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Removal of the binary tree knot

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jules last edited by

There's a binary tree, we need to remove the element.
We found the option to remove the element and place it either the right element of its left support or its left-handed right support.
Question: How is it best to organize?
It is logical that the replacement element should specify the support signs and the owner should replace the indicator with NULL, with the owner of the removed element replacing the index with the replacement element.
If you do anything in the forehead, you need two support indicators, parents ' indexes and indicators for the elements themselves.
Is there any right way to remove this hemore? Probably more streamlined.
Please see my code line 164, the function is not blinking, but it's kind of well written.
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A
Avante last edited by

In the case of a binary tree, I recommend that the correct algorithm be seen in Cormen (sighs)"Algorithms. Building and analysis." page 326 or in a simpler and more visible version http://www.tech-faq.com/binary-tree-deleting-a-node.html
In both cases, a classic storage data structure is used BSTin which the element is presented to the (parent, left, right, data)♪
In the case of a conventional binary tree, only the situation where the deleted element of the reference left and right - No. null♪ In this situation, there is a need to round the tree side by side. left or rightfind the first possible leaf in a tree and Replace the removed element with that sheet.
An additional field may be stored in each tree hub to optimize the operation leafwhich would indicate such a possible leaf to avoid the passage through the tree branch in the event of removal of the element. Although, I believe, interest in such optimization is strictly theoretical.
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Software Programming • c+++ c++11 • • authera

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std::bind is the generator of functional objects and he should know what he needs to use in the generation, so no, or use it. placeholderseither relevant or write your class (this will be quite simple) because, without clear arguments, this task is not being addressed. In your first example, FastDelegate3 He knows his arguments, that is. You're making them clear, that's the only way that C++ is done.
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Because you just had fun with the other labyrinth, you keep it. ♪ ♪ http://ideone.com/AjHRqr The way you paint yourself... Honestly, check the lazy. ♪ ♪#include <iostream> #include <iomanip> using namespace std; const int Tgt = 110; const int I = Tgt; int m[8][8]={ {1,0,0,0,0,0,0,0}, {0,2,0,0,0,0,0,0}, {0,0,0,2,0,0,2,0}, {0,0,0,0,0,0,2,0}, {0,0,0,2,2,0,0,0}, {0,0,0,0,0,0,0,2}, {0,2,2,0,0,0,0,0}, {0,0,0,0,2,0,0,0} }; struct Coord { int x = 0, y = 0; Coord(int x = 0, int y = 0):x(x),y(y){} }; const int STOP = 54; //64 - 10; Coord path[STOP]; bool step(int x, int y, int level) { if (level == STOP-1) { if (x == 0 && y == 1) { for(int i = 0; i < level; ++i) { cout << "(" << path[i].x << "," << path[i].y << ") - "; } cout << "(0,1)\n"; return true; } else return false; } m[x][y] = 1; path[level] = Coord(x,y); // Up if (y > 0 && m[x][y-1] == 0) { if (step(x,y-1,level+1)) return true; } // Dn if (y < 7 && m[x][y+1] == 0) { if (step(x,y+1,level+1)) return true; } // Lt if (x > 0 && m[x-1][y] == 0) { if (step(x-1,y,level+1)) return true; } // Rt if (x < 7 && m[x+1][y] == 0) { if (step(x+1,y,level+1)) return true; } m[x][y] = 0; return false; } int main(int argc, const char * argv[]) { step(1,0,1); } It's not enough to paint:
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Software Programming • javascript php algorithm any language • • Avante

