# Ukkonen’s Suffix Tree Construction – Part 6

This article is continuation of following five articles:

Ukkonen’s Suffix Tree Construction – Part 1

Ukkonen’s Suffix Tree Construction – Part 2

Ukkonen’s Suffix Tree Construction – Part 3

Ukkonen’s Suffix Tree Construction – Part 4

Ukkonen’s Suffix Tree Construction – Part 5

Please go through Part 1, Part 2, Part 3, Part 4 and Part 5, before looking at current article, where we have seen few basics on suffix tree, high level ukkonen’s algorithm, suffix link and three implementation tricks and activePoints along with an example string “abcabxabcd” where we went through all phases of building suffix tree.

Here, we will see the data structure used to represent suffix tree and the code implementation.

At that end of Part 5 article, we have discussed some of the operations we will be doing while building suffix tree and later when we use suffix tree in different applications.

There could be different possible data structures we may think of to fulfill the requirements where some data structure may be slow on some operations and some fast. Here we will use following in our implementation:

We will have SuffixTreeNode structure to represent each node in tree. SuffixTreeNode structure will have following members:

**children**– This will be an array of alphabet size. This will store all the children nodes of current node on different edges starting with different characters.**suffixLink**– This will point to other node where current node should point via suffix link.**start, end**– These two will store the edge label details from parent node to current node. (start, end) interval specifies the edge, by which the node is connected to its parent node. Each edge will connect two nodes, one parent and one child, and (start, end) interval of a given edge will be stored in the child node. Lets say there are two nods A (parent) and B (Child) connected by an edge with indices (5, 8) then this indices (5, 8) will be stored in node B.**suffixIndex**– This will be non-negative for leaves and will give index of suffix for the path from root to this leaf. For non-leaf node, it will be -1 .

This data structure will answer to the required queries quickly as below:

- How to check if a node is root ? — Root is a special node, with no parent and so it’s start and end will be -1, for all other nodes, start and end indices will be non-negative.
- How to check if a node is internal or leaf node ? — suffixIndex will help here. It will be -1 for internal node and non-negative for leaf nodes.
- What is the length of path label on some edge? — Each edge will have start and end indices and length of path label will be end-start+1
- What is the path label on some edge ? — If string is S, then path label will be substring of S from start index to end index inclusive, [start, end].
- How to check if there is an outgoing edge for a given character c from a node A ? — If A->children is not NULL, there is a path, if NULL, no path.
- What is the character value on an edge at some given distance d from a node A ? — Character at distance d from node A will be S[A->start + d], where S is the string.
- Where an internal node is pointing via suffix link ? — Node A will point to A->suffixLink
- What is the suffix index on a path from root to leaf ? — If leaf node is A on the path, then suffix index on that path will be A->suffixIndex

Following is C implementation of Ukkonen’s Suffix Tree Construction. The code may look a bit lengthy, probably because of a good amount of comments.

