思路
方法一： 保存中序遍历的结果，然后依次处理，这里使用vector保存中序遍历结果
1 /*
2 // Definition for a Node.
3 class Node {
4 public:
5 int val;
6 Node* left;
7 Node* right;
8
9 Node() {}
10
11 Node(int _val) {
12 val = _val;
13 left = NULL;
14 right = NULL;
15 }
16
17 Node(int _val, Node* _left, Node* _right) {
18 val = _val;
19 left = _left;
20 right = _right;
21 }
22 };
23 */
24 class Solution {
25 private:
26 vector<Node*> v;
27 public:
28 Node* treeToDoublyList(Node* root) {
29 if(root == NULL)
30 return NULL;
31 inorderTraverse(root);
32
33 Node* head = v[0];
34 Node* p = head;
35 for(int i = 1; i < v.size(); i++) {
36 p->right = v[i];
37 v[i]->left = p;
38 p = p->right;
39 }
40
41 //首尾互指
42 p->right = head;
43 head->left = p;
44
45 return head;
46 }
47
48 //中序遍历
49 void inorderTraverse(Node* root) {
50 if(root == NULL)
51 return;
52 inorderTraverse(root->left);
53 v.push_back(root);
54 inorderTraverse(root->right);
55 }
56 };
方法二：在中序遍历过程中处理每个结点（就地转换）
需要保存第一个和最后一个结点，
遍历过程中需要知道前一个结点。
1 /*
2 // Definition for a Node.
3 class Node {
4 public:
5 int val;
6 Node* left;
7 Node* right;
8
9 Node() {}
10
11 Node(int _val) {
12 val = _val;
13 left = NULL;
14 right = NULL;
15 }
16
17 Node(int _val, Node* _left, Node* _right) {
18 val = _val;
19 left = _left;
20 right = _right;
21 }
22 };
23 */
24 class Solution {
25 private:
26 Node* head = NULL;
27 Node* pre = NULL;
28 public:
29 Node* treeToDoublyList(Node* root) {
30 if(root == NULL)
31 return NULL;
32
33 inorderTraverse(root);
34 //首尾互指
35 head->left = pre;
36 pre->right = head;
37
38 return head;
39 }
40
41 //中序遍历
42 void inorderTraverse(Node* root) {
43 if(root == NULL)
44 return;
45
46 inorderTraverse(root->left);
47
48 //找到头结点
49 if(head == NULL)
50 head = root;
51 //当前和前一个结点pre相互指向对方，构成双向链表
52 if(pre != NULL) {
53 pre->right = root;
54 root->left = pre;
55 }
56 pre = root; //将前一个结点更新为当前结点
57
58 inorderTraverse(root->right);
59 }
60 };
