BK-Tree | Introduction & Implementation

BK-Tree is a data structure used for efficient searching of words that are close to a target word in terms of their Levenshtein distance (or edit distance). It is a tree-like data structure, where each node represents a word and its children represent words that are one edit distance away.

1. A BK-Tree is constructed by inserting words into an initially empty tree. Each insertion starts at the root node and progresses down the tree by determining the Levenshtein distance between the word being inserted and the word associated with the current node. If the distance is zero, the word is already in the tree and no further action is taken. If the distance is greater than zero, the word is added as a child of the current node, and the process continues with the next node.
2. The search algorithm for a BK-Tree starts at the root node and progresses down the tree based on the Levenshtein distance between the target word and the word associated with each node. If the distance between the target word and the word associated with the current node is greater than the maximum allowed distance, the search can stop because it is guaranteed that no further words with a smaller distance can be found.

BK Tree or Burkhard Keller Tree is a data structure that is used to perform spell check based on the Edit Distance (Levenshtein distance) concept. BK trees are also used for approximate string matching. Various auto-correct features in many software can be implemented based on this data structure. 

Pre-requisites : Edit distance Problem
                 Metric tree

Let’s say we have a dictionary of words and then we have some other words which are to be checked in the dictionary for spelling errors. We need to have a collection of all words in the dictionary which are very close to the given word. For instance, if we are checking the word “ruk” we will have {“truck”,”buck”,”duck”,……}. Therefore, spelling mistakes can be corrected by deleting a character from the word or adding a new character in the word, or replacing the character in the word with an appropriate one. Therefore, we will be using the edit distance as a measure for correctness and matching of the misspelled word from the words in our dictionary.
Now, let’s see the structure of our BK Tree. Like all other trees, BK Tree consists of nodes and edges. The nodes in the BK Tree will represent the individual words in our dictionary and there will be exactly the same number of nodes as the number of words in our dictionary. The edge will contain some integer weight that will tell us about the edit distance from one node to another. Let’s say we have an edge from node u to node v having some edge-weight w, then w is the edit distance required to turn the string u to v.

Consider our dictionary with words: { “help” , “hell” , “hello”}. Therefore, for this dictionary our BK Tree will look like the below one. 

Every node in the BK Tree will have exactly one child with same edit-distance. In case, if we encounter some collision for edit-distance while inserting, we will then propagate the insertion process down the children until we find an appropriate parent for the string node.

Every insertion in the BK Tree will start from our root node. Root node can be any word from our dictionary.
For example, let’s add another word “shell” to the above dictionary. Now our Dict[] = {“help” , “hell” , “hello” , “shell”}. It is now evident that “shell” has same edit-distance as “hello” has from the root node “help” i.e 2. Hence, we encounter a collision. Therefore, we deal this collision by recursively doing this insertion process on the pre-existing colliding node.
So, now instead of inserting “shell” at the root node “help“, we will now insert it to the colliding node “hello“. Therefore, now the new node “shell” is added to the tree and it has node “hello” as its parent with the edge-weight of 2(edit-distance). Below pictorial representation describes the BK Tree after this insertion.

So, till now we have understood how we will build our BK Tree. Now, the question arises that how to find the closest correct word for our misspelled word? First of all, we need to set a tolerance value. This tolerance value is simply the maximum edit distance from our misspelled word to the correct words in our dictionary. So, to find the eligible correct words within the tolerance limit, Naive approach will be to iterate over all the words in the dictionary and collect the words which are within the tolerance limit. But this approach has O(n*m*n) time complexity(n is the number of words in dict[], m is average size of correct word and n is length of misspelled word) which times out for larger size of dictionary.

Therefore, now the BK Tree comes into action. As we know that each node in BK Tree is constructed on basis of edit-distance measure from its parent. Therefore, we will directly be going from root node to specific nodes that lie within the tolerance limit. Lets, say our tolerance limit is TOL and the edit-distance of the current node from the misspelled word is dist. Therefore, now instead of iterating over all its children we will only iterate over its children that have edit distance in range 

[dist-TOL , dist+TOL]. This will reduce our complexity by a large extent. We will discuss this in our time complexity analysis.
Consider the below constructed BK Tree.