## C

`// A C program to implement Ukkonen's Suffix Tree Construction` `#include <stdio.h>` `#include <string.h>` `#include <stdlib.h>` `#define MAX_CHAR 256` `struct` `SuffixTreeNode {` ` ` `struct` `SuffixTreeNode *children[MAX_CHAR];` ` ` `//pointer to other node via suffix link` ` ` `struct` `SuffixTreeNode *suffixLink;` ` ` `/*(start, end) interval specifies the edge, by which the` ` ` `node is connected to its parent node. Each edge will` ` ` `connect two nodes, one parent and one child, and` ` ` `(start, end) interval of a given edge will be stored` ` ` `in the child node. Lets say there are two nods A and B` ` ` `connected by an edge with indices (5, 8) then this` ` ` `indices (5, 8) will be stored in node B. */` ` ` `int` `start;` ` ` `int` `*end;` ` ` `/*for leaf nodes, it stores the index of suffix for` ` ` `the path from root to leaf*/` ` ` `int` `suffixIndex;` `};` `typedef` `struct` `SuffixTreeNode Node;` `char` `text[100]; ` `//Input string` `Node *root = NULL; ` `//Pointer to root node` `/*lastNewNode will point to newly created internal node,` `waiting for it's suffix link to be set, which might get` `a new suffix link (other than root) in next extension of` `same phase. lastNewNode will be set to NULL when last` `newly created internal node (if there is any) got it's` `suffix link reset to new internal node created in next` `extension of same phase. */` `Node *lastNewNode = NULL;` `Node *activeNode = NULL;` `int` `count=0;` `/*activeEdge is represented as input string character` `index (not the character itself)*/` `int` `activeEdge = -1;` `int` `activeLength = 0;` `// remainingSuffixCount tells how many suffixes yet to` `// be added in tree` `int` `remainingSuffixCount = 0;` `int` `leafEnd = -1;` `int` `*rootEnd = NULL;` `int` `*splitEnd = NULL;` `int` `size = -1; ` `//Length of input string` `Node *newNode(` `int` `start, ` `int` `*end)` `{` ` ` `count++;` ` ` `Node *node =(Node*) ` `malloc` `(` `sizeof` `(Node));` ` ` `int` `i;` ` ` `for` `(i = 0; i < MAX_CHAR; i++)` ` ` `node->children[i] = NULL;` ` ` `/*For root node, suffixLink will be set to NULL` ` ` `For internal nodes, suffixLink will be set to root` ` ` `by default in current extension and may change in` ` ` `next extension*/` ` ` `node->suffixLink = root;` ` ` `node->start = start;` ` ` `node->end = end;` ` ` `/*suffixIndex will be set to -1 by default and` ` ` `actual suffix index will be set later for leaves` ` ` `at the end of all phases*/` ` ` `node->suffixIndex = -1;` ` ` `return` `node;` `}` `int` `edgeLength(Node *n) {` ` ` `return` `*(n->end) - (n->start) + 1;` `}` `int` `walkDown(Node *currNode)` `{` ` ` `/*activePoint change for walk down (APCFWD) using` ` ` `Skip/Count Trick (Trick 1). If activeLength is greater` ` ` `than current edge length, set next internal node as` ` ` `activeNode and adjust activeEdge and activeLength` ` ` `accordingly to represent same activePoint*/` ` ` `if` `(activeLength >= edgeLength(currNode))` ` ` `{` ` ` `activeEdge =` ` ` `(` `int` `)text[activeEdge+edgeLength(currNode)]-(` `int` `)` `' '` `;` ` ` `activeLength -= edgeLength(currNode);` ` ` `activeNode = currNode;` ` ` `return` `1;` ` ` `}` ` ` `return` `0;` `}` `void` `extendSuffixTree(` `int` `pos)` `{` ` ` `/*Extension Rule 1, this takes care of extending all` ` ` `leaves created so far in tree*/` ` ` `leafEnd = pos;` ` ` `/*Increment remainingSuffixCount indicating that a` ` ` `new suffix added to the list of suffixes yet to be` ` ` `added in tree*/` ` ` `remainingSuffixCount++;` ` ` `/*set lastNewNode to NULL while starting a new phase,` ` ` `indicating there is no internal node waiting for` ` ` `it's suffix link reset in current phase*/` ` ` `lastNewNode = NULL;` ` ` `//Add all suffixes (yet to be added) one by one in tree` ` ` `while` `(remainingSuffixCount > 0) {` ` ` `if` `(activeLength == 0) {` ` ` `//APCFALZ` ` ` `activeEdge = (` `int` `)text[pos]-(` `int` `)` `' '` `;` ` ` `}` ` ` `// There is no outgoing edge starting with` ` ` `// activeEdge from activeNode` ` ` `if` `(activeNode->children[activeEdge] == NULL)` ` ` `{` ` ` `//Extension Rule 2 (A new leaf edge gets created)` ` ` `activeNode->children[activeEdge] =` ` ` `newNode(pos, &leafEnd);` ` ` `/*A new leaf edge is created in above line starting` ` ` `from an existng node (the current activeNode), and` ` ` `if there is any internal node waiting for it's suffix` ` ` `link get reset, point the suffix link from that last` ` ` `internal node to current activeNode. Then set lastNewNode` ` ` `to NULL indicating no more node waiting for suffix link` ` ` `reset.