Trie implementation is to implement a data structure called trie. A trie is a tree-like data structure in which every node stores a character. After building the trie, strings or substrings can be retrieved by traversing down a path of the trie. Insert a word in trie, search a word and delete a word by key all takes O(n) time, at the expense of storage.
Trie’s node can be implemented with array, linked list, HashMap. Each implementation has its pros and cons. In array implementation, the character in the node will be used as index to find the pointer of the child. Array takes spaces. Linked list overcomes this problem by creating nodes only for the next character in the words. However a method is called to find the child by search each node in the list. So this implementation uses time to trade the space. This post is trie implementation using linked list.
Part 1 – Trie implementation using array
Part 2 – Trie implementation using linked list
Part 3 – Trie implementation using hash map
Table of Content
Define classes in trie implementation
The trie node class has tree variables: data stores a character, children is a data structure that points to the children (branches) nodes (here the data structure is linked list); isEnd is to mark whether this is the last node to compose a word. The trie class, there is one class level variable – root. The operation of insert, search or delete always starts from the root.
Java
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public class TrieList { static class TrieNode { char data; LinkedList<TrieNode> children = new LinkedList<>(); boolean isEnd = false; //Constructor, Time O(1), Space O(1) TrieNode(char c) { this.data = c; } //find the node by char, the same functionality as children[ch] in array implementation //Time O(k), Space O(1), k is number of children of this node TrieNode getChild(char c) { if (children != null) for (TrieNode ch : children) if (ch.data == c) return ch; return null; } } static class Trie { TrieNode root = new TrieNode(' '); } |
Javascript
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class TrieNode { //Constructor, Time O(1), Space O(1) constructor(c) { this.data = c; this.children = new Array(); //List this.isEnd = false; } //find the node by char, the same functionality as children[ch] in array implementation //Time O(k), Space O(1), k is number of children of this node getChild(c) { if (this.children != null) for (let child of this.children) if (child.data == c) return child; return null; } } class Trie { //Constructor, Time O(1), Space O(1) constructor() { this.root = new TrieNode(''); } } |
Python
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class TrieNode: #Constructor, Time O(1), Space O(1) def __init__(self, c): self.data = c self.children = [] # list self.isEnd = False #find the node by char, the same functionality as children[ch] in array implementation #Time O(k), Space O(1), k is number of children of this node def getChild(self, c): if self.children != None: for child in self.children: if child.data == c: return child return None class Trie: #Constructor, Time O(1), Space O(1) def __init__(self): self.root = TrieNode('') |
Insert in trie implementation
To insert a word, first check whether the word is in the trie so that we don’t insert duplicate word. Then we loop through each character in the word. A pointer node starts from the root node. If the character is not in the node‘s children, a child node is created. Then the node moves to the child node. When the node is at the last character of the word, we mark the node‘s isEnd to be true.
Java
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//Add a word to tire, Iteration, Time O(s), Space O(s), s is word length void insert(String word) { if (search(word) == true) { System.out.println(word + " is already in trie."); return; } TrieNode node = root; for (char ch : word.toCharArray()) { if (node.getChild(ch) == null) node.children.add(new TrieNode(ch)); node = node.getChild(ch); } node.isEnd = true; } |
Javascript
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//insert a word into the trie, Time O(s), Space O(s), s is word length insert(word) { if (this.search(word) == true) { System.out.println(word + " is already in trie."); return; } var node = this.root; for (let ch of word) { if (node.getChild(ch) == null) node.children.push(new TrieNode(ch)); node = node.getChild(ch); } node.isEnd = true; } |
Python
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#Add a word to trie, Iteration, Time O(s), Space O(s), s is word length def insert(self, word) : if self.search(word) == True : print(word + " is already in trie.") return node = self.root for ch in word: if node.getChild(ch) == None: node.children.append(TrieNode(ch)) node = node.getChild(ch) node.isEnd = True |
Doodle
Delete
To delete a word from the trie, first check whether the word exists in the trie. If not, the method can return. The same as insert, loop through each character in the word. A pointer node starts from the root node. If the character is not in the node‘s children, the method returns directly. Otherwise, the node moves to the child node. When the node is at the last character of the word, mark the node‘s isEnd to be false.
