How to print all words in a Trie?

Question

I am trying to create a Trie Implementation in C++. I cannot figure out how to print all words stored in the Trie.

This is how I've implemented the TrieNode.

struct TrieNode{
  bool isWord;
  int data; //Number of times Word Occured
  TrieNode *Child[ALPHABET_SIZE]; //defined as 26
};

I know I could store a pointer to the parent node, Depth-First Search for all nodes where isWord==True and recursively print each word from those nodes.

But I'm wondering is there a way to print out each word in the Trie with my implementation of a TrieNode.

Thanks for any help.

Answer 1

Here is a reasonably efficient version of Konrad Rudolph, without assuming data is a character. I also removed the O(n^2) total memory allocated in Konrad's version at the cost of using a std::string&. The idea is to pass down the prefix and modify it at each recursion, pushing characters onto the end and poping it afterwards, ends up being slightly more efficient than copying it madly.

void traverse(std::string& prefix, TrieNode const& node) {
  if (node.isWord)
    print(prefix);

  for (char index = 0; index < ALPHABET_SIZE; ++index) {
    char next = 'a'+index;
    TrieNode const* pChild = node.Child[index];
    if (pChild) {
      prefix.push_back(next);
      traverse(prefix, *pChild);
      prefix.pop_back();
    }
  }
}

Answer 2

You don’t need your parent node, your code readily accommodates traversal via recursion. Pseudo-code:

void traverse(string prefix, TrieNode const& n) {
    prefix += static_cast<char>(n.data);

    if (n.isWord)
        print(prefix);

    for (auto const next : n.Child)
        if (next)
            traverse(prefix, *next);
}

This is more or less valid C++. Just define print appropriately.

EDIT In response to Yakk’s comment and your clarification, here’s a version which doesn’t assume that data contains the current character (bad slip on my part!):

void traverse(string const& prefix, TrieNode const& n) {
    if (n.isWord)
        print(prefix);

    for (std::size_t i = 0; i < ALPHABET_SIZE; ++i)
        if (n.child[i])
            traverse(prefix + ('a' + i), *n.child[i]);
}

I’ll leave the more efficient implementation to Yakk’s answer.

Answer 3

void traversePrint(TrieNode* sr,char* out,int index)
{
    if(sr!=NULL)
    {
        for(int i=0;i<SIZE;i++)
        {
            if(sr->child[i]!=NULL)
            {
                out[index] = 'a'+i;
                traversePrint(sr->child[i],out,index+1);
            }
        }
        if(sr->isEnd)
        {
            out[index]='\0';
            cout << out << endl;
        }
    }
}

// Calling

char out[MAX_WORD_SIZE];
traversePrint(root,out,0);

Answer 4

I dont think isword is needed here. The existence of the pointer to children will be enough to traverse the trie for available words in the trie. to find a word, start from the root and look for the current character within the word during any recursive step.

struct trie {
  trie *children[ALPHABET_SIZE];
};


void traversal(trie *&t, string &str) {
    bool is_seen = false;
    for(int i = 0; i < ALPHABET_SIZE; i++) {
        if(t->children[i]) {
            if(!is_seen) {
               is_seen = true;
            }
            str.push_back(t[i]);
            traversal(t->children[i], str);
            str.pop_back();
        }
    }
    if(!is_seen) {
        cout << str << "\n";
    }

}