cphb/luku18.tex

\chapter{Tree queries}

\index{tree query}

This chapter discusses techniques for
efficiently performing queries for a rooted tree.
The queries are related to subtrees and paths
in the tree.
For example, possible queries are:

\begin{itemize}
\item what is the $k$th ancestor of node $x$?
\item what is the sum of values in the subtree of node $x$?
\item what is the sum of values in a path between nodes $a$ and $b$?
\item what is the lowest common ancestor of nodes $a$ and $b$?
\end{itemize}

\section{Finding ancestors}

The $k$th ancestor of node $x$ in the tree is found
when we ascend $k$ steps in the tree beginning at node $x$.
Let $f(x,k)$ denote the $k$th ancestor of node $x$.
For example, in the following tree, $f(2,1)=1$ and $f(8,2)=4$.

\begin{center}
\begin{tikzpicture}[scale=0.9]
\node[draw, circle] (1) at (0,3) {$1$};
\node[draw, circle] (2) at (2,1) {$2$};
\node[draw, circle] (3) at (-2,1) {$4$};
\node[draw, circle] (4) at (0,1) {$5$};
\node[draw, circle] (5) at (2,-1) {$6$};
\node[draw, circle] (6) at (-3,-1) {$3$};
\node[draw, circle] (7) at (-1,-1) {$7$};
\node[draw, circle] (8) at (-1,-3) {$8$};
\path[draw,thick,-] (1) -- (2);
\path[draw,thick,-] (1) -- (3);
\path[draw,thick,-] (1) -- (4);
\path[draw,thick,-] (2) -- (5);
\path[draw,thick,-] (3) -- (6);
\path[draw,thick,-] (3) -- (7);
\path[draw,thick,-] (7) -- (8);

\path[draw=red,thick,->,line width=2pt] (8) edge [bend left] (3);
\path[draw=red,thick,->,line width=2pt] (2) edge [bend right] (1);
\end{tikzpicture}
\end{center}

A straighforward way to calculate $f(x,k)$
is to move $k$ steps upwards in the tree
beginning from node $x$.
However, the time complexity of this method
is $O(n)$ because it is possible that the tree
contains a chain of $O(n)$ nodes.

As in Chapter 16.3, any value of $f(x,k)$
can be efficiently calculated in $O(\log k)$
after preprocessing.
The idea is to precalculate all values $f(x,k)$
where $k$ is a power of two.
For example, the values for the tree above
are as follows:

\begin{center}
\begin{tabular}{r|rrrrrrrrr}
$x$ & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 \\
\hline
$f(x,1)$ & 0 & 1 & 4 & 1 & 1 & 2 & 4 & 7 \\
$f(x,2)$ & 0 & 0 & 1 & 0 & 0 & 1 & 1 & 4 \\
$f(x,4)$ & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
$\cdots$ \\
\end{tabular}
\end{center}

The value $0$ means that the $k$th ancestor
of a node doesn't exist.

The preprocessing takes $O(n \log n)$ time
because each node can have at most $n$ ancestors.
After this, any value $f(x,k)$ can be calculated
in $O(\log k)$ time by representing the value $k$
as a sum where each term is a power of two.

\section{Subtrees and paths}

\index{node array}

A \key{node array} contains the nodes of a rooted tree
in the order in which a depth-first search
from the root node visits them.
For example, in the tree
\begin{center}
\begin{tikzpicture}[scale=0.9]
\node[draw, circle] (1) at (0,3) {$1$};
\node[draw, circle] (2) at (-3,1) {$2$};
\node[draw, circle] (3) at (-1,1) {$3$};
\node[draw, circle] (4) at (1,1) {$4$};
\node[draw, circle] (5) at (3,1) {$5$};
\node[draw, circle] (6) at (-3,-1) {$6$};
\node[draw, circle] (7) at (-0.5,-1) {$7$};
\node[draw, circle] (8) at (1,-1) {$8$};
\node[draw, circle] (9) at (2.5,-1) {$9$};

\path[draw,thick,-] (1) -- (2);
\path[draw,thick,-] (1) -- (3);
\path[draw,thick,-] (1) -- (4);
\path[draw,thick,-] (1) -- (5);
\path[draw,thick,-] (2) -- (6);
\path[draw,thick,-] (4) -- (7);
\path[draw,thick,-] (4) -- (8);
\path[draw,thick,-] (4) -- (9);
\end{tikzpicture}
\end{center}
a depth-first search proceeds as follows:
\begin{center}
\begin{tikzpicture}[scale=0.9]
\node[draw, circle] (1) at (0,3) {$1$};
\node[draw, circle] (2) at (-3,1) {$2$};
\node[draw, circle] (3) at (-1,1) {$3$};
\node[draw, circle] (4) at (1,1) {$4$};
\node[draw, circle] (5) at (3,1) {$5$};
\node[draw, circle] (6) at (-3,-1) {$6$};
\node[draw, circle] (7) at (-0.5,-1) {$7$};
\node[draw, circle] (8) at (1,-1) {$8$};
\node[draw, circle] (9) at (2.5,-1) {$9$};

