Question

在我的BST课程中，我使用以下函数来交换两个节点

void swapNodes(BSTNode *node1, BSTNode *node2)
  {
    if(node1->parent->left==node1)
        node1->parent->left = node2;
     else
        node1->parent->right = node2;
     if(node2->parent->left==node2)
        node2->parent->left = node1;
     else
        node2->parent->right = node1;

    BSTNode *temp;
     temp->parent = node1->parent;
     temp->left = node1->left;
     temp->right = node1->right;

     node1->parent = node2->parent;
     node1->left = node2->left;
     node1->right = node2->right;

     node2->parent = temp->parent;
     node2->left = temp->left;
     node2->right = temp->right;
  }

它需要两个节点指针并将它们交换。但是，当我调用此函数时，它会出现内存违规错误。我在这做错了什么？如果您知道在二叉树中交换两个节点的任何其他简单方法，请分享。

Answer 1

这个问题比看起来更复杂，因为涉及树的根的特殊情况，当一个是另一个父项时交换两个项，并修复左右子项的父指针。这是一些有效的代码。我怀疑它可以在某种程度上变得更整洁更短。

    void Swap(T* aP,T* aQ)
    {
    T* new_p_parent = aQ->iParent;
    T* new_p_left = aQ->iLeft;
    T* new_p_right = aQ->iRight;
    T** new_p_link = &iRoot;
    if (aQ->iParent)
        new_p_link = aQ->iParent->iLeft == aQ ? &aQ->iParent->iLeft : &aQ->iParent->iRight;

    T* new_q_parent = aP->iParent;
    T* new_q_left = aP->iLeft;
    T* new_q_right = aP->iRight;
    T** new_q_link = &iRoot;
    if (aP->iParent)
        new_q_link = aP->iParent->iLeft == aP ? &aP->iParent->iLeft : &aP->iParent->iRight;

    if (aQ->iParent == aP)
        {
        new_p_parent = aQ;
        new_p_link = nullptr;
        if (aP->iLeft == aQ)
            new_q_left = aP;
        else
            new_q_right = aP;
        }
    else if (aP->iParent == aQ)
        {
        new_q_parent = aP;
        new_q_link = nullptr;
        if (aQ->iLeft == aP)
            new_p_left = aQ;
        else
            new_p_right = aQ;
        }

    aP->iParent = new_p_parent;
    aP->iLeft = new_p_left;
    if (aP->iLeft)
        aP->iLeft->iParent = aP;
    aP->iRight = new_p_right;
    if (aP->iRight)
        aP->iRight->iParent = aP;
    if (new_p_link)
        *new_p_link = aP;

    aQ->iParent = new_q_parent;
    aQ->iLeft = new_q_left;
    if (aQ->iLeft)
        aQ->iLeft->iParent = aQ;
    aQ->iRight = new_q_right;
    if (aQ->iRight)
        aQ->iRight->iParent = aQ;
    if (new_q_link)
        *new_q_link = aQ;
    }

Answer 2

BSTNode *temp;

未初始化

如何在二叉树中交换两个节点

2 个答案: