Stack["((a($Z"]
encounter ')'
Stack["((a"], vector<"Z$">
reverse vector -> vector<"$Z">
Parent Node -> N(id='1'), children -> <N(id='$'), N(id='Z')>
push N(1) to stack

Stack["((a1]
encounter ')'
Stack["("], vector<"1a">
reverse vector -> vector<"a1">
Parent Node -> N(id='2'), children -> <N(id='a'), N(id=1)>
push N(2) to stack

Stack["(2(An"]
encounter ')'
...

Node* solve(string s) {
    stack<Node *> st;
    int idx = 0, rt = 1;
    while (idx < s.size()) {
        Node *n = get_node(s[idx]);
        st.push(n);
        ++idx;

        if (st.top()->v == ')') {
            st.pop();
            vector<Node *> child;
            while (st.size() && st.top()->v != '(') {
                child.push_back(st.top()); st.pop();
            }
            if (st.top()->v == '(') st.pop();

            Node *par = get_node('0' + rt++);
            reverse(child.begin(), child.end());
            for (Node *t : child) {
                t->parent = par;
                par->child.push_back(t);
            }
            st.push(par);
        }
    }
    return st.top();
}

input: ((a($Zk))(An))
output:
Parent: 4: 2 3
Parent: 2: a 1
Parent: 3: A n
Parent: 1: $ Z k