\chapter{Segment trees revisited}

\index{segment tree}

A segment tree is a versatile data structure
that can be used in many different situations.
However, there are many topics related to segment trees
that we haven't touched yet.
Now it's time to learn some more advanced variations
of segment trees and see their full potential.

So far, we have implemented the operations
of a segment tree by walking \emph{from the bottom to the top},
from the leaves to the root.
For example, we have calculated the sum of a range $[a,b]$
as follows (Chapter 9.3):

\begin{lstlisting}
int sum(int a, int b) {
    a += N; b += N;
    int s = 0;
    while (a <= b) {
        if (a%2 == 1) s += p[a++];
        if (b%2 == 0) s += p[b--];
        a /= 2; b /= 2;
    }
    return s;
}
\end{lstlisting}
However, in more advanced segment trees,
it's beneficial to implement the operations
in another way, \emph{from the top to the bottom},
from the root to the leaves.
Using this approach, the function becomes as follows:

\begin{lstlisting}
int sum(int a, int b, int k, int x, int y) {
    if (b < x || a > y) return 0;
    if (a == x && b == y) return p[k];
    int d = (y-x+1)/2;
    return sum(a, min(x+d-1,b), 2*k, x, x+d-1) +
           sum(max(x+d,a), b, 2*k+1, x+d, y);
}
\end{lstlisting}
Now we can calulate the sum of the range $[a,b]$
as follows:

\begin{lstlisting}
int s = sum(a, b, 1, 0, N-1);
\end{lstlisting}
The parameter $k$ is the current position
in array \texttt{p}.
Initially $k$ equals 1, because we begin
at the root of the segment tree.
The range $[x,y]$ corresponds to $k$,
and is initially $[0,N-1]$.
If $[a,b]$ is outside $[x,y]$,
the sum of the range is 0,
and if $[a,b]$ equals $[x,y]$,
the sum can be found in array \texttt{p}.
If $[a,b]$ is completely or partially inside $[x,y]$,
the search continues recursively to the
left and right half of $[x,y]$.
The size of both halves is $d=\frac{1}{2}(y-x+1)$;
the left half is $[x,x+d-1]$
and the right half is $[x+d,y]$.

The following picture shows how the search proceeds
when calculating the sum of the marked elements.
The gray nodes indicate nodes where the recursion
stops and the sum of the range can be found in array \texttt{p}.
\\
\begin{center}
\begin{tikzpicture}[scale=0.7]
\fill[color=gray!50] (5,0) rectangle (6,1);
\draw (0,0) grid (16,1);

\node[anchor=center] at (0.5, 0.5) {5};
\node[anchor=center] at (1.5, 0.5) {8};
\node[anchor=center] at (2.5, 0.5) {6};
\node[anchor=center] at (3.5, 0.5) {3};
\node[anchor=center] at (4.5, 0.5) {2};
\node[anchor=center] at (5.5, 0.5) {7};
\node[anchor=center] at (6.5, 0.5) {2};
\node[anchor=center] at (7.5, 0.5) {6};
\node[anchor=center] at (8.5, 0.5) {7};
\node[anchor=center] at (9.5, 0.5) {1};
\node[anchor=center] at (10.5, 0.5) {7};
\node[anchor=center] at (11.5, 0.5) {5};
\node[anchor=center] at (12.5, 0.5) {6};
\node[anchor=center] at (13.5, 0.5) {2};
\node[anchor=center] at (14.5, 0.5) {3};
\node[anchor=center] at (15.5, 0.5) {2};

%\node[anchor=center] at (1,2.5) {13};

\node[draw, circle] (a) at (1,2.5) {13};
\path[draw,thick,-] (a) -- (0.5,1);
\path[draw,thick,-] (a) -- (1.5,1);
\node[draw, circle,minimum size=22pt] (b) at (3,2.5) {9};
\path[draw,thick,-] (b) -- (2.5,1);
\path[draw,thick,-] (b) -- (3.5,1);
\node[draw, circle,minimum size=22pt] (c) at (5,2.5) {9};
\path[draw,thick,-] (c) -- (4.5,1);
\path[draw,thick,-] (c) -- (5.5,1);
\node[draw, circle,fill=gray!50,minimum size=22pt] (d) at (7,2.5) {8};
\path[draw,thick,-] (d) -- (6.5,1);
\path[draw,thick,-] (d) -- (7.5,1);
\node[draw, circle,minimum size=22pt] (e) at (9,2.5) {8};
\path[draw,thick,-] (e) -- (8.5,1);
\path[draw,thick,-] (e) -- (9.5,1);
\node[draw, circle] (f) at (11,2.5) {12};
\path[draw,thick,-] (f) -- (10.5,1);
\path[draw,thick,-] (f) -- (11.5,1);
\node[draw, circle,fill=gray!50,minimum size=22pt] (g) at (13,2.5) {8};
\path[draw,thick,-] (g) -- (12.5,1);
\path[draw,thick,-] (g) -- (13.5,1);
\node[draw, circle,minimum size=22pt] (h) at (15,2.5) {5};
\path[draw,thick,-] (h) -- (14.5,1);
\path[draw,thick,-] (h) -- (15.5,1);

\node[draw, circle] (i) at (2,4.5) {22};
\path[draw,thick,-] (i) -- (a);
\path[draw,thick,-] (i) -- (b);
\node[draw, circle] (j) at (6,4.5) {17};
\path[draw,thick,-] (j) -- (c);
\path[draw,thick,-] (j) -- (d);
\node[draw, circle,fill=gray!50] (k) at (10,4.5) {20};
\path[draw,thick,-] (k) -- (e);
\path[draw,thick,-] (k) -- (f);
\node[draw, circle] (l) at (14,4.5) {13};
\path[draw,thick,-] (l) -- (g);
\path[draw,thick,-] (l) -- (h);

\node[draw, circle] (m) at (4,6.5) {39};
\path[draw,thick,-] (m) -- (i);
\path[draw,thick,-] (m) -- (j);
\node[draw, circle] (n) at (12,6.5) {33};
\path[draw,thick,-] (n) -- (k);
\path[draw,thick,-] (n) -- (l);

\node[draw, circle] (o) at (8,8.5) {72};
\path[draw,thick,-] (o) -- (m);
\path[draw,thick,-] (o) -- (n);

\path[draw=red,thick,->,line width=2pt] (o) -- (m);
\path[draw=red,thick,->,line width=2pt] (o) -- (n);

\path[draw=red,thick,->,line width=2pt] (m) -- (j);
\path[draw=red,thick,->,line width=2pt] (j) -- (c);
\path[draw=red,thick,->,line width=2pt] (j) -- (d);
\path[draw=red,thick,->,line width=2pt] (c) -- (5.5,1);

\path[draw=red,thick,->,line width=2pt] (n) -- (k);
\path[draw=red,thick,->,line width=2pt] (n) -- (l);

\path[draw=red,thick,->,line width=2pt] (l) -- (g);

\draw [decoration={brace}, decorate, line width=0.5mm] (14,-0.25) -- (5,-0.25);
\end{tikzpicture}
\end{center}
Also in this implementation,
the time complexity of a range query is $O(\log n)$,
because the total number of processed nodes is $O(\log n)$.

