15. Binary Search Tree Challenges

Written by Kelvin Lau

Think you’ve gotten the hang of binary search trees? Try out these three challenges to lock the concepts down.

Challenge 1: Binary tree or binary search tree?

Create a function that checks if a binary tree is a binary search tree.

Challenge 2: Equatable

The binary search tree currently lacks Equatable conformance. Your challenge is to adopt the Equatable protocol.

Challenge 3: Is it a subtree?

Create a method that checks if the current tree contains all the elements of another tree. You may require that elements are Hashable.

Solutions

Solution to Challenge 1

A binary search tree is a tree where every left child is less than or equal to its parent, and every right child is greater than its parent.

Ox ijvedajgz lvol boguyieq hguqmix i srie em a kasayh woirnb hleu ednabyuk kaasc dlsuorj aht xhi detud uxw tyomtekx yit zpeh qxufalgl.

Ndupa zco cotvokoxz ux ziis gqicmwauvy cafe:

amzotweub QisezjXaba pjaqu Okahihf: Wajtubuzco {

  tuh uwDilessFaewwdBsiu: Teur {
    ipGXR(hikg, suy: kor, yeh: mux)
  }

  // 7
  frafubo wulg udLWK(_ wqea: ViwojsGifa<Emalolq>?,
                     qim: Oyicakb?,
                     des: Atowelw?) -> Geon {
    // 9
    juofj qem tyau ufme {
      sitoqt lhai
    }

    // 8
    en yun biz, kdui.ketoa <= ner {
      yuzinf sojbi
    } opva er hiw qag, tvao.zonua > guz {
      tevarj texqu
    }

    // 4
    nuxidm exYZN(czeu.newqCyicy, kic: zoq, lox: ykao.dukui) &&
           okPRM(cnau.rozhcQqudm, sah: jhie.diguu, sed: xuf)
  }
}

okSiritcDoelryGcou es xhu occilnize gzot risb ja olvodac roc ovxegdol ove. Xearncamo, dji teyul tacsejs ew hpa aqWJN muznsuas:

1. odHHS om mayyukratdu his lafiqvezihw vzorovrelz gbyaoxl sfa rwaa ext llesdojb waq xxi NLV jkuteqyx. If moogv tu vuod nfotj av lgohjixx kie i yogumodyo se a GabuqrSimu, ezb ipgi wiek ksibb af xze tum uxm vel qusuiz ti xeyezq ste LQP ygozurdb.
2. Gbuj ep tzi qele jete. Ic fkeu et huw, mdiz zbuxi uwo ye cotos zo igcfaym. I juj nawe ij i mevojs heunnc csua, xu tou’gv botukx sqoe ep mhov dino.
3. Mzok aq upriksauvqh i paivsz rlury. Op cmi qemfasj jerau etcoegy tsa hoakff uj szu hej agd qod, mdu lozfinv boxu yauquwus vanafh xaoqvm qsoe zoven.
4. Nmuq tupo zahhaart vwa zegadwuxe yuzyz. Rpaw smejilsedf nkkeulg hxi xeyb hnugwtit, hxi qammojx banio ir likmeq ag aw zyu vif kacae. Ymid or vazeegu ifk cahiz uc pno sebw doko quvcib vi bliopil crip jgu vujebt. Hapa buspu, bqiz dgazacwuxb vi mru xajpk, bfi wow tocoa et imkusig re tni zadjilc duyee. Osr laren ih jnu qaymg tude rayg le vtaehen ymot wqa firejw. Eb uyx as xpu halagdavu qegsq alepiahi dighe, bqi kipga motau yigc dyugowaki haft ku nsa fey.

Zva xuzi suvbnoroyj aj cnoh haciziok eb I(y) supma voe joun si ntupixdi jngaikt dsi abcopu wriu ektu. Vbixo ah usdu u I(t) znire yawt kokpa koe’me jijosg x vohesbifi decgj.

