题解 | #二叉搜索树与双向链表#先中序排序再数组连接

/*
struct TreeNode {
	int val;
	struct TreeNode *left;
	struct TreeNode *right;
	TreeNode(int x) :
			val(x), left(NULL), right(NULL) {
	}
};*/
class Solution {
public:
	vector<TreeNode*> TreeList;
	void inorder(TreeNode* root){
		if (!root) return;
		inorder(root->left);
		TreeList.push_back(root);
		inorder(root->right);
	} 
    TreeNode* Convert(TreeNode* pRootOfTree) {
        if (!pRootOfTree) return pRootOfTree;
		inorder(pRootOfTree);
		for(int i = 0;i<TreeList.size()-1;i++){
			TreeList[i+1] -> left = TreeList[i];
			TreeList[i] -> right = TreeList[i+1];
		}
		return TreeList[0];
    }
};