\path[draw,thick,-] (1) -- (2);
\path[draw,thick,-] (1) -- (3);
\path[draw,thick,-] (1) -- (4);
\path[draw,thick,-] (1) -- (5);
\path[draw,thick,-] (2) -- (6);
\path[draw,thick,-] (4) -- (7);
\path[draw,thick,-] (4) -- (8);
\path[draw,thick,-] (4) -- (9);


\path[draw=red,thick,->,line width=2pt] (1) edge [bend right=15] (2);
\path[draw=red,thick,->,line width=2pt] (2) edge [bend right=15] (6);
\path[draw=red,thick,->,line width=2pt] (6) edge [bend right=15] (2);
\path[draw=red,thick,->,line width=2pt] (2) edge [bend right=15] (1);
\path[draw=red,thick,->,line width=2pt] (1) edge [bend right=15] (3);
\path[draw=red,thick,->,line width=2pt] (3) edge [bend right=15] (1);
\path[draw=red,thick,->,line width=2pt] (1) edge [bend right=15] (4);
\path[draw=red,thick,->,line width=2pt] (4) edge [bend right=15] (7);
\path[draw=red,thick,->,line width=2pt] (7) edge [bend right=15] (4);
\path[draw=red,thick,->,line width=2pt] (4) edge [bend right=15] (8);
\path[draw=red,thick,->,line width=2pt] (8) edge [bend right=15] (4);
\path[draw=red,thick,->,line width=2pt] (4) edge [bend right=15] (9);
\path[draw=red,thick,->,line width=2pt] (9) edge [bend right=15] (4);
\path[draw=red,thick,->,line width=2pt] (4) edge [bend right=15] (1);
\path[draw=red,thick,->,line width=2pt] (1) edge [bend right=15] (5);
\path[draw=red,thick,->,line width=2pt] (5) edge [bend right=15] (1);

\end{tikzpicture}
\end{center}
Hence, the corresponding node array is as follows:
\begin{center}
\begin{tikzpicture}[scale=0.7]
\draw (0,0) grid (9,1);

\node at (0.5,0.5) {$1$};
\node at (1.5,0.5) {$2$};
\node at (2.5,0.5) {$6$};
\node at (3.5,0.5) {$3$};
\node at (4.5,0.5) {$4$};
\node at (5.5,0.5) {$7$};
\node at (6.5,0.5) {$8$};
\node at (7.5,0.5) {$9$};
\node at (8.5,0.5) {$5$};

\footnotesize
\node at (0.5,1.4) {$1$};
\node at (1.5,1.4) {$2$};
\node at (2.5,1.4) {$3$};
\node at (3.5,1.4) {$4$};
\node at (4.5,1.4) {$5$};
\node at (5.5,1.4) {$6$};
\node at (6.5,1.4) {$7$};
\node at (7.5,1.4) {$8$};
\node at (8.5,1.4) {$9$};
\end{tikzpicture}
\end{center}

\subsubsection{Subtree queries}

Each subtree of a tree corresponds to a subarray
in the node array,
where the first element is the root node.
For example, the following subarray contains the
nodes in the subtree of node $4$:
\begin{center}
\begin{tikzpicture}[scale=0.7]
\fill[color=lightgray] (4,0) rectangle (8,1);
\draw (0,0) grid (9,1);

\node at (0.5,0.5) {$1$};
\node at (1.5,0.5) {$2$};
\node at (2.5,0.5) {$6$};
\node at (3.5,0.5) {$3$};
\node at (4.5,0.5) {$4$};
\node at (5.5,0.5) {$7$};
\node at (6.5,0.5) {$8$};
\node at (7.5,0.5) {$9$};
\node at (8.5,0.5) {$5$};

\footnotesize
\node at (0.5,1.4) {$1$};
\node at (1.5,1.4) {$2$};
\node at (2.5,1.4) {$3$};
\node at (3.5,1.4) {$4$};
\node at (4.5,1.4) {$5$};
\node at (5.5,1.4) {$6$};
\node at (6.5,1.4) {$7$};
\node at (7.5,1.4) {$8$};
\node at (8.5,1.4) {$9$};
\end{tikzpicture}
\end{center}
Using this fact, we can efficiently process queries
that are related to subtrees of the 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 change the value of node $x$
\item calculate the sum of values in the subtree of node $x$
\end{itemize}

Let us consider the following tree where blue numbers
are values of nodes.
For example, the sum of values in the subtree of node $4$
is $3+4+3+1=11$.