\section{Lazy propagation}

\index{lazy propagation}
\index{lazy segment tree}

Using \key{lazy propagation}, we can construct
a segment tree that supports both range updates
and range queries in $O(\log n)$ time.
The idea is to perform the updates and queries
from the top to the bottom, and process the updates
\emph{lazily} so that they are propagated
down the tree only when it is necessary.

In a lazy segment tree, nodes contain two types of
information.
Like in a normal segment tree,
each node contains the sum or some other value
of the corresponding subarray.
In addition, the node may contain information
related to lazy updates, which has not been
propagated yet to its children.

There are two possible types for range updates:
\emph{addition} and \emph{insertion}.
In addition, each element in the range is
increased by some value,
and in insertion, each element in the range
is assigned some value.
Both operations can be implemented using
similar ideas, and it's possible to construct
a tree that supports both the operations
simultaneously.

\subsubsection{Lazy segment tree}

Let's consider an example where our goal is to
construct a segment tree that supports the following operations:

\begin{itemize}
\item increase each element in $[a,b]$ by $u$
\item calculate the sum of elements in $[a,b]$
\end{itemize}
We will construct a tree where each node
contains two values $s/z$:
$s$ denotes the sum of elements in the range,
like in a standard segment tree,
and $z$ denotes a lazy update,
which means that all elements in the range
should be increased by $z$.
In the following tree, $z=0$ for all nodes,
so there are no lazy updates.
\begin{center}
\begin{tikzpicture}[scale=0.7]
\draw (0,0) grid (16,1);

\node[anchor=center] at (0.5, 0.5) {5};
\node[anchor=center] at (1.5, 0.5) {8};
\node[anchor=center] at (2.5, 0.5) {6};
\node[anchor=center] at (3.5, 0.5) {3};
\node[anchor=center] at (4.5, 0.5) {2};
\node[anchor=center] at (5.5, 0.5) {7};
\node[anchor=center] at (6.5, 0.5) {2};
\node[anchor=center] at (7.5, 0.5) {6};
\node[anchor=center] at (8.5, 0.5) {7};
\node[anchor=center] at (9.5, 0.5) {1};
\node[anchor=center] at (10.5, 0.5) {7};
\node[anchor=center] at (11.5, 0.5) {5};
\node[anchor=center] at (12.5, 0.5) {6};
\node[anchor=center] at (13.5, 0.5) {2};
\node[anchor=center] at (14.5, 0.5) {3};
\node[anchor=center] at (15.5, 0.5) {2};

\node[draw, circle] (a) at (1,2.5) {13/0};
\path[draw,thick,-] (a) -- (0.5,1);
\path[draw,thick,-] (a) -- (1.5,1);
\node[draw, circle,minimum size=32pt] (b) at (3,2.5) {9/0};
\path[draw,thick,-] (b) -- (2.5,1);
\path[draw,thick,-] (b) -- (3.5,1);
\node[draw, circle,minimum size=32pt] (c) at (5,2.5) {9/0};
\path[draw,thick,-] (c) -- (4.5,1);
\path[draw,thick,-] (c) -- (5.5,1);
\node[draw, circle,minimum size=32pt] (d) at (7,2.5) {8/0};
\path[draw,thick,-] (d) -- (6.5,1);
\path[draw,thick,-] (d) -- (7.5,1);
\node[draw, circle,minimum size=32pt] (e) at (9,2.5) {8/0};
\path[draw,thick,-] (e) -- (8.5,1);
\path[draw,thick,-] (e) -- (9.5,1);
\node[draw, circle] (f) at (11,2.5) {12/0};
\path[draw,thick,-] (f) -- (10.5,1);
\path[draw,thick,-] (f) -- (11.5,1);
\node[draw, circle,minimum size=32pt] (g) at (13,2.5) {8/0};
\path[draw,thick,-] (g) -- (12.5,1);
\path[draw,thick,-] (g) -- (13.5,1);
\node[draw, circle,minimum size=32pt] (h) at (15,2.5) {5/0};
\path[draw,thick,-] (h) -- (14.5,1);
\path[draw,thick,-] (h) -- (15.5,1);

\node[draw, circle] (i) at (2,4.5) {22/0};
\path[draw,thick,-] (i) -- (a);
\path[draw,thick,-] (i) -- (b);
\node[draw, circle] (j) at (6,4.5) {17/0};
\path[draw,thick,-] (j) -- (c);
\path[draw,thick,-] (j) -- (d);
\node[draw, circle] (k) at (10,4.5) {20/0};
\path[draw,thick,-] (k) -- (e);
\path[draw,thick,-] (k) -- (f);
\node[draw, circle] (l) at (14,4.5) {13/0};
\path[draw,thick,-] (l) -- (g);
\path[draw,thick,-] (l) -- (h);

\node[draw, circle] (m) at (4,6.5) {39/0};
\path[draw,thick,-] (m) -- (i);
\path[draw,thick,-] (m) -- (j);
\node[draw, circle] (n) at (12,6.5) {33/0};
\path[draw,thick,-] (n) -- (k);
\path[draw,thick,-] (n) -- (l);

\node[draw, circle] (o) at (8,8.5) {72/0};
\path[draw,thick,-] (o) -- (m);
\path[draw,thick,-] (o) -- (n);
\end{tikzpicture}
\end{center}

When a range $[a,b]$ is increased by $u$,
we walk from the root towards the leaves
and modify the nodes in the tree as follows:
If the range $[x,y]$ of a node is
completely inside the range $[a,b]$,
we increase the $z$ value of the node by $u$ and stop.
However, if $[x,y]$ only partially belongs to $[a,b]$,
we increase the $s$ value of the node by $hu$,
where $h$ is the size of the intersection of $[a,b]$
and $[x,y]$, and continue our walk recursively in the tree.

