题解 | #二叉搜索树与双向链表#

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

	// 使用中序遍历
	void inorderTrave(TreeNode* root, TreeNode* left, TreeNode* right){
		if (left != nullptr) {
			inorderTrave(left, left->left, left->right);
			if (left->right != nullptr) {
				left = left->right;
			}
			root->left = left;
			left->right = root;
		}
		
		if (right != nullptr) {
			inorderTrave(right, right->left, right->right);
			if (right->left != nullptr) {
				right = right->left;
			}
			root->right = right;
			right->left = root;
		}
		
	}

    TreeNode* Convert(TreeNode* pRootOfTree) {
		if (pRootOfTree == nullptr) {
			return nullptr;
		}
		inorderTrave(pRootOfTree, pRootOfTree->left, pRootOfTree->right);
       while (pRootOfTree->left != nullptr) {
		pRootOfTree = pRootOfTree->left;
	   }
	   return pRootOfTree;
    }
};