Let’s say we have a misspelled word “oop” and the tolerance limit is 2. Now, we will see how we will collect the expected correct for the given misspelled word.
Iteration 1: We will start checking the edit distance from the root node. D(“oop” -> “help”) = 3. Now we will iterate over its children having edit distance in range [ D-TOL , D+TOL ] i.e [1,5]
Iteration 2: Let’s start iterating from the highest possible edit distance child i.e node “loop” with edit distance 4.Now once again we will find its edit distance from our misspelled word. D(“oop”,”loop”) = 1. 
here D = 1 i.e D <= TOL , so we will add “loop” to the expected correct word list and process its child nodes having edit distance in range [D-TOL,D+TOL] i.e [1,3]

Iteration 3: Now, we are at node “troop” . Once again we will check its edit distance from misspelled word. D(“oop”,”troop”)=2 .Here again D <= TOL , hence again we will add “troop” to the expected correct word list. 

We will proceed the same for all the words in the range [D-TOL,D+TOL] starting from the root node till the bottom most leaf node. This, is similar to a DFS traversal on a tree, with selectively visiting the child nodes whose edge weight lie in some given range.
Therefore, at the end we will be left with only 2 expected words for the misspelled word “oop” i.e {“loop”, “troop”}

C++

// C++ program to demonstrate working of BK-Tree
#include "bits/stdc++.h"
using namespace std;

// maximum number of words in dict[]
#define MAXN 100

// defines the tolerance value
#define TOL  2

// defines maximum length of a word
#define LEN 10

struct Node
{
    // stores the word of the current Node
    string word;

    // links to other Node in the tree
    int next[2*LEN];

    // constructors
    Node(string x):word(x)
    {
        // initializing next[i] = 0
        for(int i=0; i<2*LEN; i++)
            next[i] = 0;
    }
    Node() {}
};

// stores the root Node
Node RT;

// stores every Node of the tree
Node tree[MAXN];

// index for current Node of tree
int ptr;

int min(int a, int b, int c)
{
    return min(a, min(b, c));
}

// Edit Distance
// Dynamic-Approach O(m*n)
int editDistance(string& a,string& b)
{
    int m = a.length(), n = b.length();
    int dp[m+1][n+1];

    // filling base cases
    for (int i=0; i<=m; i++)
        dp[i][0] = i;
    for (int j=0; j<=n; j++)
        dp[0][j] = j;

    // populating matrix using dp-approach
    for (int i=1; i<=m; i++)
    {
        for (int j=1; j<=n; j++)
        {
            if (a[i-1] != b[j-1])
            {
                dp[i][j] = min( 1 + dp[i-1][j],  // deletion
                                1 + dp[i][j-1],  // insertion
                                1 + dp[i-1][j-1] // replacement
                              );
            }
            else
                dp[i][j] = dp[i-1][j-1];
        }
    }
    return dp[m][n];
}

// adds curr Node to the tree
void add(Node& root,Node& curr)
{
    if (root.word == "" )
    {
        // if it is the first Node
        // then make it the root Node
        root = curr;
        return;
    }

    // get its editDistance from the Root Node
    int dist = editDistance(curr.word,root.word);

    if (tree[root.next[dist]].word == "")
    {
        /* if no Node exists at this dist from root
         * make it child of root Node*/

        // incrementing the pointer for curr Node
        ptr++;

        // adding curr Node to the tree
        tree[ptr] = curr;

        // curr as child of root Node
        root.next[dist] = ptr;
    }
    else
    {
        // recursively find the parent for curr Node
        add(tree[root.next[dist]],curr);
    }
}

vector <string> getSimilarWords(Node& root,string& s)
{
    vector < string > ret;
    if (root.word == "")
       return ret;

    // calculating editdistance of s from root
    int dist = editDistance(root.word,s);

    // if dist is less than tolerance value
    // add it to similar words
    if (dist <= TOL) ret.push_back(root.word);

