\chapter{Tree queries}

\index{tree query}

This chapter discusses techniques for
processing queries related
to subtrees and paths of a rooted tree.
For example, such 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 a node $x$
(or $0$ if there is no such an ancestor).
For example, in the following tree, $f(2,1)=1$, $f(8,2)=4$
and $f(5,2)=0$.

\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 above tree
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 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{tree traversal array}

A \key{tree traversal array} contains the nodes of a rooted tree
in the order in which a depth-first search
from the root node visits them.
For example, in the tree
\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 tree traversal 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 tree traversal array such that
the first element in the subarray is the root node.
For example, the following subarray contains the
nodes in the subtree of node $4$:
\begin{center}
\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 tree traversal array that contains
three values for each node: the identifier of the node,
the size of the subtree, and 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[left] at (-1,0.5) {node id};
\node[left] at (-1,-0.5) {subtree size};
\node[left] at (-1,-1.5) {node value};

\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[left] at (-1,0.5) {node id};
\node[left] at (-1,-0.5) {subtree size};
\node[left] at (-1,-1.5) {node value};

\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 answer 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 tree traversal array, we can also efficiently
calculate sums of values on
paths from the root node to 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 from
the root to a node
\end{itemize}

For example, in the following tree,
the sum of values from the root node to node 7 is
$4+5+5=14$:

\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}

We can solve this problem in a similar way as before,
but now each value in the last row of the array is the sum of values
on a path from the root to the node.
For example, the following array corresponds to the above tree:
\begin{center}
\begin{tikzpicture}[scale=0.7]
\draw (0,1) grid (9,-2);

\node[left] at (-1,0.5) {node id};
\node[left] at (-1,-0.5) {subtree size};
\node[left] at (-1,-1.5) {path sum};

\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) {$4$};
\node at (1.5,-1.5) {$9$};
\node at (2.5,-1.5) {$12$};
\node at (3.5,-1.5) {$7$};
\node at (4.5,-1.5) {$9$};
\node at (5.5,-1.5) {$14$};
\node at (6.5,-1.5) {$12$};
\node at (7.5,-1.5) {$10$};
\node at (8.5,-1.5) {$6$};
% 
% \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}

When the value of a node increases by $x$,
the sums of all nodes in its subtree increase by $x$.
For example, if the value of node 4 increases by 1,
the array changes as follows:

\begin{center}
\begin{tikzpicture}[scale=0.7]
\fill[color=lightgray] (4,-1) rectangle (8,-2);
\draw (0,1) grid (9,-2);

\node[left] at (-1,0.5) {node id};
\node[left] at (-1,-0.5) {subtree size};
\node[left] at (-1,-1.5) {path sum};

\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) {$4$};
\node at (1.5,-1.5) {$9$};
\node at (2.5,-1.5) {$12$};
\node at (3.5,-1.5) {$7$};
\node at (4.5,-1.5) {$10$};
\node at (5.5,-1.5) {$15$};
\node at (6.5,-1.5) {$13$};
\node at (7.5,-1.5) {$11$};
\node at (8.5,-1.5) {$6$};
% 
% \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}

Thus, to support both the operations,
we should be able to increase all values
in a range and retrieve a single value.
This can be done in $O(\log n)$ time
using a binary indexed tree
or segment tree (see Chapter 9.4).

\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 following tree,
the lowest common ancestor of nodes 5 and 8
is node 2:
\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=40] (3);
\end{tikzpicture}
\end{center}

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.
Using this, we can divide the problem of
finding the lowest common ancestor into two parts.

We use two pointers that initially point to the
two nodes for which we should find the
lowest common ancestor.
First, we move one of the pointers upwards
so that both nodes are at the same level in the tree.

In the example case, we move from node 8 to node 6,
after which both nodes are at the same level:

\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,fill=lightgray] (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) edge [bend right] (7);
\end{tikzpicture}
\end{center}

After this, we determine the minimum number of steps
needed to move both pointers upwards so that
they will point to the same node.
This node is the lowest common ancestor of the nodes.

In the example case, it suffices to move both pointers
one step upwards to node 2,
which is 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,fill=lightgray] (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] (6) edge [bend left] (3);
\path[draw=red,thick,->,line width=2pt] (7) edge [bend right] (3);
\end{tikzpicture}
\end{center}

Since both parts of the algorithm can be performed in
$O(\log n)$ time using precomputed information,
we can find the lowest common ancestor of any two
nodes in $O(\log n)$ time using this technique.

\subsubsection{Method 2}

Another way to solve the problem is based on
a tree traversal array\footnote{This lowest common ancestor algorithm is based on \cite{ben00}.
This technique is sometimes called the \index{Euler tour technique}
\key{Euler tour technique} \cite{tar84}.}.
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 use a bit different tree
traversal array than before:
we add each node to the array \emph{always}
when the depth-first search walks through the node,
and not only at the first visit.
Hence, a node that has $k$ children appears $k+1$ times
in the array and there are a total of $2n-1$
nodes in the array.