Java
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//Remove a word from trie, Iteration, Time O(s), Space O(1), s is word length void delete(String word) { if (this.search(word) == false) { System.out.println(word + " does not exist in trie."); return; } TrieNode node = root; for (char ch : word.toCharArray()) { if (node.getChild(ch) == null) return; node = node.getChild(ch); } node.isEnd = false; } |
Javascript
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//Remove a word from trie, Iteration, Time O(s), Space O(1), s is word length delete(word) { if (this.search(word) == false){ console.log(word + " does not exist in trie."); return; } let node = this.root; for (let ch of word) { if (node.getChild(ch) == null) return; node = node.getChild(ch); } node.isEnd=false; } |
Python
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#Remove a word from trie, Iteration, Time O(s), Space O(1), s is word length def delete(self, word) : if self.search(word) == False: print(word + " does not exist in trie.") return node = self.root for ch in word: if (node.getChild(ch) == None) : return node = node.getChild(ch) node.isEnd = False |
Search
To search a word in the trie, loop through each character in the word. A pointer node starts from the root node. If the character is not in the node‘s children, the method returns false. Otherwise, the node moves to the child node. When the node is at the last character of the word, return the node‘s isEnd value, which can be true or false(deleted).
Java
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//Search a word in trie, Iteration, Time O(s), Space O(1), s is word length boolean search(String word) { TrieNode node = root; for (char c : word.toCharArray()) { if (node.getChild(c) == null) return false; node = node.getChild(c); } return node.isEnd; } |
Javascript
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//Search a word in trie, Iteration, Time O(s), Space O(1), s is word length search(word) { var node = this.root; for (let ch of word) { if (node.getChild(ch) == null) return false; node = node.getChild(ch); } return node.isEnd; } |
Python
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#Search a word in trie, Iteration, Time O(s), Space O(1), s is word length def search(self, word) : node = self.root for ch in word : if node.getChild(ch) == None: return False node = node.getChild(ch) return node.isEnd |
Print
To print all words in the trie, recursion is used to traverse all nodes in trie. This is similar to pre-order (DFS, depth first search) of the tree. When visiting the node, the method concatenates characters from previously visited nodes with the character of the current node. When the node‘s isEnd is true, the recursion reaches the last character of the word, and add the word to the result list.
Java
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//Print all words in trie, call recursion function //Time O(n), Space O(n), n is number of nodes in trie void print() { List<String> res = new ArrayList<String>(); helper(root, res, "" ); System.out.println(res); } //recursion function, Time O(n), Space O(n), n is number of nodes in trie void helper(TrieNode node, List<String> res, String prefix) { if (node.isEnd) { String word = prefix + node.data; res.add(word.substring(1)); //skip the first space from root } for (TrieNode child : node.children) helper(child, res, prefix + node.data); } |
Javascript
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//Print all words in trie, call recursion function //Time O(n), Space O(n), n is number of nodes in trie print () { let res = []; this.helper(this.root, res, ""); console.log(res); } //recursion function, Time O(n), Space O(n), n is number of nodes in trie helper (node, res, prefix) { if (node.isEnd) res.push(prefix + node.data); for (let child of node.children) //list this.helper(child, res, prefix + node.data); } |
Python
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#Print all words in trie, call recursion function #Time O(n), Space O(n), n is number of nodes in trie def print (self) : res = [] self.helper(self.root, res, "") print(res) #recursion function, Time O(n), Space O(n), n is number of nodes in trie def helper(self, node, res, prefix) : if node.isEnd: res.append(prefix + node.data); for child in node.children: self.helper(child, res, prefix + node.data) |
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