*/` ` ` `if` `(lastNewNode != NULL)` ` ` `{` ` ` `lastNewNode->suffixLink = activeNode;` ` ` `lastNewNode = NULL;` ` ` `}` ` ` `}` ` ` `// There is an outgoing edge starting with activeEdge` ` ` `// from activeNode` ` ` `else` ` ` `{` ` ` `// Get the next node at the end of edge starting` ` ` `// with activeEdge` ` ` `Node *next = activeNode->children[activeEdge];` ` ` `if` `(walkDown(next))` `//Do walkdown` ` ` `{` ` ` `//Start from next node (the new activeNode)` ` ` `continue` `;` ` ` `}` ` ` `/*Extension Rule 3 (current character being processed` ` ` `is already on the edge)*/` ` ` `if` `(text[next->start + activeLength] == text[pos])` ` ` `{` ` ` `//If a newly created node waiting for it's` ` ` `//suffix link to be set, then set suffix link` ` ` `//of that waiting node to current active node` ` ` `if` `(lastNewNode != NULL && activeNode != root)` ` ` `{` ` ` `lastNewNode->suffixLink = activeNode;` ` ` `lastNewNode = NULL;` ` ` `}` ` ` `//APCFER3` ` ` `activeLength++;` ` ` `/*STOP all further processing in this phase` ` ` `and move on to next phase*/` ` ` `break` `;` ` ` `}` ` ` `/*We will be here when activePoint is in middle of` ` ` `the edge being traversed and current character` ` ` `being processed is not on the edge (we fall off` ` ` `the tree). In this case, we add a new internal node` ` ` `and a new leaf edge going out of that new node. This` ` ` `is Extension Rule 2, where a new leaf edge and a new` ` ` `internal node get created*/` ` ` `splitEnd = (` `int` `*) ` `malloc` `(` `sizeof` `(` `int` `));` ` ` `*splitEnd = next->start + activeLength - 1;` ` ` `//New internal node` ` ` `Node *split = newNode(next->start, splitEnd);` ` ` `activeNode->children[activeEdge] = split;` ` ` `//New leaf coming out of new internal node` ` ` `split->children[(` `int` `)text[pos]-(` `int` `)` `' '` `] =` ` ` `newNode(pos, &leafEnd);` ` ` `next->start += activeLength;` ` ` `split->children[activeEdge] = next;` ` ` `/*We got a new internal node here. If there is any` ` ` `internal node created in last extensions of same` ` ` `phase which is still waiting for it's suffix link` ` ` `reset, do it now.*/` ` ` `if` `(lastNewNode != NULL)` ` ` `{` ` ` `/*suffixLink of lastNewNode points to current newly` ` ` `created internal node*/` ` ` `lastNewNode->suffixLink = split;` ` ` `}` ` ` `/*Make the current newly created internal node waiting` ` ` `for it's suffix link reset (which is pointing to root` ` ` `at present). If we come across any other internal node` ` ` `(existing or newly created) in next extension of same` ` ` `phase, when a new leaf edge gets added (i.e. when` ` ` `Extension Rule 2 applies is any of the next extension` ` ` `of same phase) at that point, suffixLink of this node` ` ` `will point to that internal node.*/` ` ` `lastNewNode = split;` ` ` `}` ` ` `/* One suffix got added in tree, decrement the count of` ` ` `suffixes yet to be added.*/` ` ` `remainingSuffixCount--;` ` ` `if` `(activeNode == root && activeLength > 0) ` `//APCFER2C1` ` ` `{` ` ` `activeLength--;` ` ` `activeEdge = (` `int` `)text[pos -` ` ` `remainingSuffixCount + 1]-(` `int` `)` `' '` `;` ` ` `}` ` ` ` ` `//APCFER2C2` ` ` `else` `if` `(activeNode != root)` ` ` `{` ` ` `activeNode = activeNode->suffixLink;` ` ` `}` ` ` `}` `}` `void` `print(` `int` `i, ` `int` `j)` `{` ` ` `int` `k;` ` ` `for` `(k=i; k<=j; k++)` ` ` `printf` `(` `"%c"` `, text[k]);` `}` `//Print the suffix tree as well along with setting suffix index` `//So tree will be printed in DFS manner` `//Each edge along with it's suffix index will be printed` `void` `setSuffixIndexByDFS(Node *n, ` `int` `labelHeight)` `{` ` ` `if` `(n == NULL) ` `return` `;` ` ` `if` `(n->start != -1) ` `//A non-root node` ` ` `{` ` ` `//Print the label on edge from parent to current node` ` ` `print(n->start, *(n->end));` ` ` `}` ` ` `int` `leaf = 1;` ` ` `int` `i;` ` ` `for` `(i = 0; i < MAX_CHAR; i++)` ` ` `{` ` ` `if` `(n->children[i] != NULL)` ` ` `{` ` ` `if` `(leaf == 1 && n->start != -1)` ` ` `printf` `(` `" [%d]\n"` `, n->suffixIndex);` ` ` `//Current node is not a leaf as it has outgoing` ` ` `//edges from it.` ` ` `leaf = 0;` ` ` `setSuffixIndexByDFS(n->children[i],` ` ` `labelHeight + edgeLength(n->children[i]));` ` ` `}` ` ` `}` ` ` `if` `(leaf == 1)` ` ` `{` ` ` `n->suffixIndex = size - labelHeight;` ` ` `printf` `(` `" [%d]\n"` `, n->suffixIndex);` ` ` `}` `}` `void` `freeSuffixTreeByPostOrder(Node *n)` `{` ` ` `if` `(n == NULL)` ` ` `return` `;` ` ` `int` `i;` ` ` `for` `(i = 0; i < MAX_CHAR; i++)` ` ` `{` ` ` `if` `(n->children[i] != NULL)` ` ` `{` ` ` `freeSuffixTreeByPostOrder(n->children[i]);` ` ` `}` ` ` `}` ` ` `if` `(n->suffixIndex == -1)` ` ` `free` `(n->end);` ` ` `free` `(n);` `}` `/*Build the suffix tree and print the edge labels along with` `suffixIndex. suffixIndex for leaf edges will be >= 0 and` `for non-leaf edges will be -1*/` `void` `buildSuffixTree()` `{` ` ` `size = ` `strlen` `(text);` ` ` `int` `i;` ` ` `rootEnd = (` `int` `*) ` `malloc` `(` `sizeof` `(` `int` `));` ` ` `*rootEnd = - 1;` ` ` `/*Root is a special node with start and end indices as -1,` ` ` `as it has no parent from where an edge comes to root*/` ` ` `root = newNode(-1, rootEnd);` ` ` `activeNode = root; ` `//First activeNode will be root` ` ` `for` `(i=0; i<size; i++)` ` ` `extendSuffixTree(i);` ` ` `int` `labelHeight = 0;` ` ` `setSuffixIndexByDFS(root, labelHeight);` ` ` `//Free the dynamically allocated memory` ` ` `freeSuffixTreeByPostOrder(root);` `}` `// driver program to test above functions` `int` `main(` `int` `argc, ` `char` `*argv[])` `{` ` ` `strcpy` `(text, ` `"abbc"` `); buildSuffixTree();` ` ` `printf` `(` `"Number of nodes in suffix tree are %d\n"` `,count);` ` ` `return` `0;` `}` |