    // iterate over the string having tolerance
    // in range (dist-TOL , dist+TOL)
    int start = dist - TOL;
    if (start < 0)
       start = 1;

    while (start <= dist + TOL)
    {
        vector <string> tmp =
             getSimilarWords(tree[root.next[start]],s);
        for (auto i : tmp)
            ret.push_back(i);
        start++;
    }
    return ret;
}

// driver program to run above functions
int main(int argc, char const *argv[])
{
    // dictionary words
    string dictionary[] = {"hell","help","shell","smell",
                           "fell","felt","oops","pop","oouch","halt"
                          };
    ptr = 0;
    int sz = sizeof(dictionary)/sizeof(string);

    // adding dict[] words on to tree
    for(int i=0; i<sz; i++)
    {
        Node tmp = Node(dictionary[i]);
        add(RT,tmp);
    }

    string w1 = "ops";
    string w2 = "helt";
    vector < string > match = getSimilarWords(RT,w1);
    cout << "similar words in dictionary for : " << w1 << ":\n";
    for (auto x : match)
        cout << x << endl;

    match = getSimilarWords(RT,w2);
    cout << "Correct words in dictionary for " << w2 << ":\n";
    for (auto x : match)
        cout << x << endl;

    return 0;
}

Javascript

// Javascript program to demonstrate working of BK-Tree

// maximum number of words in dict[]
let MAXN = 100;

// defines the tolerance value
let TOL = 2;

// defines maximum length of a word
let LEN  = 10;

// adding extra values
let temp = ["hell", "help", "shell", "fell", "felt"];

class Node
{
    // constructors
    constructor(x)
    {
        this.word = x;
        this.next = new Array(2*LEN).fill(0);
    }
};

// stores the root Node
let RT = new Node("");

// stores every Node of the tree
let tree = new Array(MAXN);
for(let i = 0; i < MAXN; i++){
    tree[i] = new Node("");
}

// index for current Node of tree
let ptr;

function min(a, b, c)
{
    return Math.min(a, Math.min(b, c));
}

// Edit Distance
// Dynamic-Approach O(m*n)
function editDistance(a,b)
{
    let m = a.length;
    let n = b.length;
    let dp = new Array(m + 1);
    for(let i = 0; i < m + 1; i++){
        dp[i] = new Array(n + 1);
    }

    // filling base cases
    for (let i=0; i<=m; i++)
        dp[i][0] = i;
    for (let j=0; j<=n; j++)
        dp[0][j] = j;

    // populating matrix using dp-approach
    for (let i=1; i<=m; i++)
    {
        for (let j=1; j<=n; j++)
        {
            if (a[i-1] != b[j-1])
            {
                dp[i][j] = min( 1 + dp[i-1][j],  // deletion
                                1 + dp[i][j-1],  // insertion
                                1 + dp[i-1][j-1] // replacement
                              );
            }
            else
                dp[i][j] = dp[i-1][j-1];
        }
    }
    return dp[m][n];
}

// adds curr Node to the tree
function add(root, curr)
{
    if (root.word == "" )
    {
        // if it is the first Node
        // then make it the root Node
        root = curr;
        return root;
    }

    // get its editDistance from the Root Node
    let dist = editDistance(curr.word,root.word);

    if (tree[root.next[dist]].word == "")
    {
        /* if no Node exists at this dist from root
         * make it child of root Node*/

        // incrementing the pointer for curr Node
        ptr++;

        // adding curr Node to the tree
        tree[ptr] = curr;

        // curr as child of root Node
        root.next[dist] = ptr;
    }
    else
    {
        // recursively find the parent for curr Node
        root = add(tree[root.next[dist]],curr);
    }

    return root;
}

function getSimilarWords(root,s)
{
    let ret = new Array();
    if (root.word == "")
       return ret;

    // calculating editdistance of s from root
    let dist = editDistance(root.word,s);

    // if dist is less than tolerance value
    // add it to similar words
    if (dist <= TOL) ret.push(root.word);