\begin{center}
\begin{tikzpicture}[scale=0.9]
\node[draw, circle] (1) at (0,3) {$1$};
\node[draw, circle] (2) at (-3,1) {$2$};
\node[draw, circle] (3) at (-1,1) {$3$};
\node[draw, circle] (4) at (1,1) {$4$};
\node[draw, circle] (5) at (3,1) {$5$};
\node[draw, circle] (6) at (-3,-1) {$6$};
\node[draw, circle] (7) at (-0.5,-1) {$7$};
\node[draw, circle] (8) at (1,-1) {$8$};
\node[draw, circle] (9) at (2.5,-1) {$9$};

\path[draw,thick,-] (1) -- (2);
\path[draw,thick,-] (1) -- (3);
\path[draw,thick,-] (1) -- (4);
\path[draw,thick,-] (1) -- (5);
\path[draw,thick,-] (2) -- (6);
\path[draw,thick,-] (4) -- (7);
\path[draw,thick,-] (4) -- (8);
\path[draw,thick,-] (4) -- (9);

\node[color=blue] at (0,3+0.65) {2};
\node[color=blue] at (-3-0.65,1) {3};
\node[color=blue] at (-1-0.65,1) {5};
\node[color=blue] at (1+0.65,1) {3};
\node[color=blue] at (3+0.65,1) {1};
\node[color=blue] at (-3,-1-0.65) {4};
\node[color=blue] at (-0.5,-1-0.65) {4};
\node[color=blue] at (1,-1-0.65) {3};
\node[color=blue] at (2.5,-1-0.65) {1};
\end{tikzpicture}
\end{center}

The idea is to construct a node array that contains
three values for each node: (1) identifier of the node,
(2) size of the subtree, and (3) value of the node.
For example, the array for the above tree is as follows:

\begin{center}
\begin{tikzpicture}[scale=0.7]
\draw (0,1) grid (9,-2);

\node at (0.5,0.5) {$1$};
\node at (1.5,0.5) {$2$};
\node at (2.5,0.5) {$6$};
\node at (3.5,0.5) {$3$};
\node at (4.5,0.5) {$4$};
\node at (5.5,0.5) {$7$};
\node at (6.5,0.5) {$8$};
\node at (7.5,0.5) {$9$};
\node at (8.5,0.5) {$5$};

\node at (0.5,-0.5) {$9$};
\node at (1.5,-0.5) {$2$};
\node at (2.5,-0.5) {$1$};
\node at (3.5,-0.5) {$1$};
\node at (4.5,-0.5) {$4$};
\node at (5.5,-0.5) {$1$};
\node at (6.5,-0.5) {$1$};
\node at (7.5,-0.5) {$1$};
\node at (8.5,-0.5) {$1$};

\node at (0.5,-1.5) {$2$};
\node at (1.5,-1.5) {$3$};
\node at (2.5,-1.5) {$4$};
\node at (3.5,-1.5) {$5$};
\node at (4.5,-1.5) {$3$};
\node at (5.5,-1.5) {$4$};
\node at (6.5,-1.5) {$3$};
\node at (7.5,-1.5) {$1$};
\node at (8.5,-1.5) {$1$};

\footnotesize
\node at (0.5,1.4) {$1$};
\node at (1.5,1.4) {$2$};
\node at (2.5,1.4) {$3$};
\node at (3.5,1.4) {$4$};
\node at (4.5,1.4) {$5$};
\node at (5.5,1.4) {$6$};
\node at (6.5,1.4) {$7$};
\node at (7.5,1.4) {$8$};
\node at (8.5,1.4) {$9$};
\end{tikzpicture}
\end{center}

Using this array, we can calculate the sum of nodes
in a subtree by first reading the size of the subtree
and then the values of the corresponding nodes.
For example, the values in the subtree of node $4$
can be found as follows:

\begin{center}
\begin{tikzpicture}[scale=0.7]
\fill[color=lightgray] (4,1) rectangle (5,0);
\fill[color=lightgray] (4,0) rectangle (5,-1);
\fill[color=lightgray] (4,-1) rectangle (8,-2);
\draw (0,1) grid (9,-2);

\node at (0.5,0.5) {$1$};
\node at (1.5,0.5) {$2$};
\node at (2.5,0.5) {$6$};
\node at (3.5,0.5) {$3$};
\node at (4.5,0.5) {$4$};
\node at (5.5,0.5) {$7$};
\node at (6.5,0.5) {$8$};
\node at (7.5,0.5) {$9$};
\node at (8.5,0.5) {$5$};

