二叉树与双向链表的转换

/*
struct TreeNode {
	int val;
	struct TreeNode *left;
	struct TreeNode *right;
	TreeNode(int x) :
			val(x), left(NULL), right(NULL) {
	}
};*/

class Solution {
public:
    TreeNode* Convert(TreeNode* pRootOfTree)
    {
        if(pRootOfTree == nullptr)    return nullptr;
        // 非递归算法，故用堆栈
        stack<TreeNode*> sck;
        TreeNode *pCur = pRootOfTree;
        TreeNode *pNode = nullptr;        // 存储前一个输出结点
        TreeNode *pHead = nullptr;        // 存储有序链表表头
        while(!sck.empty() || pCur != nullptr) {
            
            while(pCur != nullptr) {
                sck.push(pCur);
                pCur = pCur->left;
            }
            pCur = sck.top();
            // 处理输出结点
            if(pNode != nullptr) {
                pNode->right = sck.top();
                sck.top()->left = pNode;
                pNode = sck.top();
            }
            if(pNode == nullptr) {
                pNode = sck.top();
                pHead = pNode;
            }
            pNode = sck.top();
            sck.pop();
            
            if(pCur->right) {
                pCur = pCur->right;
            }
            else {
                pCur = nullptr;
            }
        }
        return pHead;
    }
};