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:
    void solotion(TreeNode* cur, TreeNode*& prev)
	{
		if (!cur)
		return;
		solotion(cur->left, prev);
        cur->left = prev;
		if (prev)
		prev->right = cur;
		prev = cur;
		solotion(cur->right, prev);	
	}

    TreeNode* Convert(TreeNode* pRootOfTree) {
         TreeNode* prev = nullptr;
		 solotion(pRootOfTree , prev);
		 while(pRootOfTree && pRootOfTree->left)
		 {
            pRootOfTree = pRootOfTree->left;
		 }        
		 return pRootOfTree;
    }
};

思路：

通过记录上一个递归的节点，然后进行修改即可。