Question

我正在制作我的删除功能，所以首先我试图找到父节点，但我一直遇到一个seg错误。当父母有2个孩子，但是当它有一个孩子时，它会崩溃。我一直试图改变现状，但我无法弄明白，有人可以帮助我。

Node* BST::getParentNode(int value){

    Node* temp = new Node;
    temp = _root;

    while (temp != NULL){

        Node* left = temp->getLeft();

        Node* right = temp->getRight();

        if (left->getData() == value)
            return temp;
        if (right->getData() == value)
            return temp;

        if (right->getData() > value){
            temp = temp->getRight();
        }
        else{
            temp = temp->getLeft();
        }

    }

    Node* empty = new Node;
    empty = NULL;
    return empty;
}

Answer 1

在访问之前，您没有检查左右指针是否为空。请参阅下面的评论

Node* BST::getParentNode(int value){

//  Node* temp = new Node; // This line isn't needed...
    Node* temp = _root;

    while (temp != NULL){

        Node* left = temp->getLeft();

        Node* right = temp->getRight();

        if(NULL != left) {    // You are missing this check.  I'm assuming this can happen
                              // because it's part of your exit condition from the while loop
            if (left->getData() == value)
                return temp;
        }
        if(NULL != right) {  // You are also missing this check. You can't call function on
                             // null pointers...
            if (right->getData() == value)
                return temp;
        }

        if (right->getData() > value){
            temp = temp->getRight();
        }
        else{
            temp = temp->getLeft();
        }

    }

    // It's unclear what you're trying to do on the next three lines
    // you create an object with new, then throw the memory away by assigning the
    // pointer to NULL      
    Node* empty = new Node;
    empty = NULL;
    return empty;  // With the code as is, you might as well return NULL;
                   // which would also get rid of the memory leak...
}

二叉树使父节点无法正常工作

1 个答案: