剑指offer 二叉搜索树与双向链表

 1 /*
 2 struct TreeNode {
 3     int val;
 4     struct TreeNode *left;
 5     struct TreeNode *right;
 6     TreeNode(int x) :
 7             val(x), left(NULL), right(NULL) {
 8     }
 9 };*/
10 class Solution {
11 public:
12     TreeNode* Convert(TreeNode* pRootOfTree)
13     {
14         if( pRootOfTree == NULL)
15             return pRootOfTree;
16         if( pRootOfTree->left == NULL && pRootOfTree->right == NULL )
17             return pRootOfTree;
18         TreeNode* head = new TreeNode(NULL);
19         //右子树递归（大于根节点的结点）
20         head = Convert(pRootOfTree->right);
21         if(head != NULL){
22             head->left = pRootOfTree;
23             pRootOfTree->right = head;
24         }
25         //左子树递归（小于根节点的结点）
26         head = Convert(pRootOfTree->left);
27         if(head != NULL){
28             //此处循环为了找到该子树下最大结点并与此子树根节点相连
29             while(head->right != NULL)
30                 head = head->right;
31             head->right = pRootOfTree;
32             pRootOfTree->left = head;
33         }
34         //当此子树的左右枝都排列完毕时，找到最小结点为与根节点相连
35         while(pRootOfTree->left != NULL)
36             pRootOfTree = pRootOfTree->left;
37         return pRootOfTree;
38     }
39 };