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1367. Linked List in Binary Tree

Difficulty Topics

Description

Given a binary tree root and a linked list with head as the first node. 

Return True if all the elements in the linked list starting from the head correspond to some downward path connected in the binary tree otherwise return False.

In this context downward path means a path that starts at some node and goes downwards.

 

Example 1:

Input: head = [4,2,8], root = [1,4,4,null,2,2,null,1,null,6,8,null,null,null,null,1,3]
Output: true
Explanation: Nodes in blue form a subpath in the binary Tree.

Example 2:

Input: head = [1,4,2,6], root = [1,4,4,null,2,2,null,1,null,6,8,null,null,null,null,1,3]
Output: true

Example 3:

Input: head = [1,4,2,6,8], root = [1,4,4,null,2,2,null,1,null,6,8,null,null,null,null,1,3]
Output: false
Explanation: There is no path in the binary tree that contains all the elements of the linked list from head.

 

Constraints:

  • The number of nodes in the tree will be in the range [1, 2500].
  • The number of nodes in the list will be in the range [1, 100].
  • 1 <= Node.val <= 100 for each node in the linked list and binary tree.

Solution

linked-list-in-binary-tree.py
# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, val=0, next=None):
#         self.val = val
#         self.next = next
# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution:
    def isSubPath(self, head: Optional[ListNode], root: Optional[TreeNode]) -> bool:

        if not head: return