算法1

(递归)

• 每次返回一个pair，分别表示已经连接双向链表的左边的节点与右边的节点
• 有几种特判情况
  • 左右节点均不存在
  • 左右节点均存在，两边进行连接
  • 左节点存在，root与左边的链表进行连接
  • 右节点存在，root与右边的链表进行连接

C++ 代码

//代码参考yxc
/**
 * 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) {
        if(!root) return nullptr;
        auto side = dfs(root);
        return side.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};
        }else
        {
            auto rside = dfs(root->right);
            rside.first->left = root,root->right = rside.first;
            return {root,rside.second};
        }

    }
};