The following is an attempt to describe the Ukkonen algorithm by first showing what it does when the string is simple (i.e. does not contain any repeated characters), and then extending it to the full algorithm.
First, a few preliminary statements.

What we are building, is basically like a search trie. So there
is a root node, edges going out of it leading to new nodes, and
further edges going out of those, and so forth 
But: Unlike in a search trie, the edge labels are not single
characters. Instead, each edge is labeled using a pair of integers
[from,to]
. These are pointers into the text. In this sense, each
edge carries a string label of arbitrary length, but takes only O(1)
space (two pointers).
Basic principle
I would like to first demonstrate how to create the suffix tree of a
particularly simple string, a string with no repeated characters:
abc
The algorithm works in steps, from left to right. There is one step for every character of the string. Each step might involve more than one individual operation, but we will see (see the final observations at the end) that the total number of operations is O(n).
So, we start from the left, and first insert only the single character
a
by creating an edge from the root node (on the left) to a leaf,
and labeling it as [0,#]
, which means the edge represents the
substring starting at position 0 and ending at the current end. I
use the symbol #
to mean the current end, which is at position 1
(right after a
).
So we have an initial tree, which looks like this:
And what it means is this:
Now we progress to position 2 (right after b
). Our goal at each step
is to insert all suffixes up to the current position. We do this
by
 expanding the existing
a
edge toab
 inserting one new edge for
b
In our representation this looks like
And what it means is:
We observe two things:
 The edge representation for
ab
is the same as it used to be
in the initial tree:[0,#]
. Its meaning has automatically changed
because we updated the current position#
from 1 to 2.  Each edge consumes O(1) space, because it consists of only two
pointers into the text, regardless of how many characters it
represents.
Next we increment the position again and update the tree by appending
a c
to every existing edge and inserting one new edge for the new
suffix c
.
In our representation this looks like
And what it means is:
We observe:
 The tree is the correct suffix tree up to the current position
after each step  There are as many steps as there are characters in the text
 The amount of work in each step is O(1), because all existing edges
are updated automatically by incrementing#
, and inserting the
one new edge for the final character can be done in O(1)
time. Hence for a string of length n, only O(n) time is required.
First extension: Simple repetitions
Of course this works so nicely only because our string does not
contain any repetitions. We now look at a more realistic string:
abcabxabcd
It starts with abc
as in the previous example, then ab
is repeated
and followed by x
, and then abc
is repeated followed by d
.
Steps 1 through 3: After the first 3 steps we have the tree from the previous example:
Step 4: We move #
to position 4. This implicitly updates all existing
edges to this:
and we need to insert the final suffix of the current step, a
, at
the root.
Before we do this, we introduce two more variables (in addition to
#
), which of course have been there all the time but we haven’t used
them so far:
 The active point, which is a triple
(active_node,active_edge,active_length)
 The
remainder
, which is an integer indicating how many new suffixes
we need to insert
The exact meaning of these two will become clear soon, but for now
let’s just say:
 In the simple
abc
example, the active point was always
(root,'\0x',0)
, i.e.active_node
was the root node,active_edge
was specified as the null character'\0x'
, andactive_length
was zero. The effect of this was that the one new edge that
we inserted in every step was inserted at the root node as a
freshly created edge. We will see soon why a triple is necessary to
represent this information.  The
remainder
was always set to 1 at the beginning of each
step. The meaning of this was that the number of suffixes we had to
actively insert at the end of each step was 1 (always just the
final character).
Now this is going to change. When we insert the current final
character a
at the root, we notice that there is already an outgoing
edge starting with a
, specifically: abca
. Here is what we do in
such a case:
 We do not insert a fresh edge
[4,#]
at the root node. Instead we
simply notice that the suffixa
is already in our
tree. It ends in the middle of a longer edge, but we are not
bothered by that. We just leave things the way they are.  We set the active point to
(root,'a',1)
. That means the active
point is now somewhere in the middle of outgoing edge of the root node that starts witha
, specifically, after position 1 on that edge. We
notice that the edge is specified simply by its first
charactera
. That suffices because there can be only one edge
starting with any particular character (confirm that this is true after reading through the entire description).  We also increment
remainder
, so at the beginning of the next step
it will be 2.
Observation: When the final suffix we need to insert is found to
exist in the tree already, the tree itself is not changed at all (we only update the active point and remainder
). The tree
is then not an accurate representation of the suffix tree up to the
current position any more, but it contains all suffixes (because the final
suffix a
is contained implicitly). Hence, apart from updating the
variables (which are all of fixed length, so this is O(1)), there was
no work done in this step.
Step 5: We update the current position #
to 5. This
automatically updates the tree to this:
And because remainder
is 2, we need to insert two final
suffixes of the current position: ab
and b
. This is basically because:
 The
a
suffix from the previous step has never been properly
inserted. So it has remained, and since we have progressed one
step, it has now grown froma
toab
.  And we need to insert the new final edge
b
.