We store two values in the array:
the identifier of the node and the level of the 
node in the tree.
The following array corresponds to the above tree:

\begin{center}
\begin{tikzpicture}[scale=0.7]

\node[left] at (-1,1.5) {node id};
\node[left] at (-1,0.5) {level};

\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) {$0$};
\node at (1.5,2.5) {$1$};
\node at (2.5,2.5) {$2$};
\node at (3.5,2.5) {$3$};
\node at (4.5,2.5) {$4$};
\node at (5.5,2.5) {$5$};
\node at (6.5,2.5) {$6$};
\node at (7.5,2.5) {$7$};
\node at (8.5,2.5) {$8$};
\node at (9.5,2.5) {$9$};
\node at (10.5,2.5) {$10$};
\node at (11.5,2.5) {$11$};
\node at (12.5,2.5) {$12$};
\node at (13.5,2.5) {$13$};
\node at (14.5,2.5) {$14$};
\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]

\node[left] at (-1,1.5) {node id};
\node[left] at (-1,0.5) {level};

\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) {$0$};
\node at (1.5,2.5) {$1$};
\node at (2.5,2.5) {$2$};
\node at (3.5,2.5) {$3$};
\node at (4.5,2.5) {$4$};
\node at (5.5,2.5) {$5$};
\node at (6.5,2.5) {$6$};
\node at (7.5,2.5) {$7$};
\node at (8.5,2.5) {$8$};
\node at (9.5,2.5) {$9$};
\node at (10.5,2.5) {$10$};
\node at (11.5,2.5) {$11$};
\node at (12.5,2.5) {$12$};
\node at (13.5,2.5) {$13$};
\node at (14.5,2.5) {$14$};
\end{tikzpicture}
\end{center}

Node 5 is at position 2, node 8 is at position 5,
and the node with lowest level between
positions $2 \ldots 5$ is node 2 at position 3
whose level is 2.
Thus, the lowest common ancestor of
nodes 5 and 8 is node 2.

Thus, to find the lowest common ancestor
of two nodes it suffices to process a range
minimum query.
Since the array is static,
we can process such queries in $O(1)$ time
after an $O(n \log n)$ time preprocessing.

\subsubsection{Distances of nodes}

Finally, let us consider the 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 root the tree arbitrarily.
After this, the distance between nodes $a$ and $b$
can be calculated using the formula
\[d(a)+d(b)-2 \cdot d(c),\]
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
first ascends from node 5 to node 2
and then descends 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
$3+4-2\cdot2=3$.

\section{Offline algorithms}

So far, we have discussed \emph{online} algorithms
that are able to efficiently process queries
one after another in a given order.
In particular, we may require that the algorithm
processes each query before receiving the next query.

However, in many problems, the online
property is not necessary.
In this section, we focus on \emph{offline} algorithms
that are given a collection of queries that can be
processed in any order.
It is often easier to design an offline algorithm
compared to an online algorithm.

\subsubsection{Merging data structures}

One method to construct an offline algorithm
is to traverse the tree recursively
and maintain data structures for processing queries.
At each node $s$, we create a data structure
$\texttt{d}[s]$ that is based on the
data structures of the children of $s$.
Then, using this data structure,
all queries related to $s$ are processed.

As an example, consider the following problem:
We are given a tree where each node has some value.
Our task is to process queries of the form
''calculate the number of nodes with value $x$
in the subtree of node $s$''.
For example, in the following tree,
the subtree of node $4$ contains two nodes
whose value is 3.

\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}

In this problem, we can use map structures
to answer the queries.
For example, the maps for node 4 and
its children are as follows:

\begin{center}
\begin{tikzpicture}[scale=0.7]

\node[draw, rectangle] (a) at (4,5.5)
{
\footnotesize
\begin{tabular}{rrr}
4 \\
\hline
1 \\
\end{tabular}};

\node[draw, rectangle] (b) at (8,5.5)
{
\footnotesize
\begin{tabular}{rrr}
3 \\
\hline
1 \\
\end{tabular}};


\node[draw, rectangle] (c) at (12,5.5)
{
\footnotesize
\begin{tabular}{rr}
1 \\
\hline
1 \\
\end{tabular}};

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

It would be too slow to create
all data structures from scratch.
Instead, at each node $s$,
we create a data structure $\texttt{d}[s]$ 
that initially only contains the value of $s$.
After this, we go through the children of $s$ and
\emph{merge} $\texttt{d}[s]$ and
all structures of the form
$\texttt{d}[u]$ where $u$ is a child of $s$.

For example, in the above tree, the map
for node $4$ is created by merging the following maps:

\begin{center}
\begin{tikzpicture}[scale=0.7]

\node[draw, rectangle] (a) at (4,5.5)
{
\footnotesize
\begin{tabular}{rrr}
4 \\
\hline
1 \\
\end{tabular}};

\node[draw, rectangle] (b) at (7,5.5)
{
\footnotesize
\begin{tabular}{rrr}
3 \\
\hline
1 \\
\end{tabular}};

\node[draw, rectangle] (c) at (10,5.5)
{
\footnotesize
\begin{tabular}{rr}
1 \\
\hline
1 \\
\end{tabular}};

\node[draw, rectangle] (d) at (1,5.5)
{
\footnotesize
\begin{tabular}{rr}
3 \\
\hline
1 \\
\end{tabular}};

\end{tikzpicture}
\end{center}

Here the first map is the initial data structure
for node 4,
and the other three maps correspond to nodes 7, 8 and 9.

The merging at node $s$ can be done as follows:
We go through the children of $s$
and at each child $u$ merge $\texttt{d}[s]$
and $\texttt{d}[u]$.
We always copy the contents from $\texttt{d}[u]$
to $\texttt{d}[s]$.
However, before this, we \emph{swap}
the contents of $\texttt{d}[s]$ and $\texttt{d}[u]$
if $\texttt{d}[s]$ is smaller than $\texttt{d}[u]$.
By doing this, each value is copied only $O(\log n)$
times during the tree traversal,
which ensures that the algorithm is efficient.

To swap the contents of two data structures $a$ and $b$
efficiently, we can just use the following code:
\begin{lstlisting}
swap(a,b);
\end{lstlisting}
It is guaranteed that the above code works in constant time
when $a$ and $b$ are C++ standard library data structures.

\subsubsection{Lowest common ancestors}