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:
    vector<TreeNode*>list;
    void dfs(TreeNode* pRootOfTree){//中序遍历
        if(!pRootOfTree)return ;
        dfs(pRootOfTree->left);
        list.push_back(pRootOfTree);
        dfs(pRootOfTree->right);

        
    }
    TreeNode* Convert(TreeNode* pRootOfTree) {
        if(!pRootOfTree)return pRootOfTree;
        dfs(pRootOfTree);
        for(int i=0;i<list.size()-1;i++){//交换指针
            list[i]->right=list[i+1];
            list[i+1]->left=list[i];
        }
        return list[0];
    }
};