Recursive function to calculate time complexity - c++

Question

I'm learning to calculate time complexity. I can calculate a simple case but I wonder how to calculate the time complexity of a recursive function. In normal function, it's clearly done knowing how many comparison, swaps,. . . In recursive function, I don't know how to count all cases that can happen because when I write a recursive function, I am basing it on some theory like "thinking recursively" that assumes that my function already works ... so I always confuse when calculating the time of the recursive function. The example function below that I wrote:
#define MAX 26 // represent 26 letter in alphabet
//--------- TRIE STRUCTURE-----------
struct Trie {
Trie* m_arr[MAX]; // leaves
bool m_isend; // check that leaf is the end of a word
string m_path; // store path from root to that leaf
Trie() { //constructor
m_path = ""; // path <empty>
for (int i = 0; i < MAX; i++) { // initialize leaves <NULL>
m_arr[i] = NULL;
}
m_isend = false; // check the end of word <false>
}
};
//-------------------
//-------INIT NODE---------
Trie* getNode() { // Dynamic a node
return(new Trie);
}
//-------------------
//----------INSERT WORD-------------
void insertWord(Trie* root, string str, string path = "") { // insert a word into trie (root<trie>, str<string want to insert>, path<store path of parent>)
if (str.length() == 0) { // if data is inputed has length equal to 0 -> return
return;
}
if (!root) { // if the trie that we want to add new word is empty, we will init it with NULL leaves
root = getNode();
}
if (!root->m_arr[str[0] - 'a']) { // the leaf that we want to jump to is NULL we have to init it
root->m_arr[str[0] - 'a'] = getNode();
}
if (root->m_arr[str[0] - 'a']->m_path.length() == 0) { // the leaf that we want to jump to has empty path -
// we have to add its parent's path to its path
root->m_arr[str[0] - 'a']->m_path += path;
root->m_arr[str[0] - 'a']->m_path.push_back(str[0]);
}
if (str.length() > 1) { //if lenght of word still great than 1, countinous with properties leaf
insertWord(root->m_arr[str[0] - 'a'], str.substr(1), root->m_arr[str[0] - 'a']->m_path);
}
else { //else finish process by turn flag (end of a word)
root->m_arr[str[0] - 'a']->m_isend = true;
}
}
//---------------
hope you can explain to me how to calculate time complexity of above function?