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:
	TreeNode* head;
	TreeNode* pre;
    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;

    }
};

使用中序递归遍历，遍历过程中更改当前节点的左为前树，前节点的右子树为当前节点