> For the complete documentation index, see [llms.txt](https://maksimdan.gitbook.io/interview-practice-problems/llms.txt). Markdown versions of documentation pages are available by appending `.md` to page URLs; this page is available as [Markdown](https://maksimdan.gitbook.io/interview-practice-problems/leetcode_sessions/114_flatten_binary_tree_to_linked_list.md).

# 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();
      }
  }
```


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