Can I horizontally align a pydot graph by distance from head?

I've created a graph as shown below. However, I'd like to arrange the graph hierarchically by shortest distance from the head (A). In other words: C, B, D, E should all be in the same level and horizontally aligned, since they're each all 1 edge (shortest path) away from A. Then, F, G, H should be in the next level since they're each 2 edges away, etc.

I like the way the graph looks, so the solution would ideally keep this visualization style.

import matplotlib.pyplot as plt
import networkx as nx
import pydot
from networkx.drawing.nx_pydot import graphviz_layout
from IPython.display import Image, display

G=nx.Graph()
G.add_edges_from([
    ('A','B'),
    ('A','C'),
    ('A','E'),
    ('A','D'),
    ('B','C'),
    ('B','F'),
    ('C','F'),
    ('D','H'),
    ('D','G'),
    ('E','H'),
    ('F','I'),
    ('G','I'),
    ('G','J'),
    ('H','J'),
    ('I','K'),
    ('J','K')
])


pdot = nx.drawing.nx_pydot.to_pydot(G)
graph = Image(pdot.create_png())
display(graph)

There is a built-in function that allows calculating the length of the shortest paths to a given node. The lengths can be used to specify one coordinate. The other coordinate is calculated based on the number of nodes that are at the same distance to A:

from collections import Counter

distances = list(nx.single_target_shortest_path_length(G, 'A'))

counts = Counter(a[1] for a in distances)

for c in counts:
    counts[c] /= 2 # divide by 2 to get symmetrical distances from the y-axis

# generate dictionary that holds node positions:
pos = {}
for (node, d) in distances:
    pos[node] = (counts[d], -d) # x-position determined by number of counts, y position determined by distance. 
    counts[d] -= 1 # decrement count to draw nodes at shifted position
    
nx.draw_networkx(G, pos=pos) # i dont have pydot, but the basic logic should work. just use the pos argument in the pydot function.