For example, the following picture shows the tree after
increasing the elements in the range marked at the bottom by 2:
\begin{center}
\begin{tikzpicture}[scale=0.7]
\fill[color=gray!50] (5,0) rectangle (6,1);
\draw (0,0) grid (16,1);

\node[anchor=center] at (0.5, 0.5) {5};
\node[anchor=center] at (1.5, 0.5) {8};
\node[anchor=center] at (2.5, 0.5) {6};
\node[anchor=center] at (3.5, 0.5) {3};
\node[anchor=center] at (4.5, 0.5) {2};
\node[anchor=center] at (5.5, 0.5) {9};
\node[anchor=center] at (6.5, 0.5) {2};
\node[anchor=center] at (7.5, 0.5) {6};
\node[anchor=center] at (8.5, 0.5) {7};
\node[anchor=center] at (9.5, 0.5) {1};
\node[anchor=center] at (10.5, 0.5) {7};
\node[anchor=center] at (11.5, 0.5) {5};
\node[anchor=center] at (12.5, 0.5) {6};
\node[anchor=center] at (13.5, 0.5) {2};
\node[anchor=center] at (14.5, 0.5) {3};
\node[anchor=center] at (15.5, 0.5) {2};

\node[draw, circle] (a) at (1,2.5) {13/0};
\path[draw,thick,-] (a) -- (0.5,1);
\path[draw,thick,-] (a) -- (1.5,1);
\node[draw, circle,minimum size=32pt] (b) at (3,2.5) {9/0};
\path[draw,thick,-] (b) -- (2.5,1);
\path[draw,thick,-] (b) -- (3.5,1);
\node[draw, circle,minimum size=32pt] (c) at (5,2.5) {11/0};
\path[draw,thick,-] (c) -- (4.5,1);
\path[draw,thick,-] (c) -- (5.5,1);
\node[draw, circle,fill=gray!50,minimum size=32pt] (d) at (7,2.5) {8/2};
\path[draw,thick,-] (d) -- (6.5,1);
\path[draw,thick,-] (d) -- (7.5,1);
\node[draw, circle,minimum size=32pt] (e) at (9,2.5) {8/0};
\path[draw,thick,-] (e) -- (8.5,1);
\path[draw,thick,-] (e) -- (9.5,1);
\node[draw, circle] (f) at (11,2.5) {12/0};
\path[draw,thick,-] (f) -- (10.5,1);
\path[draw,thick,-] (f) -- (11.5,1);
\node[draw, circle,fill=gray!50,minimum size=32pt] (g) at (13,2.5) {8/2};
\path[draw,thick,-] (g) -- (12.5,1);
\path[draw,thick,-] (g) -- (13.5,1);
\node[draw, circle,minimum size=32pt] (h) at (15,2.5) {5/0};
\path[draw,thick,-] (h) -- (14.5,1);
\path[draw,thick,-] (h) -- (15.5,1);

\node[draw, circle] (i) at (2,4.5) {22/0};
\path[draw,thick,-] (i) -- (a);
\path[draw,thick,-] (i) -- (b);
\node[draw, circle] (j) at (6,4.5) {23/0};
\path[draw,thick,-] (j) -- (c);
\path[draw,thick,-] (j) -- (d);
\node[draw, circle,fill=gray!50] (k) at (10,4.5) {20/2};
\path[draw,thick,-] (k) -- (e);
\path[draw,thick,-] (k) -- (f);
\node[draw, circle] (l) at (14,4.5) {17/0};
\path[draw,thick,-] (l) -- (g);
\path[draw,thick,-] (l) -- (h);

\node[draw, circle] (m) at (4,6.5) {45/0};
\path[draw,thick,-] (m) -- (i);
\path[draw,thick,-] (m) -- (j);
\node[draw, circle] (n) at (12,6.5) {45/0};
\path[draw,thick,-] (n) -- (k);
\path[draw,thick,-] (n) -- (l);

\node[draw, circle] (o) at (8,8.5) {90/0};
\path[draw,thick,-] (o) -- (m);
\path[draw,thick,-] (o) -- (n);

\path[draw=red,thick,->,line width=2pt] (o) -- (m);
\path[draw=red,thick,->,line width=2pt] (o) -- (n);

\path[draw=red,thick,->,line width=2pt] (m) -- (j);
\path[draw=red,thick,->,line width=2pt] (j) -- (c);
\path[draw=red,thick,->,line width=2pt] (j) -- (d);
\path[draw=red,thick,->,line width=2pt] (c) -- (5.5,1);

\path[draw=red,thick,->,line width=2pt] (n) -- (k);
\path[draw=red,thick,->,line width=2pt] (n) -- (l);

\path[draw=red,thick,->,line width=2pt] (l) -- (g);

\draw [decoration={brace}, decorate, line width=0.5mm] (14,-0.25) -- (5,-0.25);
\end{tikzpicture}
\end{center}

We also calculate the sum in a range $[a,b]$
by walking in the tree from the root towards the leaves.
If the range $[x,y]$ of a node completely belongs
to $[a,b]$, we add the $s$ value of the node to the sum.
Otherwise, we continue the search recursively
downwards in the tree.

Always before processing a node,
the value of the lazy update is propagated
to the children of the node.
This happens both in a range update
and a range query.
The idea is that the lazy update will be propagated
downwards only when it is necessary,
so that the operations are always efficient.

The following picture shows how the tree changes
when we calculate the sum in the marked range:
\begin{center}
\begin{tikzpicture}[scale=0.7]
\draw (0,0) grid (16,1);

\node[anchor=center] at (0.5, 0.5) {5};
\node[anchor=center] at (1.5, 0.5) {8};
\node[anchor=center] at (2.5, 0.5) {6};
\node[anchor=center] at (3.5, 0.5) {3};
\node[anchor=center] at (4.5, 0.5) {2};
\node[anchor=center] at (5.5, 0.5) {9};
\node[anchor=center] at (6.5, 0.5) {2};
\node[anchor=center] at (7.5, 0.5) {6};
\node[anchor=center] at (8.5, 0.5) {7};
\node[anchor=center] at (9.5, 0.5) {1};
\node[anchor=center] at (10.5, 0.5) {7};
\node[anchor=center] at (11.5, 0.5) {5};
\node[anchor=center] at (12.5, 0.5) {6};
\node[anchor=center] at (13.5, 0.5) {2};
\node[anchor=center] at (14.5, 0.5) {3};
\node[anchor=center] at (15.5, 0.5) {2};

