Display trie tree elements
In computer science, a trie is a tree-like data structure used to store a dynamic set of strings, where each node represents a single character. Tries are commonly used in applications involving text processing, such as autocomplete suggestions, spell checking, and IP routing. One common task with a trie is to display all the elements it contains.
Problem Statement
Given a Trie data structure with strings inserted into it, the task is to display all the elements (strings) present in the Trie.
Example
Consider the following Trie structure:
root
/ \ \
c d r
| | |
o o u
| \ | |
d w g n
|
e
The strings inserted into this Trie are: "run", "code", "dog", and "cow". The goal is to display these strings in lexicographic (alphabetical) order.
Idea to Solve
To solve this problem, we can perform a depth-first traversal of the Trie starting from the root node. During traversal, we can keep track of the current prefix formed by the characters encountered so far. When we reach a node that marks the end of a string, we print the current prefix to display the complete word.
Pseudocode
display(node, currentPrefix):
if node is null:
return
if node.end is true:
print currentPrefix
for i from 0 to ALPHABETS - 1:
if node.children[i] is not null:
display(node.children[i], currentPrefix + character(i))
Algorithm Explanation
- Start at the root node of the Trie.
- Initialize an empty string called
currentPrefix
. - Traverse through each child of the current node.
- For each child, recursively call the
display
function with the child node and the updatedcurrentPrefix
by adding the corresponding character. - If the current node marks the end of a string (i.e.,
end
is true), print thecurrentPrefix
. - Repeat steps 3-5 for each child of the current node.
- Continue the traversal until all paths in the Trie have been explored.
Code Solution
/*
C++ program
Trie Display Elements
*/
#include <iostream>
#include <string.h>
#define ALPHABETS 26
using namespace std;
class Node {
public:
bool end;
Node *children[ALPHABETS];
Node() {
this->end = false;
for (int i = 0; i < ALPHABETS; ++i) {
this->children[i] = NULL;
}
}
};
class Trie {
public:
Node *root;
Trie() {
this->root = new Node();
}
void insert(string text) {
int length = text.size();
int index = 0;
Node *head = this->root;
int level = 0;
while (level < length) {
index = text[level] - 'a';
if (head->children[index] == NULL) {
head->children[index] = new Node();
}
head = head->children[index];
level++;
}
if (head != NULL) {
head->end = true;
}
}
void display(Node *head, string about) {
if (head != NULL) {
if (head->end == true) {
cout << about<<endl;
}
for (int i = 0; i < ALPHABETS; i++) {
if (head->children[i] != NULL) {
this->display(head->children[i], about + ((char)(i + 97)));
}
}
}
}
};
int main() {
Trie obj;
/*
root
/ \ \
c d r
| | |
o o u
| \ | |
d w g n
|
e
*/
obj.insert("run");
obj.insert("code");
obj.insert("dog");
obj.insert("cow");
obj.display(obj.root, "");
return 0;
}
Output
code
cow
dog
run
/*
Java program
Trie Display Elements
*/
class Node {
//Indicates end of string
public boolean end;
public Node[] children;
public Node(int size) {
end = false;
children = new Node[size];
}
}
public class Trie {
public int ALPHABETS = 26;
public Node root;
public Trie() {
root = new Node(ALPHABETS);
}
public void insert(String text) {
//First get the length of text
int length = text.length();
//This variable are used to task find the index location of character
int index = 0;
//Get the root node
Node head = root;
for (int level = 0; level < length; level++) {
//Get the index
index = text.charAt(level) - 'a';
if (head.children[index] == null) {
//When need to add new Node
head.children[index] = new Node(ALPHABETS);
}
head = head.children[index];
}
if (head != null) {
//Indicates end of string
head.end = true;
}
}
public void display(Node head, String about) {
if (head != null) {
if (head.end == true) {
System.out.println(about);
}
for (int i = 0; i < ALPHABETS; i++) {
if (head.children[i] != null) {
display(head.children[i], about + ((char)(i + 97)));
}
}
}
}
public static void main(String[] args) {
//Make object
Trie obj = new Trie();
/*
root
/ \ \
c d r
| | |
o o u
| \ | |
d w g n
|
e
*/
obj.insert("run");
obj.insert("code");
obj.insert("dog");
obj.insert("cow");
obj.display(obj.root, "");
}
}
Output
code
cow
dog
run
# Python 3 Program
# Trie Display Elements
class Node :
def __init__(self, size) :
self.end = False
self.children = [None]*size
class Trie :
def __init__(self) :
self.ALPHABETS = 26
self.root = Node(self.ALPHABETS)
def insert(self, text) :
length = len(text)
index = 0
head = self.root
level = 0
while (level < length) :
index = ord(text[level]) - ord('a')
if (head.children[index] == None) :
head.children[index] = Node(self.ALPHABETS)
head = head.children[index]
level += 1
if (head != None) :
head.end = True
def display(self, head, about) :
if (head != None) :
if (head.end == True) :
print(about)
i = 0
while (i < self.ALPHABETS) :
if (head.children[i] != None) :
self.display(head.children[i], about + (chr(i + 97)))
i += 1
def main() :
obj = Trie()
# root
# / \ \
# c d r
# | | |
# o o u
# | \ | |
# d w g n
# |
# e
obj.insert("run")
obj.insert("code")
obj.insert("dog")
obj.insert("cow")
obj.display(obj.root, "")
if __name__ == "__main__":
