我的红黑树析构函数出了什么问题?

时间:2013-03-15 15:01:07

标签: c++ destructor red-black-tree

我正在构建一个红黑树,但可能是我的类RBTree的析构函数存在一些问题。我将10 ^ 7的值添加到树中,然后调用析构函数,但内存似乎没有被释放。 (我查看系统监视器,我的程序仍然使用200MB)。

你能告诉我我的析构函数有什么问题吗?这是我的源代码。

抱歉我的英语不好。

#include<cstdio>
#include<cstdlib>
#include<iostream>
using namespace std;

enum Color {RED, BLACK};

template<class Data> class RBNode;
template<class Data> class RBTree;

template<class Data> class RBNode {
    Color color; RBNode *p, *left, *right;
public:
    Data v;
    RBNode(Color color, RBNode *p, RBNode *left, RBNode *right, Data v):
        color(color), p(p), left(left), right(right), v(v) {}
    RBNode() {}
    friend class RBTree<Data>;
};

template<class Data> class RBTree {
    typedef RBNode<Data> Node;
    typedef Node * PNode;
    PNode root, nil;

    void LeftRotate(PNode x) {
        PNode y = x->right; x->right = y->left;
        if(y->left != nil) y->left->p = x;
        y->p = x->p;
        if(x->p == nil) root = y;
        else if(x == x->p->left) x->p->left = y;
        else x->p->right = y;
        y->left = x; x->p = y;
    }

    void RightRotate(PNode y) {
        PNode x = y->left; y->left = x->right;
        if(x->right != nil) x->right->p = y;
        x->p = y->p;
        if(y->p == nil) root = x;
        else if(y == y->p->left) y->p->left = x;
        else y->p->right = x;
        x->right = y; y->p = x;
    }

    void insertFixUp(PNode z) {
        while(z->p->color == RED) {
            if(z->p == z->p->p->left) {
                PNode y = z->p->p->right;
                if(y->color == RED) z->p->color = y->color = BLACK, z->p->p->color = RED, z = z->p->p;
                else {
                    if(z == z->p->right) LeftRotate(z = z->p);
                    z->p->color = BLACK; z->p->p->color = RED; RightRotate(z->p->p);
                }
            } else {
                PNode y = z->p->p->left;
                if(y->color == RED) z->p->color = y->color = BLACK, z->p->p->color = RED, z = z->p->p;
                else {
                    if(z == z->p->left) RightRotate(z = z->p);
                    z->p->color = BLACK; z->p->p->color = RED; LeftRotate(z->p->p);
                }
            }
        }
        root->color = BLACK;
    }

public:
    RBTree() {
        nil = new Node;
        nil->color = BLACK;
        nil->p = nil->left = nil->right = nil;
        nil->v = Data();
        root = nil;
    }

    ~RBTree() {
        delete root;
        delete nil;
    }

    void insert(Data v) {
        PNode y = nil, x = root;
        while(x != nil) {
            y = x;
            x = v < x->v ? x->left : x->right;
        }
        PNode z = new Node; *z = Node(RED, y, nil, nil, v);
        if(y == nil) root = z;
        else if(v < y->v) y->left = z;
        else y->right = z;
        insertFixUp(z);
    }
};

int main() {
    RBTree<int> tree;
    for(int i = 0; i < 10000000; ++i) tree.insert(i);
    tree.~RBTree();
    getchar();
    return 0;
}

5 个答案:

答案 0 :(得分:1)

您需要向RBNode添加析构函数,删除其子代:

template<class Data> class RBNode {
    ...
    ~RBNode() {
        delete left;
        delete right;
    }
    ...
};

按原样,删除树时将删除根节点,但根节点本身不会释放其资源。因此,您将丢失对root的子节点及其所有子节点等的所有引用。因为您不再具有对这些节点的引用,所以无法删除它们,您有内存泄漏。

析构函数确保当我们即将失去对节点子节点的引用时,这些子节点会被释放(以及它们的子节点等)。

答案 1 :(得分:1)

首先,它的错误在于你没有使用智能指针。其次,您没有在Node类中使用智能指针,因此当删除根时,不会删除任何其他对象。

答案 2 :(得分:0)

您的树节点似乎不会以递归方式删除其子节点。你需要在节点中使用析构函数,然后当根被破坏时,所有内容都会级联。

答案 3 :(得分:0)

你的析构函数只释放两个元素:root和nil。为了释放树的其余部分,你应该以某种方式传播释放树中的元素,如:

~RBNode() {
    if (left != nil ) delete left;
    if (right != nil) delete right;
}

(这只是一个想法,当然这个代码在字面上不起作用,因为你在析构函数中看不到nil。)

答案 4 :(得分:0)

我找到了我的析构函数,但我有更多的属性,比如大小和父指针,但我认为它会有所帮助

    ~RBTree() {
       RBNode *p(root);
       while(size!=0) {
          if(p==root && size==1) { delete root; size--;}
          else if(p->right!=0) p=p->right;
          else if(p->left!=0) p=p->left;
          else {
            RBNode *c(p);
            p=p->parent;
            if(p->left==c) {
                delete c;
                p->left=0;
            }
            else {
                delete c;
                p->right=0;
            }
            size--;
          }
       }
    }