\node[draw, circle] (a) at (1,2.5) {13/0};
\path[draw,thick,-] (a) -- (0.5,1);
\path[draw,thick,-] (a) -- (1.5,1);
\node[draw, circle,minimum size=32pt] (b) at (3,2.5) {9/0};
\path[draw,thick,-] (b) -- (2.5,1);
\path[draw,thick,-] (b) -- (3.5,1);
\node[draw, circle,minimum size=32pt] (c) at (5,2.5) {11/0};
\path[draw,thick,-] (c) -- (4.5,1);
\path[draw,thick,-] (c) -- (5.5,1);
\node[draw, circle,minimum size=32pt] (d) at (7,2.5) {8/2};
\path[draw,thick,-] (d) -- (6.5,1);
\path[draw,thick,-] (d) -- (7.5,1);
\node[draw, circle,minimum size=32pt] (e) at (9,2.5) {8/2};
\path[draw,thick,-] (e) -- (8.5,1);
\path[draw,thick,-] (e) -- (9.5,1);
\node[draw, circle,fill=gray!50,] (f) at (11,2.5) {12/2};
\path[draw,thick,-] (f) -- (10.5,1);
\path[draw,thick,-] (f) -- (11.5,1);
\node[draw, circle,fill=gray!50,minimum size=32pt] (g) at (13,2.5) {8/2};
\path[draw,thick,-] (g) -- (12.5,1);
\path[draw,thick,-] (g) -- (13.5,1);
\node[draw, circle,minimum size=32pt] (h) at (15,2.5) {5/0};
\path[draw,thick,-] (h) -- (14.5,1);
\path[draw,thick,-] (h) -- (15.5,1);

\node[draw, circle] (i) at (2,4.5) {22/0};
\path[draw,thick,-] (i) -- (a);
\path[draw,thick,-] (i) -- (b);
\node[draw, circle] (j) at (6,4.5) {23/0};
\path[draw,thick,-] (j) -- (c);
\path[draw,thick,-] (j) -- (d);
\node[draw, circle] (k) at (10,4.5) {28/0};
\path[draw,thick,-] (k) -- (e);
\path[draw,thick,-] (k) -- (f);
\node[draw, circle] (l) at (14,4.5) {17/0};
\path[draw,thick,-] (l) -- (g);
\path[draw,thick,-] (l) -- (h);

\node[draw, circle] (m) at (4,6.5) {45/0};
\path[draw,thick,-] (m) -- (i);
\path[draw,thick,-] (m) -- (j);
\node[draw, circle] (n) at (12,6.5) {45/0};
\path[draw,thick,-] (n) -- (k);
\path[draw,thick,-] (n) -- (l);

\node[draw, circle] (o) at (8,8.5) {90/0};
\path[draw,thick,-] (o) -- (m);
\path[draw,thick,-] (o) -- (n);

\path[draw=red,thick,->,line width=2pt] (o) -- (n);

\path[draw=red,thick,->,line width=2pt] (n) -- (k);
\path[draw=red,thick,->,line width=2pt] (n) -- (l);

\path[draw=red,thick,->,line width=2pt] (k) -- (f);
\path[draw=red,thick,->,line width=2pt] (l) -- (g);

\draw [decoration={brace}, decorate, line width=0.5mm] (14,-0.25) -- (10,-0.25);

\draw[color=blue,thick] (8,1.5) rectangle (12,5.5);
\end{tikzpicture}
\end{center}
The result of this query was that a lazy update was
propagated downwards in the nodes that are inside the rectangle.
It was necessary to propagate the lazy update,
because some of the updated elements were inside the range.

Note that sometimes it's necessary to combine lazy updates.
This happens when a node already has a lazy update,
and another lazy update will be added to it.
In the above tree, it's easy to combine lazy updates
because updates $z_1$ and $z_2$ combined equal to update $z_1+z_2$.

\subsubsection{Polynomial update}

A lazy update can be generalized so that it's
allowed to update a range by a polynomial
\[p(u) = t_k u^k + t_{k-1} u^{k-1} + \cdots + t_0.\]

Here, the update for the first element in the range is $p(0)$,
for the second element $p(1)$, etc., so the update
at index $i$ in range $[a,b]$ is $p(i-a)$.
For example, adding a polynomial $p(u)=u+1$
to range $[a,b]$ means that the element at index $a$
increases by 1, the element at index $a+1$
increases by 2, etc.

A polynomial update can be supported by
storing $k+2$ values to each node where $k$
equals the degree of the polynomial.
The value $s$ is the sum of the elements in the range,
and values $z_0,z_1,\ldots,z_k$ are the coefficients
of a polynomial that corresponds to a lazy update.

Now, the sum of $[x,y]$ is
\[s+\sum_{u=0}^{y-x} z_k u^k + z_{k-1} u^{k-1} + \cdots + z_0,\]
that can be efficiently calculated using sum formulas
For example, the value $z_0$ corresponds to the sum
$(y-x+1)z_0$, and the value $z_1 u$ corresponds to the sum
\[z_1(0+1+\cdots+y-x) = z_1 \frac{(y-x)(y-x+1)}{2} .\]

When propagating an update in the tree,
the indices of the polynomial $p(u)$ change,
because in each range $[x,y]$,
the values are
calculated for $x=0,1,\ldots,y-x$.
However, this is not a problem, because
$p'(u)=p(u+h)$ is a polynomial
of equal degree as $p(u)$.
For example, if $p(u)=t_2 u^2+t_1 u-t_0$, then
\[p'(u)=t_2(u+h)^2+t_1(u+h)-t_0=t_2 u^2 + (2ht_2+t_1)u+t_2h^2+t_1h-t_0.\]

\section{Dynaaminen toteutus}

\index{dynaaminen segmenttipuu@dynaaminen segmenttipuu}

Tavallinen segmenttipuu on staattinen,
eli jokaiselle solmulle on paikka taulukossa
ja puu vie kiinteän määrän muistia.
Tämä toteutus kuitenkin tuhlaa muistia,
jos suurin osa puun solmuista on tyhjiä.
\key{Dynaaminen segmenttipuu} varaa muistia vain
niille solmuille, joita todella tarvitaan.

Solmut on kätevää tallentaa tietueina tähän tapaan:

\begin{lstlisting}
struct node {
    int s;
    int x, y;
    node *l, *r;
    node(int x, int a, int b) : s(s), x(x), y(y) {}
};
\end{lstlisting}
Tässä $s$ on solmussa oleva arvo,
$[x,y]$ on solmua vastaava väli
ja $l$ ja $r$ osoittavat
solmun vasempaan ja oikeaan alipuuhun.

Tämän jälkeen solmuja voi käsitellä seuraavasti:

\begin{lstlisting}
// uuden solmun luonti
node *u = new node(0, 0, 15);
// kentän muuttaminen
u->s = 5;
\end{lstlisting}

\subsubsection{Harva segmenttipuu}

\index{harva segmenttipuu@harva segmenttipuu}

Dynaaminen segmenttipuu on hyödyllinen,
jos puun indeksialue $[0,N-1]$ on \textit{harva}
eli $N$ on suuri mutta vain
pieni osa indekseistä on käytössä.
Siinä missä tavallinen segmenttipuu 
vie muistia $O(N)$,
dynaaminen segmenttipuu vie muistia
vain $O(n \log N)$, missä $n$ on
käytössä olevien indeksien määrä.

