//面试题8：二叉树的下一个节点 BinaryTreeNode* GetNext(BinaryTreeNode* pNode) { if(pNode == nullptr) return nullptr; BinaryTreeNode*pNext=nullptr; if(pNode->m_pRight!=nullptr){ BinaryTreeNode* pRight=pNode->m_nRight; while(pRight->m_nLeft!=nullptr) pRight=pRight->m_nLeft; pNext=pRight; }else if(pNode->m_nParent!=nullptr){ BinaryTreeNode* pCurrent=pNode; BinaryTreeNode* pParent=pNode->m_pParent; while(pParent!=nullptr && pCurrent==pParent->m_pRight) { pCurrent=pParent; pParent=pParent->m_pParent; } pNext=nParent; } return pNext; }