/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    TreeNode* convert(TreeNode* root) {
        //思路，用递归思想，先用根节点，将左子树的最右边与根相连，再将右子树的最左边与根相连
        if(!root) return nullptr;

        auto x = dfs(root);
        return x.first;
    }

    pair<TreeNode*, TreeNode*> dfs(TreeNode* root){
        if(!root->left && !root->right) return {root,root};

        if(root->left && root->right){
            auto lside = dfs(root->left),rside = dfs(root->right);
            lside.second->right = root,root->left = lside.second;
            rside.first->left = root,root->right = rside.first;
            return {lside.first,rside.second};
        }

         if(root->left){
            auto lside = dfs(root->left);
            lside.second->right = root,root->left = lside.second;

            return {lside.first,root};
        }

         if(root->right){
            auto rside = dfs(root->right);

            rside.first->left = root,root->right = rside.first;
            return {root,rside.second};
        }
    }
};