给你二叉树的根结点 root ，请你将它展开为一个单链表：

展开后的单链表应该同样使用 TreeNode ，其中 right 子指针指向链表中下一个结点，而左子指针始终为 null 。
展开后的单链表应该与二叉树先序遍历顺序相同。

提示：

• 树中结点数在范围 [0, 2000] 内
• -100 <= Node.val <= 100

理解题意：前序遍历的N种写法

题解1 前序遍历

/*** Definition for a binary tree node.* struct TreeNode {*     int val;*     TreeNode *left;*     TreeNode *right;*     TreeNode() : val(0), left(nullptr), right(nullptr) {}*     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}*     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}* };*/
class Solution {vector<TreeNode*> arr;
public:void dfs(TreeNode* root){if(! root) return;arr.push_back(root);dfs(root->left);dfs(root->right);} void flatten(TreeNode* root) {if(! root) return;dfs(root);for(int i = 1; i < arr.size(); i++){root->right = arr[i];root->left = nullptr;root = root->right;}}
};

题解2 反前序遍历(代码简洁)

class Solution {TreeNode* preNode;
public:void flatten(TreeNode* root) {if (! root) return;// right先压栈（后面弹）flatten(root->right);flatten(root->left);// 第一次执行 root是前序遍历的最后访问的结点root->left = NULL;root->right = preNode;preNode = root;}
};

题解3 类似旋转的方法

class Solution {TreeNode* preNode;
public:void flatten(TreeNode* root) {TreeNode *curr = root;while (curr != nullptr) {if (curr->left != nullptr) {auto next = curr->left;auto predecessor = next;while (predecessor->right != nullptr) {predecessor = predecessor->right;}predecessor->right = curr->right;curr->left = nullptr;curr->right = next;}curr = curr->right;}}
};

题解4 迭代

class Solution {TreeNode* preNode;
public:void flatten(TreeNode* root) {auto v = vector<TreeNode*>();auto stk = stack<TreeNode*>();TreeNode *node = root;while (node != nullptr || !stk.empty()) {while (node != nullptr) {// 前序遍历： 动作放在最前面v.push_back(node);stk.push(node);node = node->left;}node = stk.top(); stk.pop();node = node->right;}int size = v.size();for (int i = 1; i < size; i++) {auto prev = v.at(i - 1), curr = v.at(i);prev->left = nullptr;prev->right = curr;}}
};

题解5 同时遍历+改左右子树

void flatten(TreeNode* root) {if (root == nullptr) {return;}auto stk = stack<TreeNode*>();stk.push(root);TreeNode *prev = nullptr;while (!stk.empty()) {TreeNode *curr = stk.top(); stk.pop();if (prev != nullptr) {prev->left = nullptr;prev->right = curr;}TreeNode *left = curr->left, *right = curr->right;// 防止脏数据，原数据先放进stack里if (right != nullptr) {stk.push(right);}if (left != nullptr) {stk.push(left);}//迭代prev = curr;}}