二叉树变双向链表

使用递归，分四种情况讨论即可

/*
struct TreeNode {
    int val;
    struct TreeNode *left;
    struct TreeNode *right;
    TreeNode(int x) :
            val(x), left(NULL), right(NULL) {
    }
};*/
class Solution {
public:
    TreeNode* Convert(TreeNode* pRootOfTree)
    {
        if (pRootOfTree == NULL) {
          return NULL;
        }  else if (pRootOfTree->left == NULL && pRootOfTree->right == NULL) {
          return pRootOfTree;
        } else if (pRootOfTree->left == NULL) {
          TreeNode* r = Convert(pRootOfTree->right);
          pRootOfTree->right = r;
          r->left = pRootOfTree;
          return pRootOfTree;
        } else if (pRootOfTree->right == NULL) {
          TreeNode* l = Convert(pRootOfTree->left);
          TreeNode* temp = l;
          while (l->right) {
            l = l->right;
          }
          pRootOfTree->left = l;
          l->right = pRootOfTree;
          return temp;
        }
      else if (pRootOfTree) {
          TreeNode* l = Convert(pRootOfTree->left);
          TreeNode* r = Convert(pRootOfTree->right);
          TreeNode* temp_l = l;
          while (l->right) {
            l = l->right;
          }
          pRootOfTree->left = l;
          l->right = pRootOfTree;
          pRootOfTree->right = r;
          r->left = pRootOfTree;
          return temp_l;
        }
    }
};