Transform the binary search tree into a sorted two-way linked list

Question:
Transform the binary search tree into a sorted two-way linked list.
do not use recursion.
do not create any new node, only adjust the pointer to point.
10
/ \
6 14
/ \ / \
4 8 12 16
-->>
4=6=8=10=12=14=16

struct BSTreeNode
{
int m_nValue; // value of node
BSTreeNode *m_pLeft; // left child of node
BSTreeNode *m_pRight; // right child of node
};

Answer:

inline BSTreeNode* eleminateMinNode(BSTreeNode** tree) {
	BSTreeNode* treeHead = *tree;
	if (!treeHead)
		return 0;
	BSTreeNode* treeHeadParent = 0;
	while (treeHead->m_pLeft) {
		treeHeadParent = treeHead;
		treeHead = treeHead->m_pLeft;
	}
	if (treeHeadParent)
		treeHeadParent->m_pLeft = treeHead->m_pRight;
	if (treeHead == *tree)
		*tree = treeHead->m_pRight;
	return treeHead;
}

inline BSTreeNode* eleminateMaxNode(BSTreeNode** tree) {
	BSTreeNode* treeHead = *tree;
	if (!treeHead)
		return 0;
	BSTreeNode* treeHeadParent = 0;
	while (treeHead->m_pRight) {
		treeHeadParent = treeHead;
		treeHead = treeHead->m_pRight;
	}
	if (treeHeadParent)
		treeHeadParent->m_pRight = treeHead->m_pLeft;
	if (treeHead == *tree)
		*tree = treeHead->m_pLeft;
	return treeHead;
}

void BSTreeToDoubleLinkChainInternal(BSTreeNode* tree, BSTreeNode** headp, BSTreeNode** tailp) {
	if (!tree)
		return;
	BSTreeNode* tail = tree;
	BSTreeNode* right = tail->m_pRight;
	BSTreeNode* minRight = eleminateMinNode(&right);
	while (minRight && minRight != tail) {
		tail->m_pRight = minRight;
		minRight->m_pLeft = tail;
		tail = minRight;
		minRight = eleminateMinNode(&right);
	}
	tail->m_pRight = 0;
	BSTreeNode* left = tree->m_pLeft;
	BSTreeNode* maxLeft = eleminateMaxNode(&left);
	BSTreeNode* head = tree;
	while (maxLeft && maxLeft != head) {
		head->m_pLeft = maxLeft;
		maxLeft->m_pRight = head;
		head = maxLeft;
		maxLeft = eleminateMaxNode(&left);
	}
	head->m_pLeft = 0;
	*headp = head;
	*tailp = tail;
}