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Anyway, I've been thinking for a long time and I haven't come up with anything better. I just did. Took every piece and wrote a logic for every step of it. It was only for the King and the Ladium. I think whoever understands that can write for other figures. Logic is simple, counting their possible steps on,y.drowChess($x, $y,$xb); $figures = [ 'King' => function ($x,$y) { if((!$x or $x < 0) or (!$y or $y < 0)) return "Poziciya $x $y naxoditsya za polem"; $xb = ['A','B','C','D','E','F','G','H']; $currentPosition = $xb[$x-1].$y; $x1 = $x-1; $y1 = $y; $x2 = $x-1; $y2 = $y-1; $x3 = $x-1; $y3 = $y+1; $x4 = $x+1; $y4 = $y; $x5 = $x+1; $y5 = $y-1; $x6 = $x+1; $y6 = $y+1; $x7 = $x; $y7 = $y-1; $x8 = $x; $y8 = $y+1; $positions = []; if(isset($xb[$x1-1]) && $y1 && $y1 <= 8 ) $positions[$xb[$x1-1].$y1] = ['x'=>$x1,'y'=>$y1]; if(isset($xb[$x2-1]) && $y2 && $y2 <= 8 ) $positions[$xb[$x2-1].$y2] = ['x'=>$x2,'y'=>$y2]; if(isset($xb[$x3-1]) && $y3 && $y3 <= 8 ) $positions[$xb[$x3-1].$y3] = ['x'=>$x3,'y'=>$y3]; if(isset($xb[$x4-1]) && $y4 && $y4 <= 8 ) $positions[$xb[$x4-1].$y4] = ['x'=>$x4,'y'=>$y4]; if(isset($xb[$x5-1]) && $y5 && $y5 <= 8 ) $positions[$xb[$x5-1].$y5] = ['x'=>$x5,'y'=>$y5]; if(isset($xb[$x6-1]) && $y6 && $y6 <= 8 ) $positions[$xb[$x6-1].$y6] = ['x'=>$x6,'y'=>$y6]; if(isset($xb[$x7-1]) && $y7 && $y7 <= 8 ) $positions[$xb[$x7-1].$y7] = ['x'=>$x7,'y'=>$y7]; if(isset($xb[$x8-1]) && $y8 && $y8 <= 8 ) $positions[$xb[$x8-1].$y8] = ['x'=>$x8,'y'=>$y8]; return [ 'Korol na' => $currentPosition, 'Dostupnie emu xodi'=>$positions ]; }, 'Rook' => function($x,$y) { if((!$x or $x < 0) or (!$y or $y < 0)) return "Poziciya $x $y naxoditsya za polem"; $positions = []; $xb = ['A','B','C','D','E','F','G','H']; $currentPosition = $xb[$x-1].$y; //V pravo vniz i vverx $yrb = $yrt = $y; for( $x1 = $x+1; $x1 <= 8 ; $x1++ ) { //Diaganal to rigth bottom if((++$yrb) <= 8) $positions[$xb[$x1-1].($yrb)] = ['x' => $x1, 'y' => $yrb]; //Diaganal to rigth top if((--$yrt) > 0) $positions[$xb[$x1-1].($yrt)] = ['x' => $x1, 'y' => $yrt]; } //V pravo vniz i vverx $ylb = $ylt = $y; for( $x2 = $x-1; $x2 >= 1 ; $x2-- ) { //Diaganal to left bottom if((++$ylb) <= 8){ $positions[$xb[$x2-1].($ylb)] = ['x' => $x2, 'y' => $ylb]; } //Diaganal to left top if((--$ylt) > 0){ $positions[$xb[$x2-1].($ylt)] = ['x' => $x2, 'y' => $ylt]; } } return [ 'Rook na' => $currentPosition, 'Dostupnie emu xodi'=>$positions ]; }, ]; echo "<pre>"; //Доступные шаги для лады print_r($figures'Rook'); //Доступные шаги для Короля print_r($figures'King'); Well, if we have accessible steps, we can write a function that will cause them with different parameters until their available steps are different.
How Binary tree removal works in C
Software Programming • c language binary tree • • morde

