Program to convert level order binary tree traversal to linked list in Python

Suppose we have a binary search tree, we have to convert it to a singly linked list using levelorder traversal.

So, if the input is like

then the output will be [5, 4, 10, 2, 7, 15, ]

To solve this, we will follow these steps −

• head := a new linked list node

• currNode := head

• q := a list with value root

• while q is not empty, do

  • curr := delete first element from q

  • if curr is not null, then

    • next of currNode := a new linked list node with value of curr

    • currNode := next of currNode

    • insert left of curr at the end of q

    • insert right curr at the end of q

• return next of head

Let us see the following implementation to get better understanding −

Example

class ListNode:
   def __init__(self, data, next = None):
      self.val = data
      self.next = next
class TreeNode:
   def __init__(self, data, left = None, right = None):
      self.val = data
      self.left = left
      self.right = right

def print_list(head):
   ptr = head
   print('[', end = "")
   while ptr:
      print(ptr.val, end = ", ")
      ptr = ptr.next
print(']')
   
class Solution:
   def solve(self, root):
      head = ListNode(None)
      currNode = head
      q = [root]
      while q:
         curr = q.pop(0)
         if curr:
            currNode.next = ListNode(curr.val)
            currNode = currNode.next
            q.append(curr.left)
            q.append(curr.right)
      return head.next

ob = Solution()
root = TreeNode(5)
root.left = TreeNode(4)
root.right = TreeNode(10)
root.left.left = TreeNode(2)
root.right.left = TreeNode(7)
root.right.right = TreeNode(15)
head = ob.solve(root)
print_list(head)

Input

root = TreeNode(5)
root.left = TreeNode(4)
root.right = TreeNode(10)
root.left.left = TreeNode(2)
root.right.left = TreeNode(7)
root.right.right = TreeNode(15)
head = ob.solve(root)

Output

[5, 4, 10, 2, 7, 15, ]

Arnab Chakraborty

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Updated on: 10-Oct-2020

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