In practice this means that we go to the active point (which points to
behind the a
on what is now the abcab
edge), and insert the
current final character b
. But: Again, it turns out that b
is
also already present on that same edge.
So, again, we do not change the tree. We simply:
 Update the active point to
(root,'a',2)
(same node and edge
as before, but now we point to behind theb
)  Increment the
remainder
to 3 because we still have not properly
inserted the final edge from the previous step, and we don’t insert
the current final edge either.
To be clear: We had to insert ab
and b
in the current step, but
because ab
was already found, we updated the active point and did
not even attempt to insert b
. Why? Because if ab
is in the tree,
every suffix of it (including b
) must be in the tree,
too. Perhaps only implicitly, but it must be there, because of the
way we have built the tree so far.
We proceed to step 6 by incrementing #
. The tree is
automatically updated to:
Because remainder
is 3, we have to insert abx
, bx
and
x
. The active point tells us where ab
ends, so we only need to
jump there and insert the x
. Indeed, x
is not there yet, so we
split the abcabx
edge and insert an internal node:
The edge representations are still pointers into the text, so
splitting and inserting an internal node can be done in O(1) time.
So we have dealt with abx
and decrement remainder
to 2. Now we
need to insert the next remaining suffix, bx
. But before we do that
we need to update the active point. The rule for this, after splitting
and inserting an edge, will be called Rule 1 below, and it applies whenever the
active_node
is root (we will learn rule 3 for other cases further
below). Here is rule 1:
After an insertion from root,
active_node
remains rootactive_edge
is set to the first character of the new suffix we
need to insert, i.e.b
active_length
is reduced by 1
Hence, the new activepoint triple (root,'b',1)
indicates that the
next insert has to be made at the bcabx
edge, behind 1 character,
i.e. behind b
. We can identify the insertion point in O(1) time and
check whether x
is already present or not. If it was present, we
would end the current step and leave everything the way it is. But x
is not present, so we insert it by splitting the edge:
Again, this took O(1) time and we update remainder
to 1 and the
active point to (root,'x',0)
as rule 1 states.
But there is one more thing we need to do. We’ll call this Rule 2:
If we split an edge and insert a new node, and if that is not the
first node created during the current step, we connect the previously
inserted node and the new node through a special pointer, a suffix
link. We will later see why that is useful. Here is what we get, the
suffix link is represented as a dotted edge:
We still need to insert the final suffix of the current step,
x
. Since the active_length
component of the active node has fallen
to 0, the final insert is made at the root directly. Since there is no
outgoing edge at the root node starting with x
, we insert a new
edge:
As we can see, in the current step all remaining inserts were made.
We proceed to step 7 by setting #
=7, which automatically appends the next character,
a
, to all leaf edges, as always. Then we attempt to insert the new final
character to the active point (the root), and find that it is there
already. So we end the current step without inserting anything and
update the active point to (root,'a',1)
.
In step 8, #
=8, we append b
, and as seen before, this only
means we update the active point to (root,'a',2)
and increment remainder
without doing
anything else, because b
is already present. However, we notice (in O(1) time) that the active point
is now at the end of an edge. We reflect this by resetting it to
(node1,'\0x',0)
. Here, I use node1
to refer to the
internal node the ab
edge ends at.
Then, in step #
=9, we need to insert ‘c’ and this will help us to
understand the final trick:
Second extension: Using suffix links
As always, the #
update appends c
automatically to the leaf edges
and we go to the active point to see if we can insert ‘c’. It turns
out ‘c’ exists already at that edge, so we set the active point to
(node1,'c',1)
, increment remainder
and do nothing else.
Now in step #
=10, remainder
is 4, and so we first need to insert
abcd
(which remains from 3 steps ago) by inserting d
at the active
point.
Attempting to insert d
at the active point causes an edge split in
O(1) time:
The active_node
, from which the split was initiated, is marked in
red above. Here is the final rule, Rule 3:
After splitting an edge from an
active_node
that is not the root
node, we follow the suffix link going out of that node, if there is
any, and reset theactive_node
to the node it points to. If there is
no suffix link, we set theactive_node
to the root.active_edge
andactive_length
remain unchanged.
So the active point is now (node2,'c',1)
, and node2
is marked in
red below:
Since the insertion of abcd
is complete, we decrement remainder
to
3 and consider the next remaining suffix of the current step,
bcd
. Rule 3 has set the active point to just the right node and edge
so inserting bcd
can be done by simply inserting its final character
d
at the active point.
Doing this causes another edge split, and because of rule 2, we
must create a suffix link from the previously inserted node to the new
one:
We observe: Suffix links enable us to reset the active point so we
can make the next remaining insert at O(1) effort. Look at the
graph above to confirm that indeed node at label ab
is linked to
the node at b
(its suffix), and the node at abc
is linked to
bc
.
