节点拒绝在二进制搜索树中删除

Question

所以我有一个对象的BST，但是由于某种原因，每当我尝试从BST删除它拒绝的Node时，都会由于该原因。 我之前使用过相同的功能，但是在BST中，节点充满了字符串和整数。 在此程序中，节点充满了对象。 此外，搜索不会产生问题，因为当我调试时，我可以看到它正确地定位了我要查找的节点，但是由于某种原因，当它到达delete调用时，它什么也没做对象中的数据。

 template <typename DATATYPE, typename KEYTYPE> Node<DATATYPE, KEYTYPE> * BSTree<DATATYPE, KEYTYPE>::deleteNode(Node<DATATYPE, KEYTYPE> * aRoot,KEYTYPE value) { /* Given a binary search tree and a key, this function deletes the key and returns the new root */ // base case if (aRoot == nullptr) return aRoot; // If the key to be deleted is smaller than the aRoot's key, // then it lies in left subtree if (value.year < aRoot->Key().year) aRoot->setLeft(deleteNode(aRoot->Left(), value)); // If the key to be deleted is greater than the root's key, // then it lies in right subtree else if (value.year > aRoot->Key().year) root->setRight(deleteNode(aRoot->Right(), value)); // if key is same as root's key, then This is the node // to be deleted else { /* first recur on left child */ deleteNode(aRoot->Left(), value); /* now check if data matches*/ if((value.film == aRoot->Key().film)&&(value.name == aRoot->Key().name)&&(value.winner == aRoot->Key().winner)&&(value.award == aRoot->Key().award)&&(value.year == aRoot->Key().year)) { // node with only one child or no child if (aRoot->Left() == nullptr) { Node<DATATYPE, KEYTYPE> *temp1 = aRoot->Right(); //this is where it ignores the delete delete aRoot; cout << "Success!\\n"; return temp1; } else if (aRoot->Right() == nullptr) { Node<DATATYPE, KEYTYPE> *temp2 = aRoot->Left(); //this is where it ignores the delete delete aRoot; cout << "Success!\\n"; return temp2; } // node with two children: Get the inorder successor (smallest // in the right subtree) Node<DATATYPE, KEYTYPE> * temp = min(aRoot->Right()); // Copy the inorder successor's content to this node aRoot->setKey(temp->Key()); aRoot->setData(temp->Data()); // Delete the inorder successor aRoot->setRight(deleteNode(aRoot->Right(), temp->Key())); cout << "Success!\\n"; } /*now recur on the right child*/ deleteNode(aRoot->Right(), value); } return aRoot; } 

 void setLeft(Node<DATATYPE, KEYTYPE> * aLeft) { left = aLeft; }; void setRight(Node<DATATYPE, KEYTYPE> * aRight) { right = aRight; };