# 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:**

![](https://assets.leetcode.com/uploads/2021/01/14/flaten.jpg)

<pre><code><strong>Input: root = [1,2,5,3,4,null,6]
</strong><strong>Output: [1,null,2,null,3,null,4,null,5,null,6]
</strong></code></pre>

**Example 2:**

<pre><code><strong>Input: root = []
</strong><strong>Output: []
</strong></code></pre>

**Example 3:**

<pre><code><strong>Input: root = [0]
</strong><strong>Output: [0]
</strong></code></pre>

&#x20;

**Constraints:**

* The number of nodes in the tree is in the range `[0, 2000]`.
* `-100 <= Node.val <= 100`

&#x20;

**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

<https://leetcode.com/problems/flatten-binary-tree-to-linked-list/description/>

### Video Links

<https://www.youtube.com/watch?v=sWf7k1x9XR4&t=587s&ab_channel=takeUforward>

### 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<TreeNode*> 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

###