Output (Each edge of Tree, along with suffix index of child node on edge, is printed in DFS order. To understand the output better, match it with the last figure no 43 in previous Part 5 article):

abbc [0] b [-1] bc [1] c [2] c [3] Number of nodes in suffix tee are 6

Now we are able to build suffix tree in linear time, we can solve many string problem in efficient way:

- Check if a given pattern P is substring of text T (Useful when text is fixed and pattern changes, KMP otherwise
- Find all occurrences of a given pattern P present in text T
- Find longest repeated substring
- Linear Time Suffix Array Creation

The above basic problems can be solved by DFS traversal on suffix tree.

We will soon post articles on above problems and others like below:

- Build Generalized suffix tree
- Linear Time Longest common substring problem
- Linear Time Longest palindromic substring

And More.**Test you understanding?**

- Draw suffix tree (with proper suffix link, suffix indices) for string “AABAACAADAABAAABAA$” on paper and see if that matches with code output.
- Every extension must follow one of the three rules: Rule 1, Rule 2 and Rule 3.

Following are the rules applied on five consecutive extensions in some Phase i (i > 5), which ones are valid:

A) Rule 1, Rule 2, Rule 2, Rule 3, Rule 3

B) Rule 1, Rule 2, Rule 2, Rule 3, Rule 2

C) Rule 2, Rule 1, Rule 1, Rule 3, Rule 3

D) Rule 1, Rule 1, Rule 1, Rule 1, Rule 1

E) Rule 2, Rule 2, Rule 2, Rule 2, Rule 2

F) Rule 3, Rule 3, Rule 3, Rule 3, Rule 3

- What are the valid sequences in above for Phase 5
- Every internal node MUST have it’s suffix link set to another node (internal or root). Can a newly created node point to already existing internal node or not ? Can it happen that a new node created in extension j, may not get it’s right suffix link in next extension j+1 and get the right one in later extensions like j+2, j+3 etc ?
- Try solving the basic problems discussed above.

We have published following articles on suffix tree applications:

- Suffix Tree Application 1 – Substring Check

- Suffix Tree Application 2 – Searching All Patterns

- Suffix Tree Application 3 – Longest Repeated Substring

- Suffix Tree Application 4 – Build Linear Time Suffix Array

- Generalized Suffix Tree 1

- Suffix Tree Application 5 – Longest Common Substring

- Suffix Tree Application 6 – Longest Palindromic Substring

**References**:

http://web.stanford.edu/~mjkay/gusfield.pdf

Ukkonen’s suffix tree algorithm in plain English

This article is contributed by **Anurag Singh**. Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above

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