Convert Linked List to Zig Zag Binary Tree in Python

Suppose we have a singly linked list, we have to convert it to a binary tree path using following rules −

• The head of the linked list is the root.
• Each subsequent node is the left child of the parent when its value is less, otherwise it will be the right child.

So, if the input is like [2,1,3,4,0,5], then the output will be

To solve this, we will follow these steps −

• Define a function solve() . This will take node
• if node is null, then
  • return null
• root := create a tree node with value same as value of node
• if next of node is not null, then
  • if value of next of node < value of node, then
    • left of root := solve(next of node)
  • otherwise,
    • right of root := solve(next of node)
• return root

Let us see the following implementation to get better understanding −

Example

 Live Demo

class ListNode:
   def __init__(self, data, next = None):
      self.val = data
      self.next = next
def make_list(elements):
   head = ListNode(elements[0])
   for element in elements[1:]:
      ptr = head
      while ptr.next:
         ptr = ptr.next
      ptr.next = ListNode(element)
   return head
class TreeNode:
   def __init__(self, data, left = None, right = None):
      self.data = data
      self.left = left
      self.right = right
def print_tree(root):
   if root is not None: print_tree(root.left)
      print(root.data, end = ', ') print_tree(root.right)
class Solution:
   def solve(self, node):
      if not node:
         return None
         root = TreeNode(node.val)
      if node.next:
         if node.next.val < node.val:
            root.left = self.solve(node.next)
         else:
            root.right = self.solve(node.next)
      return root
ob = Solution()
L = make_list([2,1,3,4,0,5])
print_tree(ob.solve(L))

Input

[2,1,3,4,0,5]

Output

1, 3, 0, 5, 4, 2,