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There was a related question here:

41236

So i will use the ad-hoc routine pointIsogeny, the curve E, and point $R$ defined in there for the start. If i understand the situation, then the line E = starting_node(E_0, 0) wants to build (randomly) the following:

p = 2^372*3^239 - 1
F = GF( p, proof=0 )
R.<x> = PolynomialRing(F)
# The quadratic extension via x^2 + 1 since p = 3 mod 4
Fp2.<j> = Fp.extension(x^2 + 1)

E0 = EllipticCurve(Fp2, [1,0])
R = E0.point( ( 9007518108646169717637829256143902727256908604612852170262845383236308734546752870948023665818612994373405439104311563180515971827416888758364379345147971116263603311381076594736408857657724917306603115510356333363208849059629*j + 8143875544876203102731118758998042519548033956324586056599014675159430178641351639084698635604334996400279245810044652728374901773305503117205094200107841651156165349992200320758569566012680521517849444975619314122642286738078, 5933067621852699133314119054291511797259450704514751366342623924502189539964363940282453138760679452407770908256363554867263728524097776132882938143129922521585626385240622607283870971683009886348379499590584807528366215593257*j + 3243060684426863808401390460086135176972427334603406644358264254553792355536601776050361287848102972929373427788393162068323805718475829130551068182582167101080584984966674872491237953303465307684436948645319932524248022850554 ) )

R_list, degrees_list = [], []
S = 3^(239-96)*R
while S != E0(0):
    R_list.append( S )
    degrees_list.append(3)
    S *= 3

X = ( E0, R_list, [], degrees_list )
count = 0
while X[1]:
    count += 1
    print count
    X = pointIsogeny( *X )

E = X[0]
E

So we have a starting reproducible elliptic curve E

Elliptic Curve defined by y^2 = x^3 + (7365844174710734349703979824267397716070848379207150714853939305480598177974648401456642446481782170109092245224349222954361643535826935062105068360613554750604758205242084022642407354099811489418896293593385968974015370653730*j+211392337950744742300479627443770128587762706472203430775009140876045287163506520387404594567561228376780127226866671410674341193885329801042247857950441952859225536775626695894662031453971315468618629095593065757279048175980)*x + (191888858730120017972798004127909028669268717193436606467893596092138461980127494443633833240727656111845218913366368409973705945852937546861824400761190208020898528637753121592516776112625907992364462314541787272525903553709*j+577381321884404637808391205760891583039563709656261523188365644721839518395090059951637547964468822095873175263496035533602755866120293764901657313023000531067504187145886064396313750664978511489728895045390110744492542111986) over Finite Field in j of size 10354717741769305252977768237866805321427389645549071170116189679054678940682478846502882896561066713624553211618840202385203911976522554393044160468771151816976706840078913334358399730952774926980235086850991501872665651576831^2

for the further walk process. (That i maybe do not understand in this second.) The walking part may be given by the following:

MAXLEVEL = 372

def walk(E, jString, level, previous_jString ):
    children_DIC = {}    # dictionary jString-value -> number of occurences
    if level >= MAXLEVEL:
        return

    for P in E(0).division_points(2):
        if P == E(0):
            continue
        # psi = E.isogeny(P)
        # EE  = psi.codomain()
        EE = pointIsogeny( E, [P], [], [2] )[0]
        jj = str( EE.j_invariant() )    # ?a string (here and elsewhere in similar places)
        level_child = level + 1
        if jj in children_DIC:
            children_DIC[jj] += 1
        else:
            children_DIC[jj]  = 1

        print "Level = %s children_DIC[%s] = %s\n" % ( level, jj, children_DIC[jj] )
        if (jj != previous_jString) or (children_DIC[jj] > 1):
            walk(EE, jj, level+1, jString)


print "start: ", str(E.j_invariant())
walk(E, str(E.j_invariant()), 0, "")

It may be that the children dictionary is / should be a global constant. (Then it should be defined outside, and declared as global inside the walk.) I inserted the print in order to see something.

At any rate, the 2-torsion isogeny problem should be solved so far.