[剑指Offer] 26.二叉搜索树与双向链表

【思路】因为二叉搜索树的中序遍历就是递增排列的，所以只要在中序遍历时将每个结点放入vector中，再分别为每个结点的左右指针赋值即可。

 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 {
12 public:
13     vector<TreeNode *> nodeVector;
14 
15     void inOrder(TreeNode* root)
16     {
17         if(root->left)
18             inOrder(root->left);
19         nodeVector.push_back(root);
20         if(root->right)
21             inOrder(root->right);
22     }
23 
24     TreeNode* Convert(TreeNode* pRootOfTree)
25     {
26         if((pRootOfTree == NULL) || (!pRootOfTree->left) && (!pRootOfTree->right))
27             return pRootOfTree;
28         inOrder(pRootOfTree);
29         nodeVector[0]->left = NULL;
30         nodeVector[0]->right = nodeVector[1];
31         int i = 1;
32         for( ; i < nodeVector.size() - 1; i ++)
33         {
34             nodeVector[i]->left = nodeVector[i - 1];
35             nodeVector[i]->right = nodeVector[i + 1];
36         }
37         nodeVector[i]->left = nodeVector[i - 1];
38         nodeVector[i]->right = NULL;
39         return nodeVector[0];
40     }
41 };

 

posted @ 2017-03-03 15:24  Strawberry丶  阅读(151)  评论(0)  编辑  收藏  举报