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This is a binary tree that keeps integers in intermediate knots, in addition to the leaves, and with the property that all numbers in the left subtree of a certain node are strictly smaller than the node number. Numbers equal or larger are in the right subtree.So let's go.void remover(No **pRaiz, int numero){ The removal function receives a pointer for a pointer for a node; this is necessary if the number to be removed (listed in the second parameter) is found in the root node itself: in this case, the address of pRaiz has to change to the function it calls remover().The first thing you do is test to see if the tree is empty, and if it is, finish: if(*pRaiz == NULL){ // esta verificacao serve para caso o numero nao exista na arvore. printf("Numero nao existe na arvore!"); getch(); // getch() é fogo, mas tudo bem return; } Although he writes a "erre" message, this is a correct behavior. If you have to remove an element that does not exist from a structure, it does nothing, and you can assume that that element no longer exists in the structure. If he existed before or not, it makes no difference.Then he compares the number to know in which sub-tree he is. For this, he tests a set of conditions in order, and will discard the possibilities methodically. Personally, I wouldn't do this kind of indentation and keep everything on the same level, but let's explain the code that exists, not what I would write:The first case is if the wanted number is smaller than the current node number: if(numero < (*pRaiz)->numero) remover(&(*pRaiz)->esquerda, numero); If it is, it must be, by construction, in the left subtree. We then execute the same algorithm recursively in the left subtree.If it's not, then we'll check if the wanted number is higher that the current number. If it is, it will be on the right subtree, and we execute the algorithm recursively on the right subtree. else if(numero > (*pRaiz)->numero) remover(&(*pRaiz)->direita, numero); Finally, having eliminated the possibility of the number being larger or smaller, we conclude that the number is located at the root of this tree. Then pRaiz is the node to be removed! We proceed therefore with the removal process: else{ // se nao eh menor nem maior, logo, eh o numero que estou procurando! :) First we create an auxiliary variable to manipulate the knots. The comment refers to the fact that compilers who do not meet the C99 standard require that all the local variables are declared at the beginning of the functions. That's what I usually do for portability, but others don't like... No *pAux = *pRaiz; // quem programar no Embarcadero vai ter que declarar o pAux no inicio do void! :[ Now we treat the case pRaiz to be a leaf, that is, has no children nor left, nor right. In this case, just delete the knot and run for the hug: if (((*pRaiz)->esquerda == NULL) && ((*pRaiz)->direita == NULL)){ // se nao houver filhos... free(pAux); (*pRaiz) = NULL; } Of course, the probability of the knot being a leaf is low; in this case, you have to deal with the subtrees. One of the nodes will have to replace that node you're deleting, and the subtrees will hang on that new node.This new node has to be either the largest number of the left subtree, or the lowest number of the right subtree. It begins by treating cases where either one, or another is null: in this case, simply replace the knot with the subtree that exists. else{ // so tem o filho da direita if ((*pRaiz)->esquerda == NULL){ (*pRaiz) = (*pRaiz)->direita; pAux->direita = NULL; free(pAux); pAux = NULL; } else{ //so tem filho da esquerda if ((*pRaiz)->direita == NULL){ (*pRaiz) = (*pRaiz)->esquerda; pAux->esquerda = NULL; free(pAux); pAux = NULL; } In this case, the end and more complicated, the item has children on both sides. There it is not enough to simply replace the knot with one of the two trees, you have to choose either the biggest knot on the left or the smallest knot on the right of the face you want to erase. In the case, the author chose the biggest knot on the left. He then "takes a step to the left" and goes down the tree to the right until he finds a node that has no right subtree. This was encapsulated in the function MaiorDireita().After that, he takes the right subtree of the node and hangs on the right of the new root; and the same with the left subtree. Then just destroy the old root and return. else{ //Escolhi fazer o maior filho direito da subarvore esquerda. pAux = MaiorDireita(&(*pRaiz)->esquerda); //se vc quiser usar o Menor da esquerda, so o que mudaria seria isso: pAux->esquerda = (*pRaiz)->esquerda; // pAux = MenorEsquerda(&(*pRaiz)->direita); pAux->direita = (*pRaiz)->direita; (*pRaiz)->esquerda = (*pRaiz)->direita = NULL; free((*pRaiz)); *pRaiz = pAux; pAux = NULL; } } } } } In the case of MaiorDireita(), what he has to do is take a tree (which will be the left subtree of a knot that we want to erase), and go down to the right until he can't get any more. The knot where we stop is the knot we want to return, but before we need to remove it from the tree.Note that it, by construction, does not have a right subtree, but may have left. In this case, we will do as we did in the analogous case in remover() and replace it with your left child. Now that it is out of the subtree and has no more children hanging on it, it can be returned to replace the root above:No *MaiorDireita(No **no){ We start walking recursively to the right until we can no longer; if((*no)->direita != NULL) return MaiorDireita(&(*no)->direita); else{ Not being able to walk further to the right, we keep a reference to the knot we will return; No *aux = *no; We take the left subtree of the node, if any, and we paste it in place of the current node; if((*no)->esquerda != NULL) // se nao houver essa verificacao, esse nó vai perder todos os seus filhos da esquerda! *no = (*no)->esquerda; else *no = NULL; And we return the knot. return aux; } } A MenorDireita() is equal, mutatis mutandis.
Implementation of the removal and insertion method in a Heap (assembly) in C++, binary tree
Software Programming • c++ poo pointers • • Laycee

