二叉搜索树与双向链表，c++

class Solution {
public:
    TreeNode* Convert(TreeNode* pRootOfTree)
    {
        if (pRootOfTree == nullptr)
            return pRootOfTree;

        stack<TreeNode*> ista;
        vector<TreeNode*> ivec;
        auto pin = pRootOfTree;
        while (!(ista.empty() && pin == nullptr)){
            while(pin != nullptr){
                ista.push(pin);
                pin = pin->left;
            }            
            if (!ista.empty()){
                pin = ista.top();
                ista.pop();
                ivec.push_back(pin);
                pin = pin->right;
            }
        }        

        ivec.front()->left = nullptr;
        ivec.back()->right = nullptr;
        int length = ivec.size();
        for (int i = 0; i != length-1; ++i){
            ivec[i]->right = ivec[i+1];
            ivec[i+1]->left = ivec[i];
        }
        return ivec.front();
    }
};

写完后看了一下，和不少同学的思路相同
1、二叉搜索树的中序遍历是递增数列，先用非递归方式中序遍历搜索树，节点存在vec容器中
2、按顺序改变个节点的指向
欢迎交流指正！！！