\node at (0.5,-0.5) {$9$};
\node at (1.5,-0.5) {$2$};
\node at (2.5,-0.5) {$1$};
\node at (3.5,-0.5) {$1$};
\node at (4.5,-0.5) {$4$};
\node at (5.5,-0.5) {$1$};
\node at (6.5,-0.5) {$1$};
\node at (7.5,-0.5) {$1$};
\node at (8.5,-0.5) {$1$};

\node at (0.5,-1.5) {$2$};
\node at (1.5,-1.5) {$3$};
\node at (2.5,-1.5) {$4$};
\node at (3.5,-1.5) {$5$};
\node at (4.5,-1.5) {$3$};
\node at (5.5,-1.5) {$4$};
\node at (6.5,-1.5) {$3$};
\node at (7.5,-1.5) {$1$};
\node at (8.5,-1.5) {$1$};

\footnotesize
\node at (0.5,1.4) {$1$};
\node at (1.5,1.4) {$2$};
\node at (2.5,1.4) {$3$};
\node at (3.5,1.4) {$4$};
\node at (4.5,1.4) {$5$};
\node at (5.5,1.4) {$6$};
\node at (6.5,1.4) {$7$};
\node at (7.5,1.4) {$8$};
\node at (8.5,1.4) {$9$};
\end{tikzpicture}
\end{center}

The remaining step is to store the values of the
nodes in a binary indexed tree or segment tree.
After this, we can both calculate the sum
of values and change a value in $O(\log n)$ time,
so we can efficiently process the queries.

\subsubsection{Path queries}

Using a node array, we can also efficiently
process paths between the root node and any other
node in the tree.
Let us next consider a problem where our task
is to support the following queries:
\begin{itemize}
\item change the value of node $x$
\item calculate the sum of values from the root to node $x$
\end{itemize}

For example, in the following tree, the sum of
values from the root to node 8 is $4+5+3=12$.

\begin{center}
\begin{tikzpicture}[scale=0.9]
\node[draw, circle] (1) at (0,3) {$1$};
\node[draw, circle] (2) at (-3,1) {$2$};
\node[draw, circle] (3) at (-1,1) {$3$};
\node[draw, circle] (4) at (1,1) {$4$};
\node[draw, circle] (5) at (3,1) {$5$};
\node[draw, circle] (6) at (-3,-1) {$6$};
\node[draw, circle] (7) at (-0.5,-1) {$7$};
\node[draw, circle] (8) at (1,-1) {$8$};
\node[draw, circle] (9) at (2.5,-1) {$9$};

\path[draw,thick,-] (1) -- (2);
\path[draw,thick,-] (1) -- (3);
\path[draw,thick,-] (1) -- (4);
\path[draw,thick,-] (1) -- (5);
\path[draw,thick,-] (2) -- (6);
\path[draw,thick,-] (4) -- (7);
\path[draw,thick,-] (4) -- (8);
\path[draw,thick,-] (4) -- (9);

\node[color=blue] at (0,3+0.65) {4};
\node[color=blue] at (-3-0.65,1) {5};
\node[color=blue] at (-1-0.65,1) {3};
\node[color=blue] at (1+0.65,1) {5};
\node[color=blue] at (3+0.65,1) {2};
\node[color=blue] at (-3,-1-0.65) {3};
\node[color=blue] at (-0.5,-1-0.65) {5};
\node[color=blue] at (1,-1-0.65) {3};
\node[color=blue] at (2.5,-1-0.65) {1};
\end{tikzpicture}
\end{center}

To solve this problem, we can use a similar
technique as we used for subtree queries,
but the values of the nodes are stored
in a special way:
if the value of a node at index $k$ 
increases by $a$,
the value at index $k$ increases by $a$
and the value at index $k+c$ decreases by $a$,
where $c$ is the size of the subtree.

\begin{samepage}
For example, the following array corresponds to the above tree:
\begin{center}
\begin{tikzpicture}[scale=0.7]
\draw (0,1) grid (10,-2);

\node at (0.5,0.5) {$1$};
\node at (1.5,0.5) {$2$};
\node at (2.5,0.5) {$6$};
\node at (3.5,0.5) {$3$};
\node at (4.5,0.5) {$4$};
\node at (5.5,0.5) {$7$};
\node at (6.5,0.5) {$8$};
\node at (7.5,0.5) {$9$};
\node at (8.5,0.5) {$5$};
\node at (9.5,0.5) {--};

\node at (0.5,-0.5) {$9$};
\node at (1.5,-0.5) {$2$};
\node at (2.5,-0.5) {$1$};
\node at (3.5,-0.5) {$1$};
\node at (4.5,-0.5) {$4$};
\node at (5.5,-0.5) {$1$};
\node at (6.5,-0.5) {$1$};
\node at (7.5,-0.5) {$1$};
\node at (8.5,-0.5) {$1$};
\node at (9.5,-0.5) {--};