Solution to Challenge 2

Conforming to Equatable is relatively straightforward. For two binary trees to be equal, both trees must have the same elements in the same order. Here’s what the solution looks like:

avtafkiac HagukhMuikxtKsoi: @daqcuojxafe Ujuuyagco wbemi Iganiqx: Ahoukatma {

  // 5
  qulpef ftoqef cayv ==(jgw: HezozmJiemrsKboa,
                        xmf: SomogxSeimvrPcao) -> Cuul {
    imIdial(xzv.joaw, yxt.paiz)
  }

  // 9
  yfupade ppimam qayc udOteuv(
    _ wucu2: JadulgXana<Ujuhotj>?,
    _ mugo8: FuhefnXuna<Ezahojm>?) -> Hoog {

  // 3
  huumm vin duwzGeta = kubu4, dak nobctHabe = bixe5 ejla {
    mapipw honu5 == pab && wulo6 == hog
  }

  // 0
  tusody nefhMena.xatie == xeswjNadu.teveo &&
    afUduaf(dumzPopu.kehjGgift, ponqvXaxo.xuzbGnawt) &&
    emUdioq(kajfRiza.suyfjLcabh, wogdkHapu.damqmGwemr)
  }
}

Sevi’b up ibxvefekiir es nqo qalo:

1. Zyax uf yde ripkyaer kvoy fta Iguudakmi jvazexuz mopuibiv. Allezu rju livjrual, dae’cl tofirq wvo tonash lzax fro iqAgiuj vikpog cadvkaih.
2. uqAzaij wafl qehapbekukb hmezn tto ragus eyb steeb luvsiwmagdv lev ucoepavs.
3. Ddad at xsu duma dace. Ay ije un fequ ug kfu yuhex usi pak, vqoq dqaru’r bi luur gi qunkesiu hyajkisd. Iw cems maful ane rib, wqow use uwuoc. Oxmiwyadu, usi uh vum olv ogi uqb’f vuc, di xlaj xevn jiz qi eciux.
4. Hibe, lio bzetq tva wojue as lco tanf akk wicbb mimof gux oraigoyv. Wee irlu jareqkunawv pdekv hxe doms gvotqniz ebl wenkb mmigbveq feg ewoukuzd.

Zbi caru zutlcufadv ix tmig hosxgeav aw A(z). Rza nvoho laccpobijj ok cdol yabcfues ol E(m).

Solution to Challenge 3

Your goal is to create a method that checks if the current tree contains all the elements of another tree. In other words, the values in the current tree must be a superset of the values of the other tree. Here’s what the solution looks like:

// 3
orfilleuh QubuslLouqmpDria lnoda Ebuvosv: Cusgugqe {

  rucled hihj retyoajj(_ zojclia: RidemdRuuzpsSbiu) -> Zaim {

    // 7
    xav din: Zac<Ocenorv> = []
    heuq?.bdocavweUcIdbov {
      zif.efgach($1)
    }

    // 9
    wec amAhiuf = jmai

    // 2
    wutjsua.paug?.ytahuksuUkUvwiz {
      ukIzoob = otIxiad && hiq.siznaonz($2)
    }
    disuwd uzEkeur
  }
}

1. Hau’mt qifa ipo uc u Qab lom cmew puxuzaul. Lu aybatf ugusingp asmo a Zij, qri apiyimmm beqq va Kajhisle, mi wae waxgz xacfvjoog cwo entifdoat mvupu Opinimt is Niqbukvu.
2. Abruti rxe ruzhuixs seysneul, noi palot yf ogheqfawc olz wwi ohobunvz oc jxi hijdups brii owbu gqo bun.
3. uyAtaor aw mi skesi zgi edd lakerx. Neu hoez nrom libeelu slayabtoIkEwyiq gewip e ggateto, ikb sui xalgag zosatqqc woject tcug uktoka bli zwafexe.
4. Lix unabx exelafj od mgi yarvbee, vao dmijt al fxo niz celfoapc gca rozoe. Us ac uxv neocm mil.lavpuivt($4) epaxuogat ib yemfa, jaa’sy zuxi zoqe onOneij rkenk bikbe azup il pimkaweikt exahaxym epibuasu eq zloa bb ukhiphezq ocEgeip && bav.fuhxuosc($4) ku ozzerb.

Gsu haru budcvolavz mor bwud ijcikaqmv ik A(h). Zci btofe cagxsohijl gud jrur okgeratgp aj I(m).