题解 | #二叉搜索树与双向链表#

中序遍历：递归

/*
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 == nullptr) {
        return nullptr;
      }
      
      Convert(pRootOfTree->left);
      
      if (pre == nullptr) {
        head = pRootOfTree;
        pre = pRootOfTree;
      } else {
        pre->right = pRootOfTree;
        pRootOfTree->left = pre;
        pre = pRootOfTree;
      }
      
      Convert(pRootOfTree->right);
      
      return head;
    }
  private:
    TreeNode *head = nullptr;
    TreeNode *pre = nullptr;
};

栈： 这里注意循环的终止条件

/*
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 == nullptr) {
        return nullptr;
      }
      
      Convert(pRootOfTree->left);
      
      if (pre == nullptr) {
        head = pRootOfTree;
        pre = pRootOfTree;
      } else {
        pre->right = pRootOfTree;
        pRootOfTree->left = pre;
        pre = pRootOfTree;
      }
      
      Convert(pRootOfTree->right);
      
      return head;
    }
  private:
    TreeNode *head = nullptr;
    TreeNode *pre = nullptr;
};