I found it a bit more easier and understandable to do this in a BFS way. Time and space complexity remains the same.

// class TrieNode {
//     TrieNode[] children = new TrieNode[26];  // Representing each character of the alphabet.
//     boolean isEnd = false;  // To determine if the current TrieNode marks the end of a word.
// }

public class Solution {
    private TrieNode root;

    public Solution() {
        // ToDo: Write Your Code Here.
        this.root = new TrieNode();
    }

    // Function to add a word into the trie structure.
    public void addWord(String word) {
        // ToDo: Write Your Code Here.
        TrieNode start = root;
        for (char c : word.toCharArray()) {
            if (start.children[c - 'a'] == null) {
                start.children[c - 'a'] = new TrieNode();
            }
            start = start.children[c - 'a'];
        }
        start.isEnd = true;
    }

    // Function to search a word, considering '.' as a wildcard.
    public boolean search(String word) {
        // ToDO: Write Your Code Here.
        TrieNode start = root;
        Queue<TrieNode> queue = new LinkedList<>();
        queue.offer(start);
        for (char c : word.toCharArray()) {
            int queueSize = queue.size();

            for (int i = 0; i < queueSize; i++) {
                TrieNode popped = queue.poll();
                if (c == '.') {
                    for (int j = 0; j < popped.children.length; j++) {
                        if (popped.children[j] != null) {
                            queue.offer(popped.children[j]);
                        }
                    }
                } else {
                    if (popped.children[c - 'a'] == null) continue;
                    queue.offer(popped.children[c - 'a']);
                }
            }
        }
        while (!queue.isEmpty()) {
            TrieNode popped = queue.poll();
            if (popped.isEnd) return true;
        }
        return false;
    }
}