# Flatten a binary tree into linked list | Set-2

Given a binary tree, flatten it into a linked list. After flattening, the left of each node should point to NULL and right should contain next node in level order.

Example:

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

Output:
1
\
2
\
3
\
4
\
5
\
6

Input:
1
/ \
3   4
/
2
\
5
Output:
1
\
3
\
4
\
2
\
5
```

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Approach: An approach using recursion has already been discussed in the previous post. A pre-order traversal of the binary tree using stack has been implied in this approach. In this traversal, every time a right child is pushed in the stack, the right child is made equal to the left child and left child is made equal to NULL. If the right child of the node becomes NULL, the stack is popped and the right child becomes the popped value from the stack. The above steps are repeated until the size of the stack is zero or root is NULL.

Below is the implementation of the above approach:

 `// C++ program to flatten the linked  ` `// list using stack | set-2  ` `#include ` `#include ` `using` `namespace` `std; ` ` `  `struct` `Node { ` `    ``int` `key; ` `    ``Node *left, *right; ` `}; ` ` `  `/* utility that allocates a new Node  ` `   ``with the given key  */` `Node* newNode(``int` `key) ` `{ ` `    ``Node* node = ``new` `Node; ` `    ``node->key = key; ` `    ``node->left = node->right = NULL; ` `    ``return` `(node); ` `} ` ` `  `// To find the inorder traversal ` `void` `inorder(``struct` `Node* root) ` `{ ` `    ``// base condition ` `    ``if` `(root == NULL) ` `        ``return``; ` `    ``inorder(root->left); ` `    ``cout << root->key << ``" "``; ` `    ``inorder(root->right); ` `} ` ` `  `// Function to convert binary tree into ` `// linked list by altering the right node ` `// and making left node point to NULL ` `Node* solution(Node* A) ` `{ ` ` `  `    ``// Declare a stack ` `    ``stack st; ` `    ``Node* ans = A; ` ` `  `    ``// Iterate till the stack is not empty ` `    ``// and till root is Null ` `    ``while` `(A != NULL || st.size() != 0) { ` ` `  `        ``// Check for NULL ` `        ``if` `(A->right != NULL) { ` `            ``st.push(A->right); ` `        ``} ` ` `  `        ``// Make the Right Left and ` `        ``// left NULL ` `        ``A->right = A->left; ` `        ``A->left = NULL; ` ` `  `        ``// Check for NULL ` `        ``if` `(A->right == NULL && st.size() != 0) { ` `            ``A->right = st.top(); ` `            ``st.pop(); ` `        ``} ` ` `  `        ``// Iterate ` `        ``A = A->right; ` `    ``} ` `    ``return` `ans; ` `} ` ` `  `// Driver Code ` `int` `main() ` `{ ` `    ``/*    1 ` `        ``/   \ ` `       ``2     5 ` `      ``/ \     \ ` `     ``3   4     6 */` ` `  `    ``// Build the tree ` `    ``Node* root = newNode(1); ` `    ``root->left = newNode(2); ` `    ``root->right = newNode(5); ` `    ``root->left->left = newNode(3); ` `    ``root->left->right = newNode(4); ` `    ``root->right->right = newNode(6); ` ` `  `    ``// Call the function to ` `    ``// flatten the tree ` `    ``root = solution(root); ` ` `  `    ``cout << ``"The Inorder traversal after "` `            ``"flattening binary tree "``; ` ` `  `    ``// call the function to print ` `    ``// inorder after flatenning ` `    ``inorder(root); ` `    ``return` `0; ` ` `  `    ``return` `0; ` `} `

Output:

```The Inorder traversal after flattening binary tree 1 2 3 4 5 6
```

Time Complexity: O(N)
Auxiliary Space: O(Log N)

My Personal Notes arrow_drop_up