The current step is not finished yet. remainder
is now 2, and we
need to follow rule 3 to reset the active point again. Since the
current active_node
(red above) has no suffix link, we reset to
root. The active point is now (root,'c',1)
.
Hence the next insert occurs at the one outgoing edge of the root node
whose label starts with c
: cabxabcd
, behind the first character,
i.e. behind c
. This causes another split:
And since this involves the creation of a new internal node,we follow
rule 2 and set a new suffix link from the previously created internal
node:
(I am using Graphviz Dot for these little
graphs. The new suffix link caused dot to rearrange the existing
edges, so check carefully to confirm that the only thing that was
inserted above is a new suffix link.)
With this, remainder
can be set to 1 and since the active_node
is
root, we use rule 1 to update the active point to (root,'d',0)
. This
means the final insert of the current step is to insert a single d
at root:
That was the final step and we are done. There are number of final
observations, though:

In each step we move
#
forward by 1 position. This automatically
updates all leaf nodes in O(1) time. 
But it does not deal with a) any suffixes remaining from previous
steps, and b) with the one final character of the current step. 
remainder
tells us how many additional inserts we need to
make. These inserts correspond onetoone to the final suffixes of
the string that ends at the current position#
. We consider one
after the other and make the insert. Important: Each insert is
done in O(1) time since the active point tells us exactly where to
go, and we need to add only one single character at the active
point. Why? Because the other characters are contained implicitly
(otherwise the active point would not be where it is). 
After each such insert, we decrement
remainder
and follow the
suffix link if there is any. If not we go to root (rule 3). If we
are at root already, we modify the active point using rule 1. In
any case, it takes only O(1) time. 
If, during one of these inserts, we find that the character we want
to insert is already there, we don’t do anything and end the
current step, even ifremainder
>0. The reason is that any
inserts that remain will be suffixes of the one we just tried to
make. Hence they are all implicit in the current tree. The fact
thatremainder
>0 makes sure we deal with the remaining suffixes
later. 
What if at the end of the algorithm
remainder
>0? This will be the
case whenever the end of the text is a substring that occurred
somewhere before. In that case we must append one extra character
at the end of the string that has not occurred before. In the
literature, usually the dollar sign$
is used as a symbol for
that. Why does that matter? –> If later we use the completed suffix tree to search for suffixes, we must accept matches only if they end at a leaf. Otherwise we would get a lot of spurious matches, because there are many strings implicitly contained in the tree that are not actual suffixes of the main string. Forcingremainder
to be 0 at the end is essentially a way to ensure that all suffixes end at a leaf node. However, if we want to use the tree to search for general substrings, not only suffixes of the main string, this final step is indeed not required, as suggested by the OP’s comment below. 
So what is the complexity of the entire algorithm? If the text is n
characters in length, there are obviously n steps (or n+1 if we add
the dollar sign). In each step we either do nothing (other than
updating the variables), or we makeremainder
inserts, each taking O(1)
time. Sinceremainder
indicates how many times we have done nothing
in previous steps, and is decremented for every insert that we make
now, the total number of times we do something is exactly n (or
n+1). Hence, the total complexity is O(n). 
However, there is one small thing that I did not properly explain:
It can happen that we follow a suffix link, update the active
point, and then find that itsactive_length
component does not
work well with the newactive_node
. For example, consider a situation
like this:
(The dashed lines indicate the rest of the tree. The dotted line is a
suffix link.)
Now let the active point be (red,'d',3)
, so it points to the place
behind the f
on the defg
edge. Now assume we made the necessary
updates and now follow the suffix link to update the active point
according to rule 3. The new active point is (green,'d',3)
. However,
the d
edge going out of the green node is de
, so it has only 2
characters. In order to find the correct active point, we obviously
need to follow that edge to the blue node and reset to (blue,'f',1)
.
In a particularly bad case, the active_length
could be as large as
remainder
, which can be as large as n. And it might very well happen
that to find the correct active point, we need not only jump over one
internal node, but perhaps many, up to n in the worst case. Does that
mean the algorithm has a hidden O(n^{2}) complexity, because
in each step remainder
is generally O(n), and the postadjustments
to the active node after following a suffix link could be O(n), too?
No. The reason is that if indeed we have to adjust the active point
(e.g. from green to blue as above), that brings us to a new node that
has its own suffix link, and active_length
will be reduced. As
we follow down the chain of suffix links we make the remaining inserts, active_length
can only
decrease, and the number of activepoint adjustments we can make on
the way can’t be larger than active_length
at any given time. Since
active_length
can never be larger than remainder
, and remainder
is O(n) not only in every single step, but the total sum of increments
ever made to remainder
over the course of the entire process is
O(n) too, the number of active point adjustments is also bounded by
O(n).