\key{Harva segmenttipuu} aluksi tyhjä
ja sen ainoa solmu on $[0,N-1]$.
Kun puu muuttuu, siihen lisätään
solmuja dynaamisesti sitä mukaa kuin niitä tarvitaan
uusien indeksien vuoksi.
Esimerkiksi jos $N=16$ ja indeksejä
3 ja 10 on muutettu,
puu sisältää seuraavat solmut:
\begin{center}
\begin{tikzpicture}[scale=0.9]
\scriptsize
\node[draw, circle,minimum size=35pt] (1) at (0,0) {$[0,15]$};
\node[draw, circle,minimum size=35pt] (2) at (-4,-2) {$[0,7]$};
\node[draw, circle,minimum size=35pt] (3) at (-6,-4) {$[0,3]$};
\node[draw, circle,minimum size=35pt] (4) at (-4,-6) {$[2,3]$};
\node[draw, circle,minimum size=35pt] (5) at (-2,-8) {$[3]$};
\node[draw, circle,minimum size=35pt] (6) at (4,-2) {$[8,15]$};
\node[draw, circle,minimum size=35pt] (7) at (2,-4) {$[8,11]$};
\node[draw, circle,minimum size=35pt] (8) at (4,-6) {$[10,11]$};
\node[draw, circle,minimum size=35pt] (9) at (2,-8) {$[10]$};

\path[draw,thick,->] (1) -- (2);
\path[draw,thick,->] (2) -- (3);
\path[draw,thick,->] (3) -- (4);
\path[draw,thick,->] (4) -- (5);

\path[draw,thick,->] (1) -- (6);
\path[draw,thick,->] (6) -- (7);
\path[draw,thick,->] (7) -- (8);
\path[draw,thick,->] (8) -- (9);
\end{tikzpicture}
\end{center}

Reitti puun juuresta lehteen sisältää
$O(\log N)$ solmua,
joten jokainen muutos puuhun lisää
enintään $O(\log N)$ uutta solmua puuhun.
Niinpä $n$ muutoksen jälkeen puussa
on enintään $O(n \log N)$ solmua.

Huomaa, että jos kaikki tarvittavat indeksit
ovat tiedossa
algoritmin alussa, dynaamisen segmenttipuun
sijasta voi käyttää tavallista segmenttipuuta
ja indeksien pakkausta (luku 9.4).
Tämä ei ole kuitenkaan mahdollista,
jos indeksit syntyvät vasta algoritmin aikana.

\subsubsection{Persistentti segmenttipuu}

\index{persistentti segmenttipuu@persistentti segmenttipuu}
\index{muutoshistoria@muutoshistoria}

Dynaamisen toteutuksen avulla on myös
mahdollista luoda \key{persistentti segmenttipuu},
joka säilyttää puun muutoshistorian.
Tällöin muistissa on jokainen
segmenttipuun vaihe, joka on esiintynyt
algoritmin suorituksen aikana.

Muutoshistorian hyötynä on,
että kaikkia vanhoja puita voi käsitellä
segmenttipuun tapaan,
koska niiden rakenne on edelleen olemassa.
Vanhoista puista voi myös johtaa uusia
puita, joita voi muokata edelleen.

Tarkastellaan esimerkiksi seuraavaa muutossarjaa,
jossa punaiset solmut muuttuvat päivityksessä
ja muut solmut säilyvät ennallaan:

\begin{center}
\begin{tikzpicture}[scale=0.8]
\node[draw, circle,minimum size=13pt] (1a) at (3,0) {};
\node[draw, circle,minimum size=13pt] (2a) at (2,-1) {};
\node[draw, circle,minimum size=13pt] (3a) at (4,-1) {};
\node[draw, circle,minimum size=13pt] (4a) at (1.5,-2) {};
\node[draw, circle,minimum size=13pt] (5a) at (2.5,-2) {};
\node[draw, circle,minimum size=13pt] (6a) at (3.5,-2) {};
\node[draw, circle,minimum size=13pt] (7a) at (4.5,-2) {};
\path[draw,thick,->] (1a) -- (2a);
\path[draw,thick,->] (1a) -- (3a);
\path[draw,thick,->] (2a) -- (4a);
\path[draw,thick,->] (2a) -- (5a);
\path[draw,thick,->] (3a) -- (6a);
\path[draw,thick,->] (3a) -- (7a);

\node[draw, circle,minimum size=13pt,fill=red] (1b) at (3+5,0) {};
\node[draw, circle,minimum size=13pt,fill=red] (2b) at (2+5,-1) {};
\node[draw, circle,minimum size=13pt] (3b) at (4+5,-1) {};
\node[draw, circle,minimum size=13pt] (4b) at (1.5+5,-2) {};
\node[draw, circle,minimum size=13pt,fill=red] (5b) at (2.5+5,-2) {};
\node[draw, circle,minimum size=13pt] (6b) at (3.5+5,-2) {};
\node[draw, circle,minimum size=13pt] (7b) at (4.5+5,-2) {};
\path[draw,thick,->] (1b) -- (2b);
\path[draw,thick,->] (1b) -- (3b);
\path[draw,thick,->] (2b) -- (4b);
\path[draw,thick,->] (2b) -- (5b);
\path[draw,thick,->] (3b) -- (6b);
\path[draw,thick,->] (3b) -- (7b);

\node[draw, circle,minimum size=13pt,fill=red] (1c) at (3+10,0) {};
\node[draw, circle,minimum size=13pt] (2c) at (2+10,-1) {};
\node[draw, circle,minimum size=13pt,fill=red] (3c) at (4+10,-1) {};
\node[draw, circle,minimum size=13pt] (4c) at (1.5+10,-2) {};
\node[draw, circle,minimum size=13pt] (5c) at (2.5+10,-2) {};
\node[draw, circle,minimum size=13pt] (6c) at (3.5+10,-2) {};
\node[draw, circle,minimum size=13pt,fill=red] (7c) at (4.5+10,-2) {};
\path[draw,thick,->] (1c) -- (2c);
\path[draw,thick,->] (1c) -- (3c);
\path[draw,thick,->] (2c) -- (4c);
\path[draw,thick,->] (2c) -- (5c);
\path[draw,thick,->] (3c) -- (6c);
\path[draw,thick,->] (3c) -- (7c);

