Balanced binary tree into a doubly linked list sorted; the children beg their parents know how

//平衡二叉树转为排序的双向链表
void func(T*& tree,T* prev = nullptr) //这里传的相当于是二级指针，因为内部的修改将会影响外部根指针的结构
{
	if (tree == nullptr)
		return;
	func(tree->left);
	tree->left = prev;
	if (prev)
		prev->right = tree;
	prev = tree;
	func(tree->right);
}

After final completion, tree becomes a point balanced binary tree, a doubly linked list after the final conversion, the maximum junction.

However, the maximum right pointer does not point to the node, so after func function completes, should make a right tree = nullptr

//知道孩子如何求双亲
Node* GetParent(Node* pRoot,Node* pNode)
{
	if(pRoot == nullptr || pNode == nullptr && pNode == pRoot)  //如果要求的结点刚好就是根节点的话，那么也是直接返回空
		return nullptr;
	if(pRoot->left == pNode || pRoot->right == pNode)
		return pRoot;
	Node* pRet = GetParent(pRoot->left,pNode);
	if(pRet)
		return pRet;
	Node* pRet = GetParent(pRoot->right,pNode);
	return pRet;
}