$};
|
|
\node at (2.5,1.5) {$5$};
|
|
\node at (3.5,1.5) {$2$};
|
|
\node at (4.5,1.5) {$6$};
|
|
\node at (5.5,1.5) {$8$};
|
|
\node at (6.5,1.5) {$6$};
|
|
\node at (7.5,1.5) {$2$};
|
|
\node at (8.5,1.5) {$1$};
|
|
\node at (9.5,1.5) {$3$};
|
|
\node at (10.5,1.5) {$1$};
|
|
\node at (11.5,1.5) {$4$};
|
|
\node at (12.5,1.5) {$7$};
|
|
\node at (13.5,1.5) {$4$};
|
|
\node at (14.5,1.5) {$1$};
|
|
|
|
|
|
\draw (0,0) grid (15,1);
|
|
\node at (0.5,0.5) {$1$};
|
|
\node at (1.5,0.5) {$2$};
|
|
\node at (2.5,0.5) {$3$};
|
|
\node at (3.5,0.5) {$2$};
|
|
\node at (4.5,0.5) {$3$};
|
|
\node at (5.5,0.5) {$4$};
|
|
\node at (6.5,0.5) {$3$};
|
|
\node at (7.5,0.5) {$2$};
|
|
\node at (8.5,0.5) {$1$};
|
|
\node at (9.5,0.5) {$2$};
|
|
\node at (10.5,0.5) {$1$};
|
|
\node at (11.5,0.5) {$2$};
|
|
\node at (12.5,0.5) {$3$};
|
|
\node at (13.5,0.5) {$2$};
|
|
\node at (14.5,0.5) {$1$};
|
|
|
|
\footnotesize
|
|
\node at (0.5,2.5) {$1$};
|
|
\node at (1.5,2.5) {$2$};
|
|
\node at (2.5,2.5) {$3$};
|
|
\node at (3.5,2.5) {$4$};
|
|
\node at (4.5,2.5) {$5$};
|
|
\node at (5.5,2.5) {$6$};
|
|
\node at (6.5,2.5) {$7$};
|
|
\node at (7.5,2.5) {$8$};
|
|
\node at (8.5,2.5) {$9$};
|
|
\node at (9.5,2.5) {$10$};
|
|
\node at (10.5,2.5) {$11$};
|
|
\node at (11.5,2.5) {$12$};
|
|
\node at (12.5,2.5) {$13$};
|
|
\node at (13.5,2.5) {$14$};
|
|
\node at (14.5,2.5) {$15$};
|
|
\end{tikzpicture}
|
|
\end{center}
|
|
|
|
Node 5 is at position 3, node 8 is at position 6,
|
|
and the node with lowest level between
|
|
positions $3 \ldots 6$ is node 2 at position 4
|
|
whose level is 2.
|
|
Thus, the lowest common ancestor of
|
|
nodes 5 and 8 is node 2.
|
|
|
|
Using a segment tree, we can find the lowest
|
|
common ancestor in $O(\log n)$ time.
|
|
Since the array is static, the time complexity
|
|
$O(1)$ is also possible, but this is rarely needed.
|
|
In both cases, the preprocessing takes $O(n \log n)$ time.
|
|
|
|
\subsubsection{Distances of nodes}
|
|
|
|
Finally, let us consider a problem of
|
|
finding the distance between
|
|
two nodes in the tree, which equals
|
|
the length of the path between them.
|
|
It turns out that this problem reduces to
|
|
finding the lowest common ancestor of the nodes.
|
|
|
|
First, we choose an arbitrary node for the
|
|
root of the tree.
|
|
After this, the distance between nodes $a$ and $b$
|
|
is $d(a)+d(b)-2 \cdot d(c)$,
|
|
where $c$ is the lowest common ancestor of $a$ and $b$
|
|
and $d(s)$ denotes the distance from the root node
|
|
to node $s$.
|
|
For example, in the tree
|
|
|
|
\begin{center}
|
|
\begin{tikzpicture}[scale=0.9]
|
|
\node[draw, circle] (1) at (0,3) {$1$};
|
|
\node[draw, circle] (2) at (2,1) {$4$};
|
|
\node[draw, circle] (3) at (-2,1) {$2$};
|
|
\node[draw, circle] (4) at (0,1) {$3$};
|
|
\node[draw, circle] (5) at (2,-1) {$7$};
|
|
\node[draw, circle] (6) at (-3,-1) {$5$};
|
|
\node[draw, circle] (7) at (-1,-1) {$6$};
|
|
\node[draw, circle] (8) at (-1,-3) {$8$};
|
|
\path[draw,thick,-] (1) -- (2);
|
|
\path[draw,thick,-] (1) -- (3);
|
|
\path[draw,thick,-] (1) -- (4);
|
|
\path[draw,thick,-] (2) -- (5);
|
|
\path[draw,thick,-] (3) -- (6);
|
|
\path[draw,thick,-] (3) -- (7);
|
|
\path[draw,thick,-] (7) -- (8);
|
|
|
|
\path[draw=red,thick,-,line width=2pt] (8) -- node[font=\small] {} (7);
|
|
\path[draw=red,thick,-,line width=2pt] (7) -- node[font=\small] {} (3);
|
|
\path[draw=red,thick,-,line width=2pt] (6) -- node[font=\small] {} (3);
|
|
\end{tikzpicture}
|
|
\end{center}
|
|
the lowest common ancestor of nodes 5 and 8 is node 2.
|
|
A path from node 5 to node 8
|
|
goes first upwards from node 5 to node 2,
|
|
and then downwards from node 2 to node 8.
|
|
The distances of the nodes from the root are
|
|
$d(5)=3$, $d(8)=4$ and $d(2)=2$,
|
|
so the distance between nodes 5 and 8 is
|
|
$3+4-2\cdot2=3$.
|
|
|
|
|