Skip to content
Related Articles

Related Articles

Flatten a binary tree into linked list | Set-2
  • Difficulty Level : Medium
  • Last Updated : 29 Dec, 2020
GeeksforGeeks - Summer Carnival Banner

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

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

Python3




# Python3 program to flatten the linked
# list using stack | set-2
class Node:
     
    def __init__(self, key):
         
        self.key = key
        self.left = None
        self.right = None
         
# Utility that allocates a new Node
# with the given key 
def newNode(key):
 
    node = Node(key)
    node.key = key
    node.left = node.right = None
    return (node)
 
# To find the inorder traversal
def inorder(root):
 
    # Base condition
    if (root == None):
        return
     
    inorder(root.left)
    print(root.key, end = ' ')
    inorder(root.right)
 
# Function to convert binary tree into
# linked list by altering the right node
# and making left node point to None
def solution(A):
  
    # Declare a stack
    st = []
    ans = A
  
    # Iterate till the stack is not empty
    # and till root is Null
    while (A != None or len(st) != 0):
  
        # Check for None
        if (A.right != None):
            st.append(A.right)
  
        # Make the Right Left and
        # left None
        A.right = A.left
        A.left = None
  
        # Check for None
        if (A.right == None and len(st) != 0):
            A.right = st[-1]
            st.pop()
  
        # Iterate
        A = A.right
         
    return ans
  
# Driver Code
if __name__=='__main__':
     
    '''    1
        /   \
       2     5
      / \     \
     3   4     6 '''
  
    # Build the tree
    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)
  
    print("The Inorder traversal after "
          "flattening binary tree ",
          end = '')
  
    # Call the function to print
    # inorder after flatenning
    inorder(root)
     
# This code is contributed by rutvik_56

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)
 

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.  To complete your preparation from learning a language to DS Algo and many more,  please refer Complete Interview Preparation Course.




My Personal Notes arrow_drop_up
Recommended Articles
Page :