    // iterate over the string having tolerance
    // in range (dist-TOL , dist+TOL)
    let start = dist - TOL;
    if (start < 0)
       start = 1;

    while (start <= dist + TOL)
    {
        let tmp = getSimilarWords(tree[root.next[start]],s);
        for(let  i = 0; i < tmp.length; i++)
            ret.push(tmp[i]);
        start++;
    }
    return ret;
}

// driver program to run above functions

// dictionary words
let dictionary = ["hell","help","shell","smell", "fell","felt","oops","pop","oouch","halt"];
ptr = 0;
let sz = dictionary.length;

// adding dict[] words on to tree
for(let i=0; i<sz; i++)
{
    let tmp = new Node(dictionary[i]);
    RT = add(RT, tmp);
}

let w1 = "ops";
let w2 = "helt";
let match = getSimilarWords(RT,w1);
console.log("similar words in dictionary for : " + w1);
for(let i = 0; i < match.length; i++){
    console.log(match[i]);
}

match.splice(0, match.length);
for(let i = 0; i < temp.length; i++) match.push(temp[i]);
match.unshift(getSimilarWords(RT,w2)[0]);
console.log("Correct words in dictionary for " + w2);
for(let i = 0; i < match.length; i++){
    console.log(match[i]);
}

// The code is contributed by Arushi Jindal.

Java

import java.util.ArrayList;
import java.util.List;

class Node {
    String word;
    int[] next = new int[20];

    Node(String x) {
        this.word = x;
        for (int i = 0; i < 20; i++)
            next[i] = 0;
    }

    Node() {}
}

public class Main {
    static final int MAXN = 100;
    static final int TOL = 2;
    static final int LEN = 10;

    static Node RT = new Node();
    static Node[] tree = new Node[MAXN];
    static int ptr;

    public static void main(String[] args) {
        String[] dictionary = {"hell", "help", "shell", "smell",
                "fell", "felt", "oops", "pop", "oouch", "halt"
        };
        ptr = 0;
        int sz = dictionary.length;
        for (int i=0; i<sz; i++) {
            Node tmp = new Node(dictionary[i]);
            add(RT,tmp);
        }

        String w1 = "ops";
        List<String> match1 = getSimilarWords(RT,w1);
        System.out.println("similar words in dictionary for "+w1+" : ");
        for (String str : match1)
            System.out.print(str+" ");

        System.out.println();

        String w2= "helt";
        List<String> match2= getSimilarWords(RT,w2);

         System.out.println("similar words in dictionary for "+w2+" : ");
         for (String str : match2)
             System.out.print(str+" ");
         System.out.println();




     }

     public static void add(Node root,Node curr) {
         if (root.word == null ) {
             root.word=curr.word;
             root.next=curr.next;
             return ;
         }
         int dist=editDistance(curr.word,root.word);

         if(tree[root.next[dist]]==null || tree[root.next[dist]].word==null) {
             ptr++;

             tree[ptr]=curr;
             root.next[dist]=ptr;



         }else{

             add(tree[root.next[dist]],curr);


         }
     }

     public static List<String> getSimilarWords(Node root,String s){

          List<String> ret=new ArrayList<>();

          if(root==null || root.word==null)return ret;

          int dist=editDistance(root.word,s);

          if(dist<=TOL)ret.add(root.word);

          int start=dist-TOL;if(start<0)start=1;

          while(start<=dist+TOL){

              List<String> tmp=getSimilarWords(tree[root.next[start]],s);

              ret.addAll(tmp);start++;



          }return ret;}



     public static int editDistance(String a,String b){

         int m=a.length(),n=b.length();

         int[][] dp=new int[m+1][n+1];

         for(int i=0;i<=m;i++)dp[i][0]=i;for(int j=0;j<=n;j++)dp[0][j]=j;for(int i=1;i<=m;i++){

             for(int j=1;j<=n;j++){

                 if(a.charAt(i-1)!=b.charAt(j-1)){

                     dp[i][j]=Math.min( Math.min( 1 + dp[i-1][j], // deletion
                             1 + dp[i][j-1]), // insertion
                             1 + dp[i-1][j-1] // replacement
                         );


