İkili arama ağacı Sorunu

oy
1

Neden arama ve ardıl ve selefi döndürür -1?

    // BST.cpp : main project file.

    #include stdafx.h
    #include <cstdlib>
    #include <iostream>
    #define SIZE 10
    using namespace std;

    struct Node {
        int value;
        Node *left;
        Node *right;
        Node *parent;
    };

    struct BST {
        Node *root;
    };

    void insert(int value, BST *tree) {
        Node *x = tree->root;
        Node *y = NULL;
        Node *z = (Node *) malloc(sizeof(Node));
        z->left = NULL;
        z->right = NULL;
        z->value = value;

        // Add your code here
        while (x!=NULL){
              y=x;
              if (z->value < x->value)
                 x= x->left;
              else x = x->right;
        }
        z->parent=y;
        if (y==NULL)
           tree->root=z;
        else if (z->value <y->value)
             y->left =z;
        else y->right =z;

    }

    Node *search(int key, Node *n) {
        if (n== NULL || key == n->value)
            return n;

        if (key < n->value)
            search(key, n->left);
        else
            search(key, n->right);
    }

    Node *min(Node *n) {
        if (n == NULL || n->left == NULL)
            return n;
        else
            return min(n->left);
    }

    Node *max(Node *n) {
        if (n == NULL || n->right == NULL)
            return n;
        else
            return max(n->right);
    }

    Node *successor(int value, Node *n) {
        Node *y = NULL;

        Node *x = search(value, n);

        if (x == NULL)
            return NULL;

        if (x->right != NULL)
            return min(x->right);

        y = x->parent;
        while (y != NULL && x == y->right) {
            x = y;
            y = y->parent;
        }
        return y;
    }

    Node *predecessor(int value, Node *n) {
        Node *x = search(value, n);
        Node *y = NULL;
        if (x == NULL)
            return NULL;

        if (x->left != NULL)
            return max(x->left);

        y = x->parent;
        while (y != NULL && x == y->left) {
            x = y;
            y = y->parent;
        }
        return y;
    }

    Node *remove(int value, BST *tree) {
        Node *z = search(value, tree->root);
        Node *y = NULL, *x = NULL;

        if (z == NULL) return NULL;

        if (z->left == NULL || z->right == NULL)
            y = z;
        else
            y = successor(value, z);

        if (y->left != NULL)
            x = y->left;
        else
            x = y->right;

        if (x != NULL)
            x->parent = y->parent;

        if (y->parent == NULL)
            tree->root = x;
        else if (y == y->parent->left)
            y->parent->left = x;
        else
            y->parent->right = x;

        if (y != z) {
            int tmp = z->value;
            z->value = y->value;
            y->value = tmp;
        }

        return y;
    }

    // ascending sort function
    void sortAsc(Node *node) {
        //Add your code here
        //inorder
        if (node->left!=NULL)
           sortAsc(node->left);
        cout<<node->value<< ;
        if (node->right!=NULL)
           sortAsc(node->right);

    }

    // descending sort function
    void sortDes(Node *node) {
        // Add your code here
        //inorder
        if (node->right!=NULL)
           sortDes(node->right);
        cout<<node->value<< ;
        if (node->left!=NULL)
           sortDes(node->left);

    }

    void clear(BST *tree) {
        Node *n = NULL;

        while (tree->root != NULL) {
            n = remove(tree->root->value, tree);
            free(n);
        }
    }


    int main() {
        int A[] = {3, 5, 10, 4, 8, 9, 1, 4, 7, 6};

        Node *node = NULL;
        BST *tree = (BST *) malloc(sizeof(BST));
        tree->root = NULL;

        // build BST tree
        cout << Input data:\n\t;
        for (int i=0; i<SIZE; i++) {
            cout << A[i] <<  ;    // by the way, print it to the console
            insert(A[i], tree);     // You need to complete TASK 1, so that it can work
        }

        // sort values in ascending order
        cout << \n\nAscending order:\n\t;
        sortAsc(tree->root);        // You need to complete TASK 2. Otherwise you see nothing in the console

        // sort values in descending order
        cout << \n\nDescending order:\n\t;
        sortDes(tree->root);        // TASK 2 also!

        // Find minimum value
        if (tree->root != NULL)
            cout << \n\nMin:  << min(tree->root)->value;

        // Find maximum value
        if (tree->root != NULL)
            cout << \n\nMax:  << max(tree->root)->value;

        // delete 4
        cout << \n\nDelete 4 and add 2;
        //free(remove(4, tree));    // You need to complete TASK 3, so that remove(int, BST *) function works properly
                                // we also need to release the resource!!!

        // insert 2
        insert(2, tree);        // It belongs to TASK 1 too.

        cout << \n\nAscending order:\n\t;
        sortAsc(tree->root);        // TASK 2!!

        // Find the successor of 5, -1 means no successor
        node = search(5, tree->root);
        cout << \n\nSearch of 5 is:  << (node != NULL?node->value:-1);


        // Find the successor of 5, -1 means no successor
        node = successor(5, tree->root);
        cout << \n\nSuccessor of 5 is:  << (node != NULL?node->value:-1);

        // Find the predecessor of 5. -1 means no predecessor
        node = predecessor(5, tree->root);
        cout << \n\nPredecessor of 5 is:  << (node != NULL?node->value:-1);

        cout << \n\n;

        // clear all elements
        clear(tree);            // delete all nodes and release resource
        free(tree);             // delte the tree too
        system(Pause);
    }
Oluştur 14/04/2011 saat 00:54
kaynak kullanıcı
Diğer dillerde...                            


1 cevaplar

oy
4

Peki tüm yolları böyle değerlere dönmek olması gerekir başlayanlar için özyinelemeli arama bir hata vardır:

Node *search(int key, Node *n) {
    if (n== NULL || key == n->value)
        return n;

    if (key < n->value)
        return search(key, n->left);
    else
        return search(key, n->right);
}

Bunun dışında ben sadece kodu gönderme ve onunla sorunun ne soran ziyade buldum gerekenler hakkında daha fazla ayrıntı önce kendi kod hata ayıklama ve vererek deneyin söylemek niyetindeyim bundan. Aksi takdirde burada bazı gerçek ukala cevap almak için sorumlu konum;)

Cevap 14/04/2011 saat 01:11
kaynak kullanıcı

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