class Solution {
private:
    Node* head, * pre = NULL;
public:
    Node* treeToDoublyList(Node* root) {//雙指針做法
        if (!root) return NULL;
        inorder(root);
        head->left = pre;
        pre->right = head;
        return head;
    }
    void inorder(Node* root)
    {
        if (root == NULL)return;
        inorder(root->left);
        root->left = pre;
        if (pre)
            pre->right = root;
        else head = root;
        pre = root;
        inorder(root->right);
    }
};

Node* treeToDoublyList(Node* root) {
        if (!root) return NULL;
        vector<Node*> nodelist;
        inorder(nodelist, root);
        int l = nodelist.size();
        if (l == 1)
        {
            nodelist[0]->right = nodelist[0];
            nodelist[0]->left = nodelist[0];
            return nodelist[0];
        }
        for (int i = 1; i < l - 1; i++) {
            nodelist[i]->left = nodelist[i - 1];
            nodelist[i]->right = nodelist[i + 1];
        }
        nodelist[0]->left = nodelist[l - 1];
        nodelist[0]->right = nodelist[1];
        nodelist[l - 1]->right = nodelist[0];
        nodelist[l - 1]->left = nodelist[l - 2];
        return nodelist[0];
    }
    void inorder(vector<Node*>& nodelist, Node* root)
    {
        if (root == NULL)return;
        inorder(nodelist, root->left);
        nodelist.push_back(root);
        inorder(nodelist, root->right);
    }

class Solution {
public:
    Node* treeToDoublyList(Node* root) {
        if(!root) return NULL;
        mid(root);
        //進(jìn)行頭節(jié)點(diǎn)和尾節(jié)點(diǎn)的相互指向，這兩句的順序也是可以顛倒的
        head->left=pre;
        pre->right=head;
        return head;
    }
    void mid(Node* cur)
    {
        if(!cur) return;
        mid(cur->left);
        //pre用于記錄雙向鏈表中位于cur左側(cè)的節(jié)點(diǎn)，即上一次迭代中的cur,當(dāng)pre==null時，cur左側(cè)沒有節(jié)點(diǎn),即此時cur為雙向鏈表中的頭節(jié)點(diǎn)
        if(pre==NULL) head=cur;
        //反之，pre!=null時，cur左側(cè)存在節(jié)點(diǎn)pre，需要進(jìn)行pre.right=cur的操作。
        else pre->right=cur;
        //pre是否為null對這句沒有影響,且這句放在上面兩句if else之前也是可以的。
        cur->left=pre;
        //pre指向當(dāng)前的cur
        pre=cur;
        //全部迭代完成后，pre指向雙向鏈表中的尾節(jié)點(diǎn)
        mid(cur->right);
    }
private:
    Node* head,*pre;
};