main()
Output
code
cow
dog
run
/*
C# program
Trie Display Elements
*/
using System;
public class Node {
//Indicates end of string
public Boolean end;
public Node[] children;
public Node(int size) {
end = false;
children = new Node[size];
}
}
public class Trie
{
public int ALPHABETS = 26;
public Node root;
public Trie() {
root = new Node(ALPHABETS);
}
public void insert(String text) {
//first get the.Length of text
int size = text.Length;
//This variable are used to task find the index location of character
int index = 0;
//get the root node
Node head = root;
for (int level = 0; level < size; level++) {
//Get the index
index = text[level] - 'a';
if (head.children[index] == null) {
//When need to add new Node
head.children[index] = new Node(ALPHABETS);
}
head = head.children[index];
}
if (head != null) {
//indicates end of string
head.end = true;
}
}
public void display(Node root, String about) {
if (root != null) {
if (root.end == true) {
Console.WriteLine(about);
}
for (int i = 0; i < ALPHABETS; i++) {
if (root.children[i] != null) {
display(root.children[i], about + ((char)(i + 97)));
}
}
}
}
public static void Main(String[] args)
{
//Make object
Trie obj = new Trie();
/*
root
/ \ \
c d r
| | |
o o u
| \ | |
d w g n
|
e
*/
obj.insert("run");
obj.insert("code");
obj.insert("dog");
obj.insert("cow");
obj.display(obj.root, "");
}
}
Output
code
cow
dog
run
# Ruby program
# Trie Display Elements
class Node
attr_reader :end, :children
attr_accessor :end, :children
def initialize(size)
@end = false
@children = Array.new(size){nil}
end
end
class Trie
attr_reader :ALPHABETS, :root
attr_accessor :ALPHABETS, :root
def initialize()
@ALPHABETS = 26
@root = Node.new(@ALPHABETS)
end
def insert(text)
size = text.length
location = 0
head = @root
level = 0
while (level < size)
location = text[level].ord - 'a'.ord
if (head.children[location] == nil)
head.children[location] = Node.new(@ALPHABETS)
end
head = head.children[location]
level += 1
end
if (head != nil)
head.end = true
end
end
def display(head, about)
if (head != nil)
if (head.end == true)
print(about,"\n")
end
i = 0
while (i < @ALPHABETS)
if (head.children[i] != nil)
self.display(head.children[i], about + ((i + 97).chr))
end
i += 1
end
end
end
end
def main()
obj = Trie.new()
# root
# / \ \
# c d r
# | | |
# o o u
# | \ | |
# d w g n
# |
# e
obj.insert("run")
obj.insert("code")
obj.insert("dog")
obj.insert("cow")
obj.display(obj.root, "")
end
main()
Output
code
cow
dog
run
<?php
/*
Php Program
Trie Display Elements
*/
class Node {
public $end;
public $children = [];
function __construct($size) {
$this->end = false;
$this->children = array_fill(0, $size, null);
}
}
class Trie {
private $ALPHABETS;
public $root;
function __construct() {
$this->ALPHABETS = 26;
$this->root = new Node($this->ALPHABETS);
}
public function insert($text) {
$size = strlen($text);
$index = 0;
$head = $this->root;
$level = 0;
while ($level < $size) {
$index = ord($text[$level]) - ord('a');
if ($head->children[$index] == null) {
$head->children[$index] = new Node($this->ALPHABETS);
}
$head = $head->children[$index];
$level++;
}
if ($head != null) {
$head->end = true;
}
}
public function display($head, $about) {
if ($head != null) {
if ($head->end == true) {
echo $about."\n";
}
$i = 0;
while ($i < $this->ALPHABETS) {
if ($head->children[$i] != null) {
$this->display($head->children[$i], $about . (chr($i + 97)));
}
$i++;
}
}
}
}
function main() {
$obj = new Trie();
/*
root
/ \ \
c d r
| | |
o o u
| \ | |
d w g n
|
e
*/
$obj->insert("run");
$obj->insert("code");
$obj->insert("dog");
$obj->insert("cow");
$obj->display($obj->root, "");
}
main();
Output
code
cow
dog
run
/*
Node Js Program
Trie Display Elements
*/
class Node {
constructor(size) {
this.end = false;
this.children = new Array(size).fill(null);
}
}
class Trie {
constructor() {
this.ALPHABETS = 26;
this.root = new Node(this.ALPHABETS);
}
insert(text) {
var size = text.length;
var index = 0;
var head = this.root;
var level = 0;
while (level < size) {
index = text.charCodeAt(level) - 'a'.charCodeAt(0);
if (head.children[index] == null) {
head.children[index] = new Node(this.ALPHABETS);
}
head = head.children[index];
level++;
}
if (head != null) {
head.end = true;
}
}
display(head, about) {
if (head != null) {
if (head.end == true) {
console.log(about);
}
var i = 0;
while (i < this.ALPHABETS) {
if (head.children[i] != null) {
this.display(head.children[i], about + (String.fromCharCode(i + 97)));
}
i++;
}
}
}
}
function main() {
var obj = new Trie();
/*
root
/ \ \
c d r
| | |
o o u
| \ | |
d w g n
|
e
*/
obj.insert("run");
obj.insert("code");
obj.insert("dog");
obj.insert("cow");
obj.display(obj.root, "");
}
main();
Output
code
cow
dog
run
Time Complexity Analysis
In the Trie, each character is stored in a separate node, and each node has a fixed number of children (in this case,
26 for the English alphabet). Therefore, the time complexity of the display
function depends on the
number of nodes in the Trie. Let's denote the total number of nodes as N
.
During the traversal, each node is visited once, and each visit involves a constant amount of work. Hence, the time
complexity of the display
function is O(N), where N
is the total number of nodes in the
Trie.
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