Identifying root parents and all their children in trees

I suggest you use networkx, as this is a graph problem. In particular the descendants function:

import networkx as nx
import pandas as pd

data = [['A', 'B', 0, 1],
        ['B', 'C', 1, 2],
        ['B', 'D', 1, 2],
        ['X', 'Y', 0, 2],
        ['X', 'D', 0, 2],
        ['Y', 'Z', 2, 3]]

df = pd.DataFrame(data=data, columns=['parent', 'child', 'parent_level', 'child_level'])

roots = df.parent[df.parent_level.eq(0)].unique()
dg = nx.from_pandas_edgelist(df, source='parent', target='child', create_using=nx.DiGraph)

result = pd.DataFrame(data=[[root, nx.descendants(dg, root)] for root in roots], columns=['root', 'children'])
print(result)

Output

  root   children
0    A  {D, B, C}
1    X  {Z, Y, D}

With Recursion

def find_root(tree, child):
    if child in tree:
        return {p for x in tree[child] for p in find_root(tree, x)}
    else:
        return {child}

tree = {}
for parent, child in zip(df.parent, df.child):
    tree.setdefault(child, set()).add(parent)

descendents = {}
for child in tree:
    for parent in find_root(tree, child):
        descendents.setdefault(parent, set()).add(child)

pd.DataFrame(descendents.items(), columns=['root', 'children'])

  root   children
0    A  {B, D, C}
1    X  {Z, D, Y}

---

You could alternatively set up find_root as a generator

def find_root(tree, child):
    if child in tree:
        for x in tree[child]:
            yield from find_root(tree, x)
    else:
        yield child

Further, if you want to avoid recursion depth issues, you can use the "stack of iterators" pattern to define find_root

def find_root(tree, child):
    stack = [iter([child])]
    while stack:
        for node in stack[-1]:
            if node in tree:
                stack.append(iter(tree[node]))
            else:
                yield node
            break
        else:  # yes!  that is an `else` clause on a for loop
            stack.pop()