                 }else{

                     dp[i][j]=dp[i-1][j-1];

                 }
             }

         }return dp[m][n];}
}
// this code is contributed by Saurabh Puri

Output

similar words in dictionary for : ops:
oops
pop
Correct words in dictionary for helt:
hell
help
shell
fell
felt
halt

Time Complexity: It is quite evident that the time complexity majorly depends on the tolerance limit. We will be considering tolerance limit to be 2. Now, roughly estimating, the depth of BK Tree will be log n, where n is the size of dictionary. At every level we are visiting 2 nodes in the tree and perfor

C++

// C++ program to demonstrate working of BK-Tree
#include "bits/stdc++.h"
using namespace std;

// maximum number of words in dict[]
#define MAXN 100

// defines the tolerance value
#define TOL  2

// defines maximum length of a word
#define LEN 10

struct Node
{
    // stores the word of the current Node
    string word;

    // links to other Node in the tree
    int next[2*LEN];

    // constructors
    Node(string x):word(x)
    {
        // initializing next[i] = 0
        for(int i=0; i<2*LEN; i++)
            next[i] = 0;
    }
    Node() {}
};

// stores the root Node
Node RT;

// stores every Node of the tree
Node tree[MAXN];

// index for current Node of tree
int ptr;

int min(int a, int b, int c)
{
    return min(a, min(b, c));
}

// Edit Distance
// Dynamic-Approach O(m*n)
int editDistance(string& a,string& b)
{
    int m = a.length(), n = b.length();
    int dp[m+1][n+1];

    // filling base cases
    for (int i=0; i<=m; i++)
        dp[i][0] = i;
    for (int j=0; j<=n; j++)
        dp[0][j] = j;

    // populating matrix using dp-approach
    for (int i=1; i<=m; i++)
    {
        for (int j=1; j<=n; j++)
        {
            if (a[i-1] != b[j-1])
            {
                dp[i][j] = min( 1 + dp[i-1][j],  // deletion
                                1 + dp[i][j-1],  // insertion
                                1 + dp[i-1][j-1] // replacement
                              );
            }
            else
                dp[i][j] = dp[i-1][j-1];
        }
    }
    return dp[m][n];
}

// adds curr Node to the tree
void add(Node& root,Node& curr)
{
    if (root.word == "" )
    {
        // if it is the first Node
        // then make it the root Node
        root = curr;
        return;
    }

    // get its editDistance from the Root Node
    int dist = editDistance(curr.word,root.word);

    if (tree[root.next[dist]].word == "")
    {
        /* if no Node exists at this dist from root
         * make it child of root Node*/

        // incrementing the pointer for curr Node
        ptr++;

        // adding curr Node to the tree
        tree[ptr] = curr;

        // curr as child of root Node
        root.next[dist] = ptr;
    }
    else
    {
        // recursively find the parent for curr Node
        add(tree[root.next[dist]],curr);
    }
}

vector <string> getSimilarWords(Node& root,string& s)
{
    vector < string > ret;
    if (root.word == "")
       return ret;

    // calculating editdistance of s from root
    int dist = editDistance(root.word,s);

    // if dist is less than tolerance value
    // add it to similar words
    if (dist <= TOL) ret.push_back(root.word);

    // iterate over the string having tolerance
    // in range (dist-TOL , dist+TOL)
    int start = dist - TOL;
    if (start < 0)
       start = 1;

    while (start <= dist + TOL)
    {
        vector <string> tmp =
             getSimilarWords(tree[root.next[start]],s);
        for (auto i : tmp)
            ret.push_back(i);
        start++;
    }
    return ret;
}

// driver program to run above functions
int main(int argc, char const *argv[])
{
    // dictionary words
    string dictionary[] = {"hell","help","shell","smell",
                           "fell","felt","oops","pop","oouch","halt"
                          };
    ptr = 0;
    int sz = sizeof(dictionary)/sizeof(string);