\node at (3,-3) {vaihe 1};
\node at (3+5,-3) {vaihe 2};
\node at (3+10,-3) {vaihe 3};
\end{tikzpicture}
\end{center}
Jokaisen muutoksen jälkeen suurin osa puun
solmuista säilyy ennallaan,
joten muistia säästävä tapa tallentaa muutoshistoria
on käyttää mahdollisimman paljon hyväksi
puun vanhoja osia muutoksissa.
Tässä tapauksessa muutoshistorian voi
tallentaa seuraavasti:
\begin{center}
\begin{tikzpicture}[scale=0.8]
\path[use as bounding box] (0, 1) rectangle (16, -3.5);

\node[draw, circle,minimum size=13pt] (1a) at (3,0) {};
\node[draw, circle,minimum size=13pt] (2a) at (2,-1) {};
\node[draw, circle,minimum size=13pt] (3a) at (4,-1) {};
\node[draw, circle,minimum size=13pt] (4a) at (1.5,-2) {};
\node[draw, circle,minimum size=13pt] (5a) at (2.5,-2) {};
\node[draw, circle,minimum size=13pt] (6a) at (3.5,-2) {};
\node[draw, circle,minimum size=13pt] (7a) at (4.5,-2) {};
\path[draw,thick,->] (1a) -- (2a);
\path[draw,thick,->] (1a) -- (3a);
\path[draw,thick,->] (2a) -- (4a);
\path[draw,thick,->] (2a) -- (5a);
\path[draw,thick,->] (3a) -- (6a);
\path[draw,thick,->] (3a) -- (7a);

\node[draw, circle,minimum size=13pt,fill=red] (1b) at (3+5,0) {};
\node[draw, circle,minimum size=13pt,fill=red] (2b) at (2+5,-1) {};
\node[draw, circle,minimum size=13pt,fill=red] (5b) at (2.5+5,-2) {};
\path[draw,thick,->] (1b) -- (2b);

\draw[thick,->] (1b) .. controls (3+5+2,0-1) and (3+5,2.5) .. (3a);

\draw[thick,->] (2b) .. controls (2+5-0.5,-1-0.5) and (2,4.5) .. (4a);


\path[draw,thick,->] (2b) -- (5b);

\node[draw, circle,minimum size=13pt,fill=red] (1c) at (3+10,0) {};
\node[draw, circle,minimum size=13pt,fill=red] (3c) at (4+10,-1) {};
\node[draw, circle,minimum size=13pt,fill=red] (7c) at (4.5+10,-2) {};
\path[draw,thick,->] (1c) -- (2b);
\path[draw,thick,->] (1c) -- (3c);

\draw[thick,->] (3c) .. controls (2.5+5,-3) and (3.5,-3) .. (6a);

\path[draw,thick,->] (3c) -- (7c);

\node at (3,-3) {vaihe 1};
\node at (3+5,-3) {vaihe 2};
\node at (3+10,-3) {vaihe 3};
\end{tikzpicture}
\end{center}

Nyt muistissa on jokaisesta puun vaiheesta
puun juuri, jonka avulla pystyy selvittämään
koko puun rakenteen kyseisellä hetkellä.
Jokainen muutos tuo vain $O(\log N)$ uutta solmua puuhun,
kun puun indeksialue on $[0,N-1]$,
joten koko muutoshistorian pitäminen muistissa on mahdollista.

\section{Tietorakenteet}

Segmenttipuun solmussa voi olla
yksittäisen arvon
sijasta myös jokin tietorakenne,
joka pitää yllä tietoa solmua vastaavasta välistä.
Tällöin segmenttipuun operaatiot vievät aikaa
$O(f(n) \log n)$, missä $f(n)$ on
yksittäisen solmun tietorakenteen
käsittelyyn kuluva aika.

Tarkastellaan esimerkkinä segmenttipuuta,
jonka avulla voi laskea, montako kertaa
luku $x$ esiintyy taulukon välillä $[a,b]$.
Esimerkiksi seuraavassa taulukossa
luku 1 esiintyy kolme kertaa
merkityllä välillä:

\begin{center}
\begin{tikzpicture}[scale=0.7]
\fill[lightgray] (1,0) rectangle (6,1);
\draw (0,0) grid (8,1);

\node[anchor=center] at (0.5, 0.5) {3};
\node[anchor=center] at (1.5, 0.5) {1};
\node[anchor=center] at (2.5, 0.5) {2};
\node[anchor=center] at (3.5, 0.5) {3};
\node[anchor=center] at (4.5, 0.5) {1};
\node[anchor=center] at (5.5, 0.5) {1};
\node[anchor=center] at (6.5, 0.5) {1};
\node[anchor=center] at (7.5, 0.5) {2};
\end{tikzpicture}
\end{center}

Ideana on toteuttaa segmenttipuu, jonka
jokaisessa solmussa on tietorakenne,
josta voi kysyä,
montako kertaa luku $x$ esiintyy solmun välillä.
Tällaisen segmenttipuun avulla
vastaus kyselyyn syntyy laskemalla yhteen
esiintymismäärät väleiltä, joista $[a,b]$ muodostuu.

Esimerkiksi yllä olevasta taulukosta syntyy
seuraava segmenttipuu:
\begin{center}
\begin{tikzpicture}[scale=0.7]

\node[draw, rectangle] (a) at (1,2.5)
{
\footnotesize
\begin{tabular}{r}
3 \\
\hline
1 \\
\end{tabular}};
\node[draw, rectangle] (b) at (3,2.5)
{
\footnotesize
\begin{tabular}{r}
1 \\
\hline
1 \\
\end{tabular}};
\node[draw, rectangle] (c) at (5,2.5)
{
\footnotesize
\begin{tabular}{r}
2 \\
\hline
1 \\
\end{tabular}};
\node[draw, rectangle] (d) at (7,2.5)
{
\footnotesize
\begin{tabular}{r}
3 \\
\hline
1 \\
\end{tabular}};
\node[draw, rectangle] (e) at (9,2.5)
{
\footnotesize
\begin{tabular}{r}
1 \\
\hline
1 \\
\end{tabular}};
\node[draw, rectangle] (f) at (11,2.5)
{
\footnotesize
\begin{tabular}{r}
1 \\
\hline
1 \\
\end{tabular}};
\node[draw, rectangle] (g) at (13,2.5)
{
\footnotesize
\begin{tabular}{r}
1 \\
\hline
1 \\
\end{tabular}};
\node[draw, rectangle] (h) at (15,2.5)
{
\footnotesize
\begin{tabular}{r}
2 \\
\hline
1 \\
\end{tabular}};

