# 114. Flatten Binary Tree to Linked List
## Problem Statement
\
Given the `root` of a binary tree, flatten the tree into a "linked list":
* The "linked list" should use the same `TreeNode` class where the `right` child pointer points to the next node in the list and the `left` child pointer is always `null`.
* The "linked list" should be in the same order as a [**pre-order traversal**](https://en.wikipedia.org/wiki/Tree_traversal#Pre-order,_NLR) of the binary tree.
**Example 1:**
Input: root = [1,2,5,3,4,null,6]
Output: [1,null,2,null,3,null,4,null,5,null,6]
**Example 2:**
Input: root = []
Output: []
**Example 3:**
Input: root = [0]
Output: [0]
**Constraints:**
* The number of nodes in the tree is in the range `[0, 2000]`.
* `-100 <= Node.val <= 100`
**Follow up:** Can you flatten the tree in-place (with `O(1)` extra space)?
## Intuition
```
Approach 1:
Using recursion, maintain a prev node to attach left tree to right part
And right part to null
Approach 2:
Iterative, same as above
Approach 3:
Morris traversal
```
### Links
### Video Links
### Approach 1:
```
Recursion
```
{% code title="C++" lineNumbers="true" %}
```cpp
class Solution {
public:
TreeNode *prev = nullptr;
void flatten(TreeNode* root) {
if(root == nullptr)
return ;
flatten(root->right);
flatten(root->left);
root->right = prev;
root->left = nullptr;
prev = root;
}
};
```
{% endcode %}
### Approach 2:
```
Iterative
```
{% code title="C++" lineNumbers="true" %}
```cpp
class Solution {
public:
void flatten(TreeNode* root) {
if(root == NULL) return;
stack st;
st.push(root);
while(!st.empty()){
TreeNode* curr = st.top();
st.pop();
if(curr->right)
st.push(curr->right);
if(curr->left)
st.push(curr->left);
if(!st.empty())
curr->right = st.top();
curr->left = nullptr;
}
}
};
```
{% endcode %}
### Approach 3:
```
```
{% code title="C++" lineNumbers="true" %}
```cpp
```
{% endcode %}
### Approach 4:
```
```
{% code title="C++" lineNumbers="true" %}
```cpp
```
{% endcode %}
### Similar Problems
###
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