LeetCode 208. Implement Trie (Prefix Tree) (javascript solution)

Description:

A trie (pronounced as "try") or prefix tree is a tree data structure used to efficiently store and retrieve keys in a dataset of strings. There are various applications of this data structure, such as autocomplete and spellchecker.

Implement the Trie class:

Trie() Initializes the trie object.
void insert(String word) Inserts the string word into the trie.
boolean search(String word) Returns true if the string word is in the trie (i.e., was inserted before), and false otherwise.
boolean startsWith(String prefix) Returns true if there is a previously inserted string word that has the prefix prefix, and false otherwise.

/**
 * Initialize your data structure here.
 */
var Trie = function() {
    this.root = {};
};

/**
 * Inserts a word into the trie. 
 * @param {string} word
 * @return {void}
 */
Trie.prototype.insert = function(word) {
    let node = this.root;
    for(let letter of word) {
        if (node[letter] === undefined) node[letter] = {};
        node = node[letter]
    }
    node.isEnd = true;
};

/**
 * Returns if the word is in the trie. 
 * @param {string} word
 * @return {boolean}
 */
Trie.prototype.search = function(word) {
    let node = this.root
    for(let letter of word) {
        // check if current letter is in the node
        if(!node[letter]) {
            return false;
        } else {
            node = node[letter];
        }
    }

    return node && node.isEnd === true;
};

/**
 * Returns if there is any word in the trie that starts with the given prefix. 
 * @param {string} prefix
 * @return {boolean}
 */
Trie.prototype.startsWith = function(prefix) {
    let node = this.root;
    for(let letter of prefix) {
        if(!node[letter]) {
            return false;
        } else {
            node = node[letter];
        }
    }
    return true;
};

Comments

[Siva Rama Krishna Bathula]

Woow!! Simply woow. Amazed at your solution. Its like cascading nodes of letters. Performs significantly faster than my own solution.