\node[draw, rectangle] (i) at (2,4.5)
{
\footnotesize
\begin{tabular}{rr}
1 & 3 \\
\hline
1 & 1 \\
\end{tabular}};
\path[draw,thick,-] (i) -- (a);
\path[draw,thick,-] (i) -- (b);
\node[draw, rectangle] (j) at (6,4.5)
{
\footnotesize
\begin{tabular}{rr}
2 & 3 \\
\hline
1 & 1 \\
\end{tabular}};
\path[draw,thick,-] (j) -- (c);
\path[draw,thick,-] (j) -- (d);
\node[draw, rectangle] (k) at (10,4.5)
{
\footnotesize
\begin{tabular}{r}
1 \\
\hline
2 \\
\end{tabular}};
\path[draw,thick,-] (k) -- (e);
\path[draw,thick,-] (k) -- (f);
\node[draw, rectangle] (l) at (14,4.5)
{
\footnotesize
\begin{tabular}{rr}
1 & 2 \\
\hline
1 & 1 \\
\end{tabular}};
\path[draw,thick,-] (l) -- (g);
\path[draw,thick,-] (l) -- (h);

\node[draw, rectangle] (m) at (4,6.5)
{
\footnotesize
\begin{tabular}{rrr}
1 & 2 & 3 \\
\hline
1 & 1 & 2 \\
\end{tabular}};
\path[draw,thick,-] (m) -- (i);
\path[draw,thick,-] (m) -- (j);
\node[draw, rectangle] (n) at (12,6.5)
{
\footnotesize
\begin{tabular}{rr}
1 & 2 \\
\hline
3 & 1 \\
\end{tabular}};
\path[draw,thick,-] (n) -- (k);
\path[draw,thick,-] (n) -- (l);

\node[draw, rectangle] (o) at (8,8.5)
{
\footnotesize
\begin{tabular}{rrr}
1 & 2 & 3 \\
\hline
4 & 2 & 2 \\
\end{tabular}};
\path[draw,thick,-] (o) -- (m);
\path[draw,thick,-] (o) -- (n);
\end{tikzpicture}
\end{center}


Sopiva tietorakenne segmenttipuun toteuttamiseen on
hakemistorakenne, joka pitää kirjaa välillä esiintyvien
lukujen määrästä.
Esimerkiksi \texttt{map}-ra\-ken\-net\-ta käyttäen
yhden solmun käsittely vie aikaa $O(\log n)$,
minkä seurauksena kyselyn aikavaativuus on $O(\log^2 n)$.

Solmuissa olevat tietorakenteet kasvattavat
segmenttipuun muistinkäyttöä.
Tässä tapauksessa
segmenttipuu vie tilaa $O(n \log n)$,
koska siinä on $O(\log n)$ tasoa, joista
jokaisella hakemistorakenteet sisältävät $O(n)$ lukua.

\section{Kaksiulotteisuus}

\index{kaksiulotteinen segmenttipuu@kaksiulotteinen segmenttipuu}

\key{Kaksiulotteinen segmenttipuu} mahdollistaa
kaksiulotteisen taulukon
suorakulmaisia alueita koskevat kyselyt.
Tällaisen segmenttipuun voi toteuttaa
sisäkkäisinä segmenttipuina:
suuri puu vastaa taulukon rivejä
ja sen kussakin solmussa on pieni puu,
joka vastaa taulukon sarakkeita.

Esimerkiksi taulukon
\begin{center}
\begin{tikzpicture}[scale=0.7]
\draw (0,0) grid (4,4);

\node[anchor=center] at (0.5, 0.5) {8};
\node[anchor=center] at (1.5, 0.5) {5};
\node[anchor=center] at (2.5, 0.5) {3};
\node[anchor=center] at (3.5, 0.5) {8};

\node[anchor=center] at (0.5, 1.5) {3};
\node[anchor=center] at (1.5, 1.5) {9};
\node[anchor=center] at (2.5, 1.5) {7};
\node[anchor=center] at (3.5, 1.5) {1};

\node[anchor=center] at (0.5, 2.5) {8};
\node[anchor=center] at (1.5, 2.5) {7};
\node[anchor=center] at (2.5, 2.5) {5};
\node[anchor=center] at (3.5, 2.5) {2};

\node[anchor=center] at (0.5, 3.5) {7};
\node[anchor=center] at (1.5, 3.5) {6};
\node[anchor=center] at (2.5, 3.5) {1};
\node[anchor=center] at (3.5, 3.5) {6};
\end{tikzpicture}
\end{center}
alueiden summia voi laskea seuraavasta segmenttipuusta:
\begin{center}
\begin{tikzpicture}[scale=0.4]
\footnotesize
\begin{scope}[shift={(-12,0)}]
\draw (-1,-1) rectangle (5,6);
\draw (0,0) grid (4,1);
\node[anchor=center,scale=0.8] at (0.5, 0.5) {7};
\node[anchor=center,scale=0.8] at (1.5, 0.5) {6};
\node[anchor=center,scale=0.8] at (2.5, 0.5) {1};
\node[anchor=center,scale=0.8] at (3.5, 0.5) {6};

\node[draw, circle,scale=0.8,inner sep=1pt] (a) at (1,2.5) {13};
\path[draw,thick,-] (a) -- (0.5,1);
\path[draw,thick,-] (a) -- (1.5,1);
\node[draw, circle,scale=0.8,inner sep=2.5pt] (b) at (3,2.5) {7};
\path[draw,thick,-] (b) -- (2.5,1);
\path[draw,thick,-] (b) -- (3.5,1);

\node[draw, circle,scale=0.8,inner sep=1pt] (i) at (2,4.5) {20};
\path[draw,thick,-] (i) -- (a);
\path[draw,thick,-] (i) -- (b);
\end{scope}
\begin{scope}[shift={(-4,0)}]
\draw (-1,-1) rectangle (5,6);
\draw (0,0) grid (4,1);
\node[anchor=center,scale=0.8] at (0.5, 0.5) {8};
\node[anchor=center,scale=0.8] at (1.5, 0.5) {7};
\node[anchor=center,scale=0.8] at (2.5, 0.5) {5};
\node[anchor=center,scale=0.8] at (3.5, 0.5) {2};

\node[draw, circle,scale=0.8,inner sep=1pt] (a) at (1,2.5) {15};
\path[draw,thick,-] (a) -- (0.5,1);
\path[draw,thick,-] (a) -- (1.5,1);
\node[draw, circle,scale=0.8,inner sep=2.5pt] (b) at (3,2.5) {7};
\path[draw,thick,-] (b) -- (2.5,1);
\path[draw,thick,-] (b) -- (3.5,1);

