[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。

 /*
 struct TreeNode {
     int val;
     struct TreeNode *left;
     struct TreeNode *right;
     TreeNode(int x) :
             val(x), left(NULL), right(NULL) {
     }
 };*/
 class Solution
 {
 public:
     vector<TreeNode *> nodeVector;

     void inOrder(TreeNode* root)
     {
         if(root->left)
             inOrder(root->left);
         nodeVector.push_back(root);
         if(root->right)
             inOrder(root->right);
     }

     TreeNode* Convert(TreeNode* pRootOfTree)
     {
         if((pRootOfTree == NULL) || (!pRootOfTree->left) && (!pRootOfTree->right))
             return pRootOfTree;
         inOrder(pRootOfTree);
         nodeVector[]->left = NULL;
         nodeVector[]->right = nodeVector[];
         int i = ;
         for( ; i < nodeVector.size() - ; i ++)
         {
             nodeVector[i]->left = nodeVector[i - ];
             nodeVector[i]->right = nodeVector[i + ];
         }
         nodeVector[i]->left = nodeVector[i - ];
         nodeVector[i]->right = NULL;
         return nodeVector[];
     }
 };