\node at (0.5,-1.5) {$4$};
\node at (1.5,-1.5) {$5$};
\node at (2.5,-1.5) {$3$};
\node at (3.5,-1.5) {$-5$};
\node at (4.5,-1.5) {$2$};
\node at (5.5,-1.5) {$5$};
\node at (6.5,-1.5) {$-2$};
\node at (7.5,-1.5) {$-2$};
\node at (8.5,-1.5) {$-4$};
\node at (9.5,-1.5) {$-4$};

\footnotesize
\node at (0.5,1.4) {$1$};
\node at (1.5,1.4) {$2$};
\node at (2.5,1.4) {$3$};
\node at (3.5,1.4) {$4$};
\node at (4.5,1.4) {$5$};
\node at (5.5,1.4) {$6$};
\node at (6.5,1.4) {$7$};
\node at (7.5,1.4) {$8$};
\node at (8.5,1.4) {$9$};
\node at (9.5,1.4) {$10$};
\end{tikzpicture}
\end{center}
\end{samepage}

For example, the value of node $3$ is $-5$,
because it is the next node after the subtrees
of nodes $2$ and $6$ and its own value is $3$.
So the value decreases by $5+3$ and increases by $3$.
Note that the array contains an extra index 10
that only has the opposite number of the value
of the root node.

Using this array, the sum of values in a path
from the root to node $x$ equals the sum
of values in the array from the beginning to node $x$.
For example, the sum from the root to node $8$
can be calculated as follows:

\begin{center}
\begin{tikzpicture}[scale=0.7]
\fill[color=lightgray] (6,1) rectangle (7,0);
\fill[color=lightgray] (0,-1) rectangle (7,-2);
\draw (0,1) grid (10,-2);

\node at (0.5,0.5) {$1$};
\node at (1.5,0.5) {$2$};
\node at (2.5,0.5) {$6$};
\node at (3.5,0.5) {$3$};
\node at (4.5,0.5) {$4$};
\node at (5.5,0.5) {$7$};
\node at (6.5,0.5) {$8$};
\node at (7.5,0.5) {$9$};
\node at (8.5,0.5) {$5$};
\node at (9.5,0.5) {--};

\node at (0.5,-0.5) {$9$};
\node at (1.5,-0.5) {$2$};
\node at (2.5,-0.5) {$1$};
\node at (3.5,-0.5) {$1$};
\node at (4.5,-0.5) {$4$};
\node at (5.5,-0.5) {$1$};
\node at (6.5,-0.5) {$1$};
\node at (7.5,-0.5) {$1$};
\node at (8.5,-0.5) {$1$};
\node at (9.5,-0.5) {--};

\node at (0.5,-1.5) {$4$};
\node at (1.5,-1.5) {$5$};
\node at (2.5,-1.5) {$3$};
\node at (3.5,-1.5) {$-5$};
\node at (4.5,-1.5) {$2$};
\node at (5.5,-1.5) {$5$};
\node at (6.5,-1.5) {$-2$};
\node at (7.5,-1.5) {$-2$};
\node at (8.5,-1.5) {$-4$};
\node at (9.5,-1.5) {$-4$};

\footnotesize
\node at (0.5,1.4) {$1$};
\node at (1.5,1.4) {$2$};
\node at (2.5,1.4) {$3$};
\node at (3.5,1.4) {$4$};
\node at (4.5,1.4) {$5$};
\node at (5.5,1.4) {$6$};
\node at (6.5,1.4) {$7$};
\node at (7.5,1.4) {$8$};
\node at (8.5,1.4) {$9$};
\node at (9.5,1.4) {$10$};
\end{tikzpicture}
\end{center}
The sum is
\[4+5+3-5+2+5-2=12,\]
that equals the sum $4+5+3=12$.
This method works because the value of each node
is added to the sum when the depth-first search
visits it for the first time, and correspondingly,
the value is removed from the sum when the subtree of the
node has been processed.

Once again, we can store the values of the nodes
in a binary indexed tree or a segment tree,
so it is possible to both calculate the sum of values and
change a value efficiently in $O(\log n)$ time.

