直接通过中序遍历来解。但是需要设置两个节点记录每个子树排序后的第一和最后一个节点，尤其是最后一个，因为最后一个节点是由于连接所需。

上代码：

class Solution {public:       TreeNode* Convert(TreeNode* pRootOfTree)    {        TreeNode* head=NULL;//每个子树排序后的第一个节点        TreeNode* tail=NULL;//每个子树排序后的最后一个节点        ConvertHelp(pRootOfTree,head,tail);        return head;    } void ConvertHelp(TreeNode* pRootOfTree,TreeNode* &head,TreeNode* &tail) {     if(pRootOfTree==NULL) return;     ConvertHelp(pRootOfTree->left,head,tail);     if (tail == NULL) {
    //说明找到了中序遍历的第一个节点，把其作为链表头节点         tail = pRootOfTree;         head = pRootOfTree;     } else {         tail->right = pRootOfTree;         pRootOfTree->left = tail;         tail = pRootOfTree;     }     ConvertHelp(pRootOfTree->right,head,tail);    }};

法二：（思路同上，但是更简洁）

class Solution {public:    TreeNode* Convert(TreeNode* pRootOfTree)    {        if(pRootOfTree == nullptr) return nullptr;        TreeNode* pre = nullptr;                 convertHelper(pRootOfTree, pre);                 TreeNode* res = pRootOfTree;        while(res ->left)            res = res ->left;        return res;    }         void convertHelper(TreeNode* cur, TreeNode*& pre)    {        if(cur == nullptr) return;                 convertHelper(cur ->left, pre);                 cur ->left = pre;        if(pre) pre ->right = cur;        pre = cur;                 convertHelper(cur ->right, pre);                               }};