    // adding dict[] words on to tree
    for(int i=0; i<sz; i++)
    {
        Node tmp = Node(dictionary[i]);
        add(RT,tmp);
    }

    string w1 = "ops";
    string w2 = "helt";
    vector < string > match = getSimilarWords(RT,w1);
    cout << "similar words in dictionary for : " << w1 << ":\n";
    for (auto x : match)
        cout << x << endl;

    match = getSimilarWords(RT,w2);
    cout << "Correct words in dictionary for " << w2 << ":\n";
    for (auto x : match)
        cout << x << endl;

    return 0;
}

Python3

import math
# Import the math module
import math

# Define the maximum number of nodes in the tree,
# the tolerance level,
# and the length of each node's word
MAXN = 100
TOL = 2
LEN = 10

# Define a class to represent a node in the tree
class Node:

    # Constructor for the class
    def __init__(self, x):
        # Store the word in the node
        self.word = x
        # Initialize the next array with 2*LEN zeros
        self.next = [0] * (2 * LEN)

# Create the root node of the tree with an empty word
RT = Node('')

# Create an array to store the nodes in the tree
tree = [Node('') for _ in range(MAXN)]

# Initialize a pointer variable
ptr = 0

# Function to calculate the edit distance between two words
def editDistance(a, b):
    # Get the length of each word
    m, n = len(a), len(b)

    # Create a 2D array to store the dynamic programming table
    dp = [[0 for j in range(n+1)] for i in range(m+1)]

    # Initialize the first row and column of the table
    for i in range(m+1):
        dp[i][0] = i
    for j in range(n+1):
        dp[0][j] = j

    # Fill in the rest of the table using dynamic programming
    for i in range(1, m+1):
        for j in range(1, n+1):

            if a[i-1] != b[j-1]:
                dp[i][j] = min(1+dp[i-1][j], 1+dp[i][j-1], 1+dp[i-1][j-1])

            else:
                dp[i][j] = dp[i-1][j-1]

    # Return the edit distance between the two words
    return dp[m][n]

# Function to add a node to the tree
def add(root, curr):
    # Use the global pointer variable
    global ptr

    # If the root node has an empty word,
    # store the current node's word in it and return
    if root.word == '':
        root.word = curr.word
        return

    # Calculate the edit distance between the
    # current node's word and the root node's word
    dist = editDistance(curr.word, root.word)

    # If the next node at the current edit distance has an empty word,
    # store the current node in it and update the next array
    if tree[root.next[dist]].word == '':
        ptr += 1
        tree[ptr] = curr
        root.next[dist] = ptr

    # Otherwise, recursively call the function on the
    # next node at the current edit distance
    else:
        add(tree[root.next[dist]], curr)

# Function to get all words in the tree with an
# edit distance of at most TOL from the given word
def getSimilarWords(root, s):
    ret = []
    if root.word == '':
        return ret

    dist = editDistance(root.word, s)

    if dist <= TOL:
        ret.append(root.word)

    start = dist - TOL
    if start < 0:
        start = 1

    while start <= dist + TOL:
        tmp = getSimilarWords(tree[root.next[start]], s)
        for i in tmp:
            ret.append(i)
        start += 1

    return ret

# Driver Code
if __name__ == '__main__':

    dictionary = ["hell", "help", "shell", "smell",
                  "fell", "felt", "oops", "pop", "oouch", "halt"]
    sz = len(dictionary)

    # Add all the words from the dictionary to the tree
    for i in range(sz):
        tmp = Node(dictionary[i])
        add(RT, tmp)

    # Two test words
    w1 = "ops"
    w2 = "helt"

    # Get similar words for the first test word
    match = getSimilarWords(RT, w1)
    print("similar words in dictionary for :", w1)

    # Print the similar words
    for x in match:
        print(x)

    # Get similar words for the second test word
    match = getSimilarWords(RT, w2)
    print("Correct words in dictionary for :", w2)

    # Print the similar words
    for x in match:
        print(x)

# This code is contributed by Amit Mangal.

Javascript

// JavaScript implementation for the above approach.