\section{Lowest common ancestor}

\index{lowest common ancestor}

The \key{lowest common ancestor}
of two nodes is a the lowest node in the tree
whose subtree contains both the nodes.
A typical problem is to efficiently process
queries where the task is to find the lowest
common ancestor of two nodes.
For example, in the tree
\begin{center}
\begin{tikzpicture}[scale=0.9]
\node[draw, circle] (1) at (0,3) {$1$};
\node[draw, circle] (2) at (2,1) {$4$};
\node[draw, circle] (3) at (-2,1) {$2$};
\node[draw, circle] (4) at (0,1) {$3$};
\node[draw, circle] (5) at (2,-1) {$7$};
\node[draw, circle] (6) at (-3,-1) {$5$};
\node[draw, circle] (7) at (-1,-1) {$6$};
\node[draw, circle] (8) at (-1,-3) {$8$};
\path[draw,thick,-] (1) -- (2);
\path[draw,thick,-] (1) -- (3);
\path[draw,thick,-] (1) -- (4);
\path[draw,thick,-] (2) -- (5);
\path[draw,thick,-] (3) -- (6);
\path[draw,thick,-] (3) -- (7);
\path[draw,thick,-] (7) -- (8);
\end{tikzpicture}
\end{center}
the lowest common ancestor of nodes 5 and 8 is node 2,
and the lowest common ancestor of nodes 3 and 4 is node 1.

Next we will discuss two efficient techniques for
finding the lowest common ancestor of two nodes.

\subsubsection{Method 1}

One way to solve the problem is use the fact
that we can efficiently find the $k$th
ancestor of any node in the tree.
Using this idea, we can first ensure that
both nodes are at the same level in the tree,
and then find the smallest value of $k$
where the $k$th ancestor of both nodes is the same.

As an example, let's find the lowest common
ancestor of nodes $5$ and $8$ in the following tree:

\begin{center}
\begin{tikzpicture}[scale=0.9]
\node[draw, circle] (1) at (0,3) {$1$};
\node[draw, circle] (2) at (2,1) {$4$};
\node[draw, circle] (3) at (-2,1) {$2$};
\node[draw, circle] (4) at (0,1) {$3$};
\node[draw, circle] (5) at (2,-1) {$7$};
\node[draw, circle,fill=lightgray] (6) at (-3,-1) {$5$};
\node[draw, circle] (7) at (-1,-1) {$6$};
\node[draw, circle,fill=lightgray] (8) at (-1,-3) {$8$};
\path[draw,thick,-] (1) -- (2);
\path[draw,thick,-] (1) -- (3);
\path[draw,thick,-] (1) -- (4);
\path[draw,thick,-] (2) -- (5);
\path[draw,thick,-] (3) -- (6);
\path[draw,thick,-] (3) -- (7);
\path[draw,thick,-] (7) -- (8);
\end{tikzpicture}
\end{center}

Node $5$ is at level $3$, while node $8$ is at level $4$.
Thus, we first move one step upwards from node $8$ to node $6$.
After this, it turns out that the parent of both node $5$
and node $6$ is node $2$, so we have found the lowest common ancestor.

\begin{center}
\begin{tikzpicture}[scale=0.9]
\node[draw, circle] (1) at (0,3) {$1$};
\node[draw, circle] (2) at (2,1) {$4$};
\node[draw, circle] (3) at (-2,1) {$2$};
\node[draw, circle] (4) at (0,1) {$3$};
\node[draw, circle] (5) at (2,-1) {$7$};
\node[draw, circle,fill=lightgray] (6) at (-3,-1) {$5$};
\node[draw, circle] (7) at (-1,-1) {$6$};
\node[draw, circle,fill=lightgray] (8) at (-1,-3) {$8$};
\path[draw,thick,-] (1) -- (2);
\path[draw,thick,-] (1) -- (3);
\path[draw,thick,-] (1) -- (4);
\path[draw,thick,-] (2) -- (5);
\path[draw,thick,-] (3) -- (6);
\path[draw,thick,-] (3) -- (7);
\path[draw,thick,-] (7) -- (8);

\path[draw=red,thick,->,line width=2pt] (6) edge [bend left] (3);
\path[draw=red,thick,->,line width=2pt] (8) edge [bend right] (7);
\path[draw=red,thick,->,line width=2pt] (7) edge [bend right] (3);
\end{tikzpicture}
\end{center}

Using this method, we can find the lowest common ancestor
of any two nodes in $O(\log n)$ time after an $O(n \log n)$ time
preprocessing, because both steps can be
done in $O(\log n)$ time.

\subsubsection{Method 2}

Another way to solve the problem is based on
a node array.
Again, the idea is to traverse the nodes
using a depth-first search:

\begin{center}
\begin{tikzpicture}[scale=0.9]
\node[draw, circle] (1) at (0,3) {$1$};
\node[draw, circle] (2) at (2,1) {$4$};
\node[draw, circle] (3) at (-2,1) {$2$};
\node[draw, circle] (4) at (0,1) {$3$};
\node[draw, circle] (5) at (2,-1) {$7$};
\node[draw, circle] (6) at (-3,-1) {$5$};
\node[draw, circle] (7) at (-1,-1) {$6$};
\node[draw, circle] (8) at (-1,-3) {$8$};
\path[draw,thick,-] (1) -- (2);
\path[draw,thick,-] (1) -- (3);
\path[draw,thick,-] (1) -- (4);
\path[draw,thick,-] (2) -- (5);
\path[draw,thick,-] (3) -- (6);
\path[draw,thick,-] (3) -- (7);
\path[draw,thick,-] (7) -- (8);

