\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{Dynamic trees}

\index{dynamic segment tree}

A regular segment tree is static,
which means that each node has a fixed position
in the array and storing the tree requires
a fixed amount of memory.
However, if most nodes are empty, such an
implementation wastes memory.
In a \key{dynamic segment tree},
memory is reserved only for nodes that
are actually needed.

The nodes can be represented as structs as follows:

\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}
Here $s$ is the value of the node,
$[x,y]$ is the corresponding range,
and $l$ and $r$ point to the left
and right subtree.

After this, nodes can be manipulated as follows:

\begin{lstlisting}
// create new node
node *u = new node(0, 0, 15);
// change value
u->s = 5;
\end{lstlisting}

\subsubsection{Sparse segment tree}

\index{sparse segment tree}

A dynamic segment tree is useful if
the range $[0,N-1]$ covered by the tree is \emph{sparse},
which means that $N$ is large but only a
small portion of the indices are used.
While a regular segment tree uses $O(n)$ memory,
a dynamic segment tree only uses $O(n \log N)$ memory,
where $n$ is the number of indices used.

A \key{sparse segment tree} is initially empty
and its only node is $[0,N-1]$.
When the tree changes, new nodes are added dynamically
always when they are needed because of new indices.
For example, if $N=16$, and the elements
in indices 3 and 10 have been changes,
the tree contains the following nodes:
\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}

Any path from the root to a leaf contains
$O(\log N)$ nodes,
so each change adds at most $O(\log n)$
new nodes to the tree.
Thus, after $n$ changes, the tree contains
at most $O(n \log N)$ nodes.

Note that if all indices of the elements
are known at the beginning of the algorithm,
a dynamic segment tree is not needed,
but we can use a regular segment tree with
index compression (Chapter 9.4).
However, this is not possible if the indices
are generated during the algorithm.

\subsubsection{Persistent segment tree}

\index{persistent segment tree}
\index{version history}

Using a dynamic implementation,
it is also possible to create a
\key{persistent segment tree} that stores
the \key{version history} of the tree.
In such an implementation, we can
efficiently access
all versions of the tree that have been
existed during the algorithm.

When the version history is available,
we can access all versions of the tree
like a regular segment tree, because their
structure is stored.
We can also derive new trees from the history
and further manipulate them.

Consider the following sequence of updates,
where red nodes change in an update
and other nodes remain the same:

\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) {step 1};
\node at (3+5,-3) {step 2};
\node at (3+10,-3) {step 3};
\end{tikzpicture}
\end{center}
After each update, most nodes in the tree
remain the same,
so an efficient way to store the version history
is to represent each tree in the history as a combination
of new nodes and subtrees of previous trees.
In this case, the version history can be
stored as follows:
\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) {step 1};
\node at (3+5,-3) {step 2};
\node at (3+10,-3) {step 3};
\end{tikzpicture}
\end{center}

The structure of each version of the tree in the history can be
reconstructed by following the pointers from the root.
Each update only adds $O(\log N)$ new nodes to the tree
when the indices are $[0,N-1]$,
so it is possible to store the full version history
of the tree.

\section{Data structures}

Insted of a single value, a node in a segment tree
can also contain a data structure that maintains information
about the corresponding range.
In this case, the operations of the tree take
$O(f(n) \log n)$ time, where $f(n)$ is
the time needed for retrieving or updating the
information in a single node.

As an example, consider a segment tree that
supports queries of the form
''how many times does element $x$ appear
in range $[a,b]$?''
For example, element 1 appears three times
in the following subarray:

\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}

The idea is to construct a segment tree
where each node has a data structure
that can return the number of any element $x$
in the range.
Using such a segment tree,
the answer for a query can be calculated
by combining the results from the nodes
that belong to the range.

For example, the following segment tree
corresponds to the above array:
\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}

Each node in the tree should contain
an appropriate data structure, for example a
\texttt{map} structure.
In this case, the time needed for accessing
a node is $O(\log n)$, so the total time complexity
of a query is $O(\log^2 n)$.

Data structures in nodes increase the memory usage
of the tree.
In this example, $O(n \log n)$ memory is needed,
because the tree consists of $O(\log n)$ levels,
and the map structures contain a total
of $O(n)$ values at each level.

\section{Two-dimensionality}

\index{two-dimensional segment tree}

A \key{two-dimensional segment tree} supports
queries about rectangles in a two-dimensional array.
Such a segment tree can be implemented as
nested segmenet trees: a big tree corresponds to the
rows in the array, and each node contains a small tree
that corresponds to a column.

For example, in the array
\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}
sums of rectangles can be calculated
from the following segment tree:
\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}

The operations in a two-dimensional segment tree
take $O(\log^2 n)$ time, because the big tree
and each small tree contain $O(\log n)$ levels.
The tree uses $O(n^2)$ memory, because each
small tree uses $O(n)$ memory.

Using a similar idea, it is also possible to create
segment trees with more dimensions,
but this is rarely needed.