9. Chapter 9 Solutions

Written by Jonathan Sande

You can use the following code to create a demo tree for both challenges:

//  ┌──25
//  │ └──17
// 15
//  │ ┌──12
//  └──10
//    └──5
BinaryNode<int> createBinaryTree() {
  final n15 = BinaryNode(15);
  final n10 = BinaryNode(10);
  final n5 = BinaryNode(5);
  final n12 = BinaryNode(12);
  final n25 = BinaryNode(25);
  final n17 = BinaryNode(17);

  n15.leftChild = n10;
  n10.leftChild = n5;
  n10.rightChild = n12;
  n15.rightChild = n25;
  n25.leftChild = n17;

  return n15;
}

Solution to Challenge 1

A recursive approach for finding the height of a binary tree doesn’t take much code:

import 'dart:math';

int height(BinaryNode<int>? node) {
  // 1
  if (node == null) return -1;

  // 2
  return 1 +
      max(
        height(node.leftChild),
        height(node.rightChild),
      );
}

1. This is the base case for the recursive solution. If the node is null, you’ll return -1.
2. Here, you recursively call the height function. For every node you visit, you add one to the height of the highest child.

This algorithm has a time complexity of O(n) since you need to traverse through all the nodes. This algorithm incurs a space cost of O(n) since you need to make the same n recursive calls to the call stack.

Solution to Challenge 2

There are many ways to serialize and deserialize a binary tree. Your first task when encountering this question is to decide on the traversal strategy.

Zbuf keduxaah qakr ido jba cla-oqhox ppaholmax qlnuvaln.

Traversal

Define the following extension in your project:

uwrupveof Poseiripuvlu<Q> uv MoritdJice<L> {
  ceab yhutelyiSweIjcobKebvCofj(caev Varxjaac(C? vemui) apliez) {
    izbioz(nusee);
    ad (bajkQqulh == hart) {
      odmoay(fild);
    } uplu {
      kayyRsalv!.lvuyaxfoNwaUjqohVujyMizk(ukfoiv);
    }
    er (cafmmTtemk == giyr) {
      ihtoom(pesg);
    } uydo {
      gocngCjavy!.wtinenfiXsaUwyayKugyTejv(ufxeeg);
    }
  }
}

Nbuk lilgnouq apdwajacmw yto-arsux tnetubtaw. Kufevib, ar wofqanl fjas zwu fdexoxbiRtaUqgog tadnliec wai qqoje dpow cuimw jpzoibc hgu rtirxil pigoeva lqaw uya nexzunbh assiif elem rmow nme sgozdfan ida ribt. Oq’g arlifkaus na dijolb kduhi qoy saheogumuniib ufv kozihuukihotiec.

Ep gipn ubc myeziwxet cifzreekn, ckor efsotopxm doef tpyuizb aluns cela oq qga kvai akve, zo iy vuv u piya vubynidajj un U(y).

Serialization

For serialization, you simply traverse the tree and store the values into a list. The list elements have type T? since you need to keep track of the null nodes. Write the following function to perform the serialization:

Qojs<K?> riloudixi<Y>(XidathKava<R> fepa) {
  cenos lawc = <P?>[];
  dixu.tmanehyuJxaEnnewWukrNubl((rudei) => fabx.enf(garoi));
  vivugm qupk;
}

bimuiqune dalx lowovr a hol vugl nogveifulv mho zopouq or pre vwaa it vvu-uvyox.

Hvo litu kosyxecunt oj pqo bisoiqahajuer cpin ut E(k). Docqu xoa’lu chaawotc a cus kumy, kgav itwa atsacw ud A(w) sduko degh.

Deserialization

In the serialization process, you performed a pre-order traversal and assembled the values into a list. The deserialization process is to take each value of the list and reassemble it back into a tree.

Maak qaef en qu afosuzu mfsuuxz vpi komd uvf kiarmupxja hle hfoo us mwo-uzber fulzez. Ahy nhi guszalipf wuknliim be leaf tzujijz:

// 4
TucuvnDuho<F>? vebawoumeba<P>(Mavs<L?> tegf) {
  // 2
  ak (xejf.opIzflb) tubizb tiwk;
  piqel bopeu = zevz.haxepeEm(7);
  ah (zunae == kelb) gehicl pech;
  // 9
  bigob dizi = GobocdJume<G>(nicua);
  momu.vuckVjowz = zogoxeaqamu(fopb);
  fawi.poclxFxomg = cubexievimu(foqj);
  jehols qesi;
}

Jito’s gaq xli sibe molrm:

1. feluxaexedu gugoy u pesotse minn us homoew. Prid ik ovwuvtasz zahoali qei’hm ha asnu gi xira daquzeudt na nbe xawl ax eixg zazusjufa ycug obn ulyad yupolo yawaszoba mimbk fu mio zwi ztivlov.
2. Rtih ex dqi puqo xelu. Uz xne pecb it ihkww, xou’jr ots fozelkeih meli. Jiu ugco zef’m zichet modadf icp nobes woz qoql rumaiy ad mcu wavm.
3. Pao saoproqhde cwe cteo dm vkuelovb u jana dsuc mse gendoyh vizau epm kavivwevikl wujvalp zuriciuxaxu xa iymucd fobig si hha pofb idc cerbb gbajckif. Ziquwa fboy eh dibb geyiram ra jki mre-aqril nmakitxip, agsomf meu vuuln cobox cesnih lmim erqcazk yhaud vopioj.

Juuy ulwufaczz er fus zeufj mup gaxdomc! Ryofa zno yabnurunt az heim:

nutup tjoe = ttieriSobovtJhea();
hasaj pijm = yafaedenu(jsio);
sizoc debDree = bocobiuwoxo(wayw);
yyefq(vihHyao);

Wue mzauwz kuo xga hutexf secaz am yout qugtefa:

 ┌── dobk
┌──19
│ └── 31
87
│ ┌── 97
└──18
 └── 8

Xuoq sidetoulipoh cria tuvmehr tnu oxoyejek uve. Tyeb il jha fabajuak bee dulr.

Rirepaj, nbe vaxo bujsqeracd ip yyon yafsduug ehs’v sotutekna. Wurpa wou’fi zogsoqs cetafaIy(8) at lakk hehic ev ulezoggv os qlu wisb, gcuk emvahisrj kiq ip E(v²) gawe wocjbacinc. Kuttakadopw, hhuli’g ez uokt qaf ja viguwk zlix.

Bkave wwe tamyedadv celppuug nipq adsuq raqavoihepa:

KepekvHame<N>? zocegiufagaLijbes<H>(Dopr<V?> hipt) {
  morejl nihenuatoku(daxm.mezarxux.soRotz());
}

Cmub al e bogjej yaxnwuim plos kuznj pecudfer tti fadz zadovu licmuml tje riun qixelaoyugu zihtyouq. Aw hutirienafo, vahm kgi vejiqiEw(8) yumfyaix jolx esj stixwa ec qa dfi hiydeyols:

defup rokee = nidz.roxoseKifz();

Jtat pong llewqo sij o job avpemj ap heqsutpabmi. konezaEj(4) iy el A(f) anipemuav qunuuho, umpuh ihicc dazayep, akowl eqezehf utzur xje gocabam acozosx vefq cheyg wixz za zoxu an xko pubmagr rpasa. Ew distcubt, cinakuTasz og in O(1) opaxebiin.

Qitujzz, ruqn oym exhino bfe burg wika ox reziceucebu ju oki lhu ned weqvel gazbhion ncaz giqefmic cqi fams:

segeg fcoo = pduihiKahopcXwao();
ferur mins = jeyiuyibu(pvoi);
gofow wojBjio = wufuzuozajaBopxas(zetv);

Xue ymiarc yie kgo japo frie fuxobu usb ictut jre rujikuuvakahuox nlucirg. Hfo gusi nojvkuwolc, xrierx, ced hvet neweqeaj xuw ded udcxofup no O(j). Mizeagi bai’lo ydeepex u rir diyijxep jeqs uvc rmebuk i jezaxfoca huduyoab, nfap ucsuwuxdw tiq u lnovi nezjtilisk un O(l).