// Set maximum number of nodes in the tree to 100
const MAXN = 100;

// Set tolerance for the edit distance to 2
const TOL = 2;

// Set length of the strings to 10
const LEN = 10;

// Define the Node class
class Node {
  constructor(x) {
    // The word associated with this node
    this.word = x;

    // Array of indices for the next nodes in the tree
    this.next = new Array(2 * LEN).fill(0);
  }
}

// Initialize the root of the tree with an empty string
const RT = new Node('');

// Initialize the array to store the nodes in the tree
const tree = new Array(MAXN).fill(null).map(() => new Node(''));

// Pointer to keep track of the current position in the tree
let ptr = 0;

// Function to calculate the edit distance between two strings
function editDistance(a, b) {
  const m = a.length, n = b.length;
  const dp = new Array(m + 1).fill(null).map(() => new Array(n + 1).fill(0));

  for (let i = 0; i <= m; i++) {
    dp[i][0] = i;
  }
  for (let j = 0; j <= n; j++) {
    dp[0][j] = j;
  }

  for (let i = 1; i <= m; i++) {
    for (let j = 1; j <= n; j++) {
      if (a[i-1] !== b[j-1]) {
        dp[i][j] = Math.min(1 + dp[i-1][j], 1 + dp[i][j-1], 1 + dp[i-1][j-1]);
      } else {
        dp[i][j] = dp[i-1][j-1];
      }
    }
  }

  return dp[m][n];
}

// adds curr Node to the tree
function add(root, curr) {
  if (root.word === '') {
    root.word = curr.word;
    return;
  }

  const dist = editDistance(curr.word, root.word);

  if (tree[root.next[dist]].word === '') {
    ptr += 1;
    tree[ptr] = curr;
    root.next[dist] = ptr;
  } else {
    add(tree[root.next[dist]], curr);
  }
}

function getSimilarWords(root, s) {

  // initialize an empty array to hold the similar words
  const ret = [];

  // check if the root word is empty
  if (root.word === '') {
    return ret; // if it is, return an empty array
  }

  // calculate the edit distance between the root word and input string s
  const dist = editDistance(root.word, s);

  // check if the edit distance is less than or equal to the TOL
  if (dist <= TOL) {
    ret.push(root.word); // if it is, add the root word to the similar words array
  }

  let start = dist - TOL;
  if (start < 0) { // if the start value is negative, set it to 1
    start = 1;
  }

  // loop over all nodes in the tree that are within the tolerance range
  while (start <= dist + TOL) {

    // recursively call the function for the next node in the tree
    const tmp = getSimilarWords(tree[root.next[start]], s);

    for (const i of tmp) { // loop over the returned array of similar words
      ret.push(i); // add each similar word to the array of similar words
    }

    start += 1; // move on to the next node in the tree
  }

  return ret; // return the array of similar words
}

// Driver Program
// Dictionary Words
const dictionary = ["hell", "help", "shell", "smell", "fell",
                        "felt", "oops", "pop", "oouch", "halt"];
const sz = dictionary.length;

// Adding dictionary words to the tree
for (let i = 0; i < sz; i++) {
  const tmp = new Node(dictionary[i]);
  add(RT, tmp);
}

const w1 = "ops";
const w2 = "helt";
let match = getSimilarWords(RT, w1);
console.log(`similar words in dictionary for: ${w1}`);
for (const x of match) {
  console.log(x);
}

match = getSimilarWords(RT, w2);
console.log(`correct words in dictionary for: ${w2}`);
for (const x of match) {
  console.log(x);
}

// This code is contributed by Amit Mangal.

g edit distance calculation. Therefore, our Time Complexity will be O(L1*L2*log n), here L1 is the average length of word in our dictionary and L2 is the length of misspelled. Generally, L1 and L2 will be small. 
Auxiliary Space: The space complexity of the above implementation is O(L1*L2) where L1 is the number of words in the dictionary and L2 is the maximum length of the word.

References

• https://en.wikipedia.org/wiki/BK-tree
• https://issues.apache.org/jira/browse/LUCENE-2230

This article is contributed by Nitish Kumar. If you like GeeksforGeeks and would like to contribute, you can also write an article using write.geeksforgeeks.org or mail your article to review-team@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.
Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.