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template <class T> class node { public: virtual ~node(); }; If the class isn't going to be hereable, it doesn't make sense for the destroyer to be virtual. Does that matter? In part, yes. Defining virtual methods forces the compiler to create a virtual table (known as vtable) and this has the following impact:Each instance of the object will have a pointer to that table. The pointer will be implied and you won't have to define it... the compiler will do it for you. This makes the classes a little bigger (the size of a pointer) and you have to be very careful if you try to serialize the object in the C-style (by attaching to the object with a type pointer) char).Invoking a virtual call, it cannot be called directly, but it have to go to vtablewhich adds a level of indirection and its corresponding cost (which can be despicable or not).By creating the vtable RTTI mechanisms are enabled for this class (e.g. dynamic_cast). That these mechanisms remain disabled for non-heirable classes helps to prevent people from screwing with conversions (a compilation error occurs).On the other hand, if all the members of the class are public is more clean and beautiful to define a structure (whose default visibility is public(c):template <class T> struct node { node<T> L, R; T date; node(T date); ~node(); }; We follow:template <class T> node<T>::~node() { delete (this->L); delete (this->R); } The class node pretends to be the base link of a double-linked list. Suppose then we have 2 items on that list, A and B, such that:NULL <-- A <--> B --> NULL What happens if we destroy A? Seeing the destroyer, A will invoke the destroyer of B... And what does the B destroyer do? He'll try to destroy A... again. Double delete = error in execution time.This error can be treated in several ways and each has its implications. Perhaps the easiest thing is that every node concerns itself only and forgets its neighbors... that's why you already have class. max_heap:template <class T> node<T>::~node() { } More things:template <class T> node<T>::node(T date) { This part is a bit of trache... you are creating a template to store any element in the same... and the variable that stores that value is called date? And if what I want to store is integer. Does it make sense that it has to int valor = nodo->dateI let you answer the question.Once you have seen this, how to add an item to the list?void insert(T date) { } Well, the first thing I'd have to check is if the list is empty, that's, yeah. root==0. If so the operation is simple: Create a new node in root:node<T> nuevo = new node<T>(date); // en serio, cambia el nombre de esta if( root == 0 ) root = nuevo; else { } Yeah. root is not null then you have to look for a place to add the new element. What's your place? It will depend on the nature of the list:If the list is not ordered the normal is to add the new element at the end of the list... although it could also be at the beginningIf the list is ordered, you will have to go through the list to find the position to add the element based on a comparison function.I will assume that the element is added at the end of the list:node<T> prev = root; while( prev->R ) prev = prev->R; prev->R = nuevo; nuevo->P = prev; And to get an element out of the list? The process is similar. First you have to look for the element and then make the previous and next node point to each other:for( nodo<T>* nodo = root; nodo != 0; nodo=nodo->R ) { if( nodo->date == date ) { if( nodo == root ) root = nodo->P; if( nodo->R ) nodo->R->P = nodo->P; if( nodo->P ) nodo->P->R = nodo->R; delete nodo; break; } }

Removal of the binary tree knot

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