Program to find out if a linked list is present in a given binary tree in Python

Suppose we are given a binary tree that has a root node 'root' and a linked list that has a head node 'head'. We have to find out if that linked list exists in that binary tree. If a set of nodes in the tree have links with each other in order as a linked list, and if that order is similar to that of the provided linked list, then we return 'True' or otherwise, we return 'False'.

So, if the input is like

Tree

Linked List

then the output will be True.

To solve this, we will follow these steps −

• arr := a new list
• size := size of arr
• temp_arr := an array of size (size + 1) initialized with -1
• Define a function helper() . This will take root, val
  • if val >= size, then
    • return True
  • if root is similar to None, then
    • return False
  • val := val + 1
  • while val > 0 and value of root is not same as arr[val - 1], do
    • val := temp_arr[val - 1] + 1
  • if helper(left of root, val) or helper(right of root, val) is True, then
    • return True
  • return False
• start := head
• while start is not same as null, do
  • add the value of start at the end of arr
  • start := next of start
• for node in range 1 to size + 1, do
  • temp_arr[node] := temp_arr[node - 1] + 1
  • while temp_arr[node] > 0 and arr[node - 1] is not same as arr[temp_arr[node] - 1], do
    • temp_arr[node] := temp_arr[temp_arr[node] - 1] + 1
• return helper(root, 0)

Example

Let us see the following implementation to get better understanding −

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

class ListNode:
   def __init__(self, val, next=None):
      self.val = val
      self.next = next

def insert(temp,data):
   que = []
   que.append(temp)
   while (len(que)):
      temp = que[0]
      que.pop(0)
      if (not temp.left):
         if data is not None:
            temp.left = TreeNode(data)
         else:
            temp.left = TreeNode(0)
         break
      else:
         que.append(temp.left)

      if (not temp.right):
         if data is not None:
            temp.right = TreeNode(data)
         else:
            temp.right = TreeNode(0)
         break
      else:
         que.append(temp.right)

def make_tree(elements):
   node = TreeNode(elements[0])
   for element in elements[1:]:
      insert(node, element)
   return node

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

def solve(root, head):
   arr = []
   start = head
   while start:
      arr += (start.val,)
      start = start.next
   size = len(arr)
   temp_arr = [-1] * (size + 1)
   for node in range(1, size + 1):
      temp_arr[node] = temp_arr[node - 1] + 1
      while temp_arr[node] > 0 and arr[node - 1] != arr[temp_arr[node] - 1]:
         temp_arr[node] = temp_arr[temp_arr[node] - 1] + 1
   def helper(root, val):
      if val >= size:
         return True
      if not root:
         return False
      val += 1
      while val > 0 and root.val != arr[val - 1]:
         val = temp_arr[val - 1] + 1
      if helper(root.left, val) or helper(root.right, val):
         return True
      return False
   return helper(root, 0)

root = make_tree([6, 7, 8, 9, 10])
head = make_list([6, 7, 10])
print(solve(root, head))

Input

root = make_tree([6, 7, 8, 9, 10])
head = make_list([6, 7, 10])
print(solve(root, head))

Output

True