Get Rid Off Suffix Tree !!!!! An Online Compilation of Helpful Tutorial

Excellent Study Material For Understanding Algorithms & implementing Suffix Tree

1.Suffix Tree Wikipedia

2. A Good Explication

3. Its Awesome Suffix Tree By Sahni

4. Check These Strings

5. String Search Algorithms

6.http://apps.topcoder.com/forums/?module=RevisionHistory&messageID=1171511

7.http://www.stanford.edu/class/cs97si/suffix-array.pdf

Application of Suffix Tree

1.Longest repeated substring problem:

http://en.wikipedia.org/wiki/Longest_repeated_substring_problem

The longest repeated substring problem is finding the longest substring of a string that occurs at least twice. This problem can be solved in linear time and space by building a suffix tree for the string, and finding the deepest internal node in the tree. The string spelled by the edges from the root to such a node is a longest repeated substring. The problem of finding the longest substring with at least k occurrences can be found by first preprocessing the tree to count the number of leaf descendants for each internal node, and then finding the deepest node with at least k descendants.

By from Sartaj Sahni

Find the longest substring of S that appears at least m > 1 times. This query can be answered in O(|S|) time in the following way:
(a) Traverse the suffix tree labeling the branch nodes with the sum of the label lengths from the root and also with the number of information nodes in the subtrie. (b) Traverse the suffix tree visiting branch nodes with information node count >= m. Determine the visited branch node with longest label length.
Note that step (a) needs to be done only once. Following this, we can do step (b) for as many values of m as is desired. Also, note that when m = 2 we can avoid determining the number of information nodes in subtries. In a compressed trie, every subtrie rooted at a branch node has at least two information nodes in it.
Note that step (a) needs to be done only once. Following this, we can do step (b) for as many values of m as is desired. Also, note that when m = 2 we can avoid determining the number of information nodes in subtries. In a compressed trie, every subtrie rooted at a branch node has at least two information nodes in it.
2. Pattern Searching/Sub-String Searching 
Searching for a substring, pat[1..m], in txt[1..n], can be solved in O(m) time (after the suffix tree for txt has been built in O(n) time).
3. Longest Repeated Substring 
4. Longest Common Subsequence 
5. Longest Palindrome
http://www.cs.helsinki.fi/u/ukkonen/SuffixT1withFigs.pdf

http://www.allisons.org/ll/AlgDS/Tree/Suffix/

Add a special ``end of string'' character, e.g. `$', to txt[1..n] and build a suffix tree; the longest repeated substring oftxt[1..n] is indicated by the deepest fork node in the suffix tree, where depth is measured by the number of characters traversed from the root, i.e., `issi' in the case of `mississippi'. The longest repeated substring can be found in O(n) time using a suffix tree.
The longest common substring of two strings, txt1 and txt2, can be found by building a generalized suffix tree for txt1 andtxt2: Each node is marked to indicate if it represents a suffix of txt1 or txt2 or both. The deepest node marked for both txt1and txt2 represents the longest common substring.
Equivalently, one can build a (basic) suffix tree for the string txt1$txt2#, where `$' is a special terminator for txt1 and `#' is a special terminator for txt2. The longest common substring is indicated by the deepest fork node that has both `...$...' and `...#...' (no $) beneath it.
(Try it using the HTML FORM above.)

Note that the `longest common substring problem' is different to the `longest common subsequence problem' which is closely related to the `edit-distance problem': An instance of a subsequence can have gaps where it appears in txt1 and in txt2, but an instance of a substring cannot have gaps.
A palindrome is a string, P, such that P=reverse(P). e.g. `abba'=reverse(`abba'). e.g. `ississi' is the longest palindrome in `mississippi'. The longest palindrome of txt[1..n] can be found in O(n) time, e.g. by building the suffix tree for txt$reverse(txt)# or by building the generalized suffix tree for txt and reverse(txt).
(Try it.)

Others 

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Some Awesome Algorithms/Codes For Suffix Tree Implementation

1.by Dr. Sartaj Sahni (CISE Department Chair at University of Florida)

2. by Lloyd Allison

3. NIST Dictionary

4.  ANSI C implementation of a Suffix Tree

5.  Generic suffix tree library written in C

6.Perl binding to libstree

7.# a faster generic suffix tree library written in C (uses arrays instead of linked lists)

8. Python binding to Strmat