\node[draw, circle,scale=0.8,inner sep=1pt] (i) at (2,4.5) {22};
\path[draw,thick,-] (i) -- (a);
\path[draw,thick,-] (i) -- (b);
\end{scope}
\begin{scope}[shift={(4,0)}]
\draw (-1,-1) rectangle (5,6);
\draw (0,0) grid (4,1);
\node[anchor=center,scale=0.8] at (0.5, 0.5) {3};
\node[anchor=center,scale=0.8] at (1.5, 0.5) {9};
\node[anchor=center,scale=0.8] at (2.5, 0.5) {7};
\node[anchor=center,scale=0.8] at (3.5, 0.5) {1};

\node[draw, circle,scale=0.8,inner sep=1pt] (a) at (1,2.5) {12};
\path[draw,thick,-] (a) -- (0.5,1);
\path[draw,thick,-] (a) -- (1.5,1);
\node[draw, circle,scale=0.8,inner sep=2.5pt] (b) at (3,2.5) {8};
\path[draw,thick,-] (b) -- (2.5,1);
\path[draw,thick,-] (b) -- (3.5,1);

\node[draw, circle,scale=0.8,inner sep=1pt] (i) at (2,4.5) {20};
\path[draw,thick,-] (i) -- (a);
\path[draw,thick,-] (i) -- (b);
\end{scope}
\begin{scope}[shift={(12,0)}]
\draw (-1,-1) rectangle (5,6);
\draw (0,0) grid (4,1);
\node[anchor=center,scale=0.8] at (0.5, 0.5) {8};
\node[anchor=center,scale=0.8] at (1.5, 0.5) {5};
\node[anchor=center,scale=0.8] at (2.5, 0.5) {3};
\node[anchor=center,scale=0.8] at (3.5, 0.5) {8};

\node[draw, circle,scale=0.8,inner sep=1pt] (a) at (1,2.5) {13};
\path[draw,thick,-] (a) -- (0.5,1);
\path[draw,thick,-] (a) -- (1.5,1);
\node[draw, circle,scale=0.8,inner sep=1pt] (b) at (3,2.5) {11};
\path[draw,thick,-] (b) -- (2.5,1);
\path[draw,thick,-] (b) -- (3.5,1);

\node[draw, circle,scale=0.8,inner sep=1pt] (i) at (2,4.5) {24};
\path[draw,thick,-] (i) -- (a);
\path[draw,thick,-] (i) -- (b);
\end{scope}
\begin{scope}[shift={(-8,10)}]
\draw (-1,-1) rectangle (5,6);
\draw (0,0) grid (4,1);
\node[anchor=center,scale=0.8] at (0.5, 0.5) {15};
\node[anchor=center,scale=0.8] at (1.5, 0.5) {13};
\node[anchor=center,scale=0.8] at (2.5, 0.5) {6};
\node[anchor=center,scale=0.8] at (3.5, 0.5) {8};

\node[draw, circle,scale=0.8,inner sep=1pt] (a) at (1,2.5) {28};
\path[draw,thick,-] (a) -- (0.5,1);
\path[draw,thick,-] (a) -- (1.5,1);
\node[draw, circle,scale=0.8,inner sep=1pt] (b) at (3,2.5) {14};
\path[draw,thick,-] (b) -- (2.5,1);
\path[draw,thick,-] (b) -- (3.5,1);

\node[draw, circle,scale=0.8,inner sep=1pt] (i) at (2,4.5) {42};
\path[draw,thick,-] (i) -- (a);
\path[draw,thick,-] (i) -- (b);
\end{scope}
\begin{scope}[shift={(8,10)}]
\draw (-1,-1) rectangle (5,6);
\draw (0,0) grid (4,1);
\node[anchor=center,scale=0.8] at (0.5, 0.5) {11};
\node[anchor=center,scale=0.8] at (1.5, 0.5) {14};
\node[anchor=center,scale=0.8] at (2.5, 0.5) {10};
\node[anchor=center,scale=0.8] at (3.5, 0.5) {9};

\node[draw, circle,scale=0.8,inner sep=1pt] (a) at (1,2.5) {25};
\path[draw,thick,-] (a) -- (0.5,1);
\path[draw,thick,-] (a) -- (1.5,1);
\node[draw, circle,scale=0.8,inner sep=1pt] (b) at (3,2.5) {19};
\path[draw,thick,-] (b) -- (2.5,1);
\path[draw,thick,-] (b) -- (3.5,1);

\node[draw, circle,scale=0.8,inner sep=1pt] (i) at (2,4.5) {44};
\path[draw,thick,-] (i) -- (a);
\path[draw,thick,-] (i) -- (b);
\end{scope}
\begin{scope}[shift={(0,20)}]
\draw (-1,-1) rectangle (5,6);
\draw (0,0) grid (4,1);
\node[anchor=center,scale=0.8] at (0.5, 0.5) {26};
\node[anchor=center,scale=0.8] at (1.5, 0.5) {27};
\node[anchor=center,scale=0.8] at (2.5, 0.5) {16};
\node[anchor=center,scale=0.8] at (3.5, 0.5) {17};

\node[draw, circle,scale=0.8,inner sep=1pt] (a) at (1,2.5) {53};
\path[draw,thick,-] (a) -- (0.5,1);
\path[draw,thick,-] (a) -- (1.5,1);
\node[draw, circle,scale=0.8,inner sep=1pt] (b) at (3,2.5) {33};
\path[draw,thick,-] (b) -- (2.5,1);
\path[draw,thick,-] (b) -- (3.5,1);

\node[draw, circle,scale=0.8,inner sep=1pt] (i) at (2,4.5) {86};
\path[draw,thick,-] (i) -- (a);
\path[draw,thick,-] (i) -- (b);
\end{scope}
\path[draw,thick,-] (2,19) -- (-6,16);
\path[draw,thick,-] (2,19) -- (10,16);
\path[draw,thick,-] (-6,9) -- (-10,6);
\path[draw,thick,-] (-6,9) -- (-2,6);
\path[draw,thick,-] (10,9) -- (6,6);
\path[draw,thick,-] (10,9) -- (14,6);
\end{tikzpicture}
\end{center}

Kaksiulotteisen
segmenttipuun operaatiot vievät aikaa
$O(\log^2 n)$,
koska suuressa puussa ja kussakin
pienessä puussa on $O(\log n)$ tasoa.
Segmenttipuu vie muistia $O(n^2)$,
koska jokainen pieni puu
vie muistia $O(n)$.

Vastaavalla tavalla voi luoda myös segmenttipuita,
joissa on vielä enemmän ulottuvuuksia,
mutta tälle on harvoin tarvetta.

