# 114 Flatten Binary Tree to Linked List

Given a binary tree, flatten it to a linked list in-place.

For example, Given

```
         1
        / \
       2   5
      / \   \
     3   4   6
```

The flattened tree should look like:

```
   1
    \
     2
      \
       3
        \
         4
          \
           5
            \
             6
```

// runs on my comp

* Approach:
  * Create a LL through traversal, and then append to root.
  * Technically O(1) space, since I make to delete the left subtree.
  * O(n), since I just append to root

```cpp
    Node<T> *head = nullptr;
    Node<T> *front = nullptr;

    void addLL(T val) {
        if (!head) {
            head = new Node<T>(val);
            front = head;
            front->right = nullptr;
            front->left = nullptr;
        }
        else {
            Node<T> *temp = new Node<T>(val);
            front->right = temp;
            front = front->right;
            front->left = nullptr;
            front->right = nullptr;
        }
    }

    void flatten(Node<T> *myroot, Node<T> *parent) {
        if (myroot) {
            flatten(myroot->left, myroot);
            addLL(myroot->value);
            flatten(myroot->right, myroot);

            // recusively remove nodes from bottom - top
            delete myroot;
            myroot = nullptr;
        }

    }

    void printLL() {
        Node<T> *temp = head;
        while (temp) {
            cout << temp->value << " ";
            temp = temp->right;
        }
        cout << endl;
    }

    void flatten(Node<T> *root) {
        flatten(root, nullptr);
        root = head;
        printLL();
        pause();
    }
```

* Working O(N) space trivial solution

```cpp
  void flatten(TreeNode* root) {
      if (!root) return;

      TreeNode *t;
      stack<TreeNode*> s;
      s.push(root);

      while (!s.empty()) {
          t = s.top();
          s.pop();
          if (t->right) s.push(t->right);
          if (t->left) s.push(t->left);
          t->left = nullptr;
          t->right = s.empty() ? nullptr : s.top();
      }
  }
```


---

# Agent Instructions: Querying This Documentation

If you need additional information that is not directly available in this page, you can query the documentation dynamically by asking a question.

Perform an HTTP GET request on the current page URL with the `ask` query parameter:

```
GET https://maksimdan.gitbook.io/interview-practice-problems/leetcode_sessions/114_flatten_binary_tree_to_linked_list.md?ask=<question>
```

The question should be specific, self-contained, and written in natural language.
The response will contain a direct answer to the question and relevant excerpts and sources from the documentation.

Use this mechanism when the answer is not explicitly present in the current page, you need clarification or additional context, or you want to retrieve related documentation sections.