Flatten a binary tree into linked list

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++

// C++ program to flatten the linked  
// list using stack | set-2  
#include <iostream> 
#include <stack> 
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<Node*> 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; 
} 

Java

// Java program to flatten the linked  
// list using stack | set-2  
import java.util.Stack; 
  
class GFG  
{ 
      
static class Node 
{ 
    int key; 
    Node left, right; 
} 
  
/* utility that allocates a new Node  
with the given key */
static Node newNode(int key) 
{ 
    Node node = new Node(); 
    node.key = key; 
    node.left = node.right = null; 
    return (node); 
} 
  
// To find the inorder traversal 
static void inorder(Node root) 
{ 
    // base condition 
    if (root == null) 
        return; 
    inorder(root.left); 
    System.out.print(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 
static Node solution(Node A) 
{ 
  
    // Declare a stack 
    Stack<Node> st = new Stack<>(); 
    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.peek(); 
            st.pop(); 
        } 
  
        // Iterate 
        A = A.right; 
    } 
    return ans; 
} 
  
// Driver Code 
public static void main(String[] args)  
{ 
    /* 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); 
  
    System.out.print("The Inorder traversal after "
            +"flattening binary tree "); 
  
    // call the function to print 
    // inorder after flatenning 
    inorder(root); 
} 
} 
  
// This code has been contributed by 29AjayKumar 

C#

// C# program to flatten the linked  
// list using stack | set-2  
using System; 
using System.Collections.Generic; 
  
class GFG  
{ 
    public class Node 
    { 
        public int key; 
        public Node left, right; 
    } 
  
    /* utility that allocates a new Node  
    with the given key */
    static Node newNode(int key) 
    { 
        Node node = new Node(); 
        node.key = key; 
        node.left = node.right = null; 
        return (node); 
    } 
  
    // To find the inorder traversal 
    static void inorder(Node root) 
    { 
        // base condition 
        if (root == null) 
            return; 
        inorder(root.left); 
        Console.Write(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 
    static Node solution(Node A) 
    { 
  
        // Declare a stack 
        Stack<Node> st = new Stack<Node>(); 
        Node ans = A; 
  
        // Iterate till the stack is not empty 
        // and till root is Null 
        while (A != null || st.Count != 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.Count != 0) 
            { 
                A.right = st.Peek(); 
                st.Pop(); 
            } 
  
            // Iterate 
            A = A.right; 
        } 
        return ans; 
    } 
  
    // Driver Code 
    public static void Main(String[] args)  
    { 
        /* 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); 
  
        Console.Write("The Inorder traversal after "
                +"flattening binary tree "); 
  
        // call the function to print 
        // inorder after flatenning 
        inorder(root); 
    } 
} 
  
// This code contributed by Rajput-Ji 

Output:

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

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

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My Personal Notes

Flatten a binary tree into linked list | Set-2

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

C++

Java

C#

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