\path[draw=red,thick,->,line width=2pt] (1) edge [bend right=15] (3);
\path[draw=red,thick,->,line width=2pt] (3) edge [bend right=15] (6);
\path[draw=red,thick,->,line width=2pt] (6) edge [bend right=15] (3);
\path[draw=red,thick,->,line width=2pt] (3) edge [bend right=15] (7);
\path[draw=red,thick,->,line width=2pt] (7) edge [bend right=15] (8);
\path[draw=red,thick,->,line width=2pt] (8) edge [bend right=15] (7);
\path[draw=red,thick,->,line width=2pt] (7) edge [bend right=15] (3);
\path[draw=red,thick,->,line width=2pt] (3) edge [bend right=15] (1);
\path[draw=red,thick,->,line width=2pt] (1) edge [bend right=15] (4);
\path[draw=red,thick,->,line width=2pt] (4) edge [bend right=15] (1);
\path[draw=red,thick,->,line width=2pt] (1) edge [bend right=15] (2);
\path[draw=red,thick,->,line width=2pt] (2) edge [bend right=15] (5);
\path[draw=red,thick,->,line width=2pt] (5) edge [bend right=15] (2);
\path[draw=red,thick,->,line width=2pt] (2) edge [bend right=15] (1);
\end{tikzpicture}
\end{center}

However, we add each node to the node array \emph{always}
when the depth-first search visits the node,
and not only at the first visit.
Thus, a node that has $k$ children appears $k+1$ times
in the node array, and there are a total of $2n-1$
nodes in the array.

We store two values in the array:
(1) identifier of the node, and (2) the level of the 
node in the tree.
The following array corresponds to the above tree:

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

\draw (0,1) grid (15,2);
%\node at (-1.1,1.5) {\texttt{node}};
\node at (0.5,1.5) {$1$};
\node at (1.5,1.5) {$2$};
\node at (2.5,1.5) {$5$};
\node at (3.5,1.5) {$2$};
\node at (4.5,1.5) {$6$};
\node at (5.5,1.5) {$8$};
\node at (6.5,1.5) {$6$};
\node at (7.5,1.5) {$2$};
\node at (8.5,1.5) {$1$};
\node at (9.5,1.5) {$3$};
\node at (10.5,1.5) {$1$};
\node at (11.5,1.5) {$4$};
\node at (12.5,1.5) {$7$};
\node at (13.5,1.5) {$4$};
\node at (14.5,1.5) {$1$};


\draw (0,0) grid (15,1);
%\node at (-1.1,0.5) {\texttt{depth}};
\node at (0.5,0.5) {$1$};
\node at (1.5,0.5) {$2$};
\node at (2.5,0.5) {$3$};
\node at (3.5,0.5) {$2$};
\node at (4.5,0.5) {$3$};
\node at (5.5,0.5) {$4$};
\node at (6.5,0.5) {$3$};
\node at (7.5,0.5) {$2$};
\node at (8.5,0.5) {$1$};
\node at (9.5,0.5) {$2$};
\node at (10.5,0.5) {$1$};
\node at (11.5,0.5) {$2$};
\node at (12.5,0.5) {$3$};
\node at (13.5,0.5) {$2$};
\node at (14.5,0.5) {$1$};

\footnotesize
\node at (0.5,2.5) {$1$};
\node at (1.5,2.5) {$2$};
\node at (2.5,2.5) {$3$};
\node at (3.5,2.5) {$4$};
\node at (4.5,2.5) {$5$};
\node at (5.5,2.5) {$6$};
\node at (6.5,2.5) {$7$};
\node at (7.5,2.5) {$8$};
\node at (8.5,2.5) {$9$};
\node at (9.5,2.5) {$10$};
\node at (10.5,2.5) {$11$};
\node at (11.5,2.5) {$12$};
\node at (12.5,2.5) {$13$};
\node at (13.5,2.5) {$14$};
\node at (14.5,2.5) {$15$};
\end{tikzpicture}
\end{center}

Using this array, we can find the lowest common ancestor
of nodes $a$ and $b$ by locating the node with lowest level
between nodes $a$ and $b$ in the array.
For example, the lowest common ancestor of nodes $5$ and $8$
can be found as follows:

\begin{center}
\begin{tikzpicture}[scale=0.7]
\fill[color=lightgray] (2,1) rectangle (3,2);
\fill[color=lightgray] (5,1) rectangle (6,2);
\fill[color=lightgray] (2,0) rectangle (6,1);

\node at (3.5,-0.5) {$\uparrow$};

\draw (0,1) grid (15,2);
\node at (0.5,1.5) {$1$};
\node at (1.5,1.5) {$2$};
\node at (2.5,1.5) {$5$};
\node at (3.5,1.5) {$2$};
\node at (4.5,1.5) {$6$};
\node at (5.5,1.5) {$8$};
\node at (6.5,1.5) {$6$};
\node at (7.5,1.5) {$2$};
\node at (8.5,1.5) {$1$};
\node at (9.5,1.5) {$3$};
\node at (10.5,1.5) {$1$};
\node at (11.5,1.5) {$4$};
\node at (12.5,1.5) {$7$};
\node at (13.5,1.5) {$4$};
\node at (14.5,1.5) {$1$};


\draw (0,0) grid (15,1);
\node at (0.5,0.5) {$1$};
\node at (1.5,0.5) {$2$};
\node at (2.5,0.5) {$3$};
\node at (3.5,0.5) {$2$};
\node at (4.5,0.5) {$3$};
\node at (5.5,0.5) {$4$};
\node at (6.5,0.5) {$3$};
\node at (7.5,0.5) {$2$};
\node at (8.5,0.5) {$1$};
\node at (9.5,0.5) {$2$};
\node at (10.5,0.5) {$1$};
\node at (11.5,0.5) {$2$};
\node at (12.5,0.5) {$3$};
\node at (13.5,0.5) {$2$};
\node at (14.5,0.5) {$1$};

\footnotesize
\node at (0.5,2.5) {$1$};
\node at (1.5,2.5) {$2$};
\node at (2.5,2.5) {$3$};
\node at (3.5,2.5) {$4$};
\node at (4.5,2.5) {$5$};
\node at (5.5,2.5) {$6$};
\node at (6.5,2.5) {$7$};
\node at (7.5,2.5) {$8$};
\node at (8.5,2.5) {$9$};
\node at (9.5,2.5) {$10$};
\node at (10.5,2.5) {$11$};
\node at (11.5,2.5) {$12$};
\node at (12.5,2.5) {$13$};
\node at (13.5,2.5) {$14$};
\node at (14.5,2.5) {$15$};
\end{tikzpicture}
\end{center}

Node 5 is at index 3, node 8 is at index 6,
and the node with lowest level between
indices $3 \ldots 6$ is node 2 at index 4
whose level is 2.
Thus, the lowest common ancestor of
nodes 5 and 8 is node 2.

Using a segment tree, we can find the lowest
common ancestor in $O(\log n)$ time.
Since the array is static, the time complexity
$O(1)$ is also possible, but this is rarely needed.
In both cases, preprocessing takes $O(n \log n)$ time.

\subsubsection{Distances of nodes}

Finally, let's consider a problem where
each query asks to find the distance between
two nodes in the tree, i.e., the length of the
path between them.
It turns out that this problem reduces to
finding the lowest common ancestor.

First, we choose an arbitrary node for the
root of the tree.
After this, the distance between nodes $a$ and $b$
is $d(a)+d(b)-2 \cdot d(c)$,
where $c$ is the lowest common ancestor,
and $d(s)$ is the distance from the root node
to node $s$.
For example, in the tree

\begin{center}
\begin{tikzpicture}[scale=0.9]
\node[draw, circle] (1) at (0,3) {$1$};
\node[draw, circle] (2) at (2,1) {$4$};
\node[draw, circle] (3) at (-2,1) {$2$};
\node[draw, circle] (4) at (0,1) {$3$};
\node[draw, circle] (5) at (2,-1) {$7$};
\node[draw, circle] (6) at (-3,-1) {$5$};
\node[draw, circle] (7) at (-1,-1) {$6$};
\node[draw, circle] (8) at (-1,-3) {$8$};
\path[draw,thick,-] (1) -- (2);
\path[draw,thick,-] (1) -- (3);
\path[draw,thick,-] (1) -- (4);
\path[draw,thick,-] (2) -- (5);
\path[draw,thick,-] (3) -- (6);
\path[draw,thick,-] (3) -- (7);
\path[draw,thick,-] (7) -- (8);

\path[draw=red,thick,-,line width=2pt] (8) -- node[font=\small] {} (7);
\path[draw=red,thick,-,line width=2pt] (7) -- node[font=\small] {} (3);
\path[draw=red,thick,-,line width=2pt] (6) -- node[font=\small] {} (3);
\end{tikzpicture}
\end{center}
the lowest common ancestor of nodes 5 and 8 is node 2.
A path from node 5 to node 8
goes first upwards from node 5 to node 2,
and then downwards from node 2 to node 8.
The distances of the nodes from the root are
$d(5)=3$, $d(8)=4$ and $d(2)=2$,
so the distance between nodes 5 and 8 is
$3+4-2\cdot2=3$.