![\"\"](\"https:\/\/demensdeum.com\/blog\/wp-content\/uploads\/2022\/08\/tree.png\")
<\/p>\nClassifica\u00e7\u00e3o em \u00e1rvore \u2013 classifica\u00e7\u00e3o usando uma \u00e1rvore de pesquisa bin\u00e1ria. Complexidade de tempo – O(n\u00b2). Nessa \u00e1rvore, cada n\u00f3 da esquerda tem n\u00fameros menores que o n\u00f3, \u00e0 direita h\u00e1 mais que o n\u00f3, ao vir da raiz e imprimir os valores da esquerda para a direita, obtemos uma lista ordenada de n\u00fameros . Surpreendente, certo?<\/p>\n
Considere o diagrama de \u00e1rvore de pesquisa bin\u00e1ria:<\/p>\n
<\/p>\n
Derrick Coetzee<\/a> (dom\u00ednio p\u00fablico)<\/span><\/p>\n Tente ler manualmente os n\u00fameros come\u00e7ando pelo pen\u00faltimo n\u00f3 esquerdo do canto inferior esquerdo, para cada n\u00f3 \u00e0 esquerda – um n\u00f3 – \u00e0 direita.<\/p>\n Ficar\u00e1 assim:<\/p>\n Para implementar o algoritmo em c\u00f3digo, voc\u00ea precisar\u00e1 de duas fun\u00e7\u00f5es:<\/p>\n A \u00e1rvore bin\u00e1ria de busca \u00e9 montada da mesma forma que \u00e9 lida, um n\u00famero \u00e9 anexado a cada n\u00f3 \u00e0 esquerda ou \u00e0 direita, dependendo se \u00e9 menor ou maior.<\/p>\n Exemplo em Lua:<\/p>\n \u0412\u0430\u0436\u043d\u044b\u0439 \u043d\u044e\u0430\u043d\u0441 \u0447\u0442\u043e \u0434\u043b\u044f \u0447\u0438\u0441\u0435\u043b \u043a\u043e\u0442\u043e\u0440\u044b\u0435 \u0440\u0430\u0432\u043d\u044b \u0432\u0435\u0440\u0448\u0438\u043d\u0435 \u043f\u0440\u0438\u0434\u0443\u043c\u0430\u043d\u043e \u043c\u043d\u043e\u0436\u0435\u0441\u0442\u0432\u043e \u0438\u043d\u0442\u0435\u0440\u0435\u0441\u043d\u044b\u0445 \u043c\u0435\u0445\u0430\u043d\u0438\u0437\u043c\u043e\u0432 \u043f\u043e\u0434\u0446\u0435\u043f\u043b\u0435\u043d\u0438\u044f \u043a \u043d\u043e\u0434\u0435, \u044f \u0436\u0435 \u043f\u0440\u043e\u0441\u0442\u043e \u0434\u043e\u0431\u0430\u0432\u0438\u043b \u0441\u0447\u0435\u0442\u0447\u0438\u043a \u043a \u043a\u043b\u0430\u0441\u0441\u0443 \u0432\u0435\u0440\u0448\u0438\u043d\u044b, \u043f\u0440\u0438 \u0440\u0430\u0441\u043f\u0435\u0447\u0430\u0442\u043a\u0435 \u0447\u0438\u0441\u043b\u0430 \u0432\u043e\u0437\u0432\u0440\u0430\u0449\u0430\u044e\u0442\u0441\u044f \u043f\u043e \u0441\u0447\u0435\u0442\u0447\u0438\u043a\u0443.<\/p>\n https:\/\/gitlab.com\/demensdeum\/algorithms\/-\/tree\/master\/sortAlgorithms\/treesort<\/a><\/p>\n TreeSort Algorithm Explained and Implemented with Examples in Java | Sorting Algorithms | Geekific – YouTube<\/a><\/p>\n Tree sort – YouTube<\/a><\/p>\n Convert Sorted Array to Binary Search Tree (LeetCode 108. Algorithm Explained) – YouTube<\/a><\/p>\n Sorting algorithms\/Tree sort on a linked list – Rosetta Code<\/a><\/p>\n Tree Sort – GeeksforGeeks<\/a><\/p>\n Tree sort – Wikipedia<\/a><\/p>\n How to handle duplicates in Binary Search Tree? – GeeksforGeeks<\/a><\/p>\n Tree Sort | GeeksforGeeks – YouTube<\/a><\/p>\n","protected":false},"excerpt":{"rendered":" Classifica\u00e7\u00e3o em \u00e1rvore \u2013 classifica\u00e7\u00e3o usando uma \u00e1rvore de pesquisa bin\u00e1ria. Complexidade de tempo – O(n\u00b2). Nessa \u00e1rvore, cada n\u00f3 da esquerda tem n\u00fameros menores que o n\u00f3, \u00e0 direita h\u00e1 mais que o n\u00f3, ao vir da raiz e imprimir os valores da esquerda para a direita, obtemos uma lista ordenada de n\u00fameros .Continue reading “Tipo de \u00e1rvore”<\/span><\/a><\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"_monsterinsights_skip_tracking":false,"_monsterinsights_sitenote_active":false,"_monsterinsights_sitenote_note":"","_monsterinsights_sitenote_category":0,"_jetpack_memberships_contains_paid_content":false,"footnotes":""},"categories":[61,52],"tags":[131,190,207],"class_list":["post-3409","post","type-post","status-publish","format-standard","hentry","category-techie","category-tutorials","tag-algorithms","tag-sorting","tag-tree-sort","entry"],"translation":{"provider":"WPGlobus","version":"3.0.2","language":"pt","enabled_languages":["en","ru","zh","de","fr","ja","pt","hi"],"languages":{"en":{"title":true,"content":true,"excerpt":false},"ru":{"title":true,"content":true,"excerpt":false},"zh":{"title":true,"content":true,"excerpt":false},"de":{"title":true,"content":true,"excerpt":false},"fr":{"title":true,"content":true,"excerpt":false},"ja":{"title":true,"content":true,"excerpt":false},"pt":{"title":true,"content":true,"excerpt":false},"hi":{"title":false,"content":false,"excerpt":false}}},"jetpack_featured_media_url":"","jetpack_sharing_enabled":true,"_links":{"self":[{"href":"https:\/\/demensdeum.com\/blog\/pt\/wp-json\/wp\/v2\/posts\/3409","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/demensdeum.com\/blog\/pt\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/demensdeum.com\/blog\/pt\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/demensdeum.com\/blog\/pt\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/demensdeum.com\/blog\/pt\/wp-json\/wp\/v2\/comments?post=3409"}],"version-history":[{"count":7,"href":"https:\/\/demensdeum.com\/blog\/pt\/wp-json\/wp\/v2\/posts\/3409\/revisions"}],"predecessor-version":[{"id":3870,"href":"https:\/\/demensdeum.com\/blog\/pt\/wp-json\/wp\/v2\/posts\/3409\/revisions\/3870"}],"wp:attachment":[{"href":"https:\/\/demensdeum.com\/blog\/pt\/wp-json\/wp\/v2\/media?parent=3409"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/demensdeum.com\/blog\/pt\/wp-json\/wp\/v2\/categories?post=3409"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/demensdeum.com\/blog\/pt\/wp-json\/wp\/v2\/tags?post=3409"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}\n
\n
\n\n\nfunction Node:new(value, lhs, rhs)\n output = {}\n setmetatable(output, self)\n self.__index = self \n output.value = value\n output.lhs = lhs\n output.rhs = rhs\n output.counter = 1\n return output \nend\n\nfunction Node:Increment()\n self.counter = self.counter + 1\nend\n\nfunction Node:Insert(value)\n if self.lhs ~= nil and self.lhs.value > value then\n self.lhs:Insert(value)\n return\n end\n\n if self.rhs ~= nil and self.rhs.value < value then\n self.rhs:Insert(value)\n return\n end\n\n if self.value == value then\n self:Increment()\n return\n elseif self.value > value then\n if self.lhs == nil then\n self.lhs = Node:new(value, nil, nil)\n else\n self.lhs:Insert(value)\n end\n return\n else\n if self.rhs == nil then\n self.rhs = Node:new(value, nil, nil)\n else\n self.rhs:Insert(value)\n end\n return\n end\nend\n\nfunction Node:InOrder(output)\n if self.lhs ~= nil then\n output = self.lhs:InOrder(output)\n end\n output = self:printSelf(output)\n if self.rhs ~= nil then\n output = self.rhs:InOrder(output)\n end\n return output\nend\n\nfunction Node:printSelf(output)\n for i=0,self.counter-1 do\n output = output .. tostring(self.value) .. \" \"\n end\n return output\nend\n\nfunction PrintArray(numbers)\n output = \"\"\n for i=0,#numbers do\n output = output .. tostring(numbers[i]) .. \" \"\n end \n print(output)\nend\n\nfunction Treesort(numbers)\n rootNode = Node:new(numbers[0], nil, nil)\n for i=1,#numbers do\n rootNode:Insert(numbers[i])\n end\n print(rootNode:InOrder(\"\"))\nend\n\n\nnumbersCount = 10\nmaxNumber = 9\n\nnumbers = {}\n\nfor i=0,numbersCount-1 do\n numbers[i] = math.random(0, maxNumber)\nend\n\nPrintArray(numbers)\nTreesort(numbers)<\/code><\/pre>\n<\/div>\n\u0421\u0441\u044b\u043b\u043a\u0438<\/h3>\n
\u0418\u0441\u0442\u043e\u0447\u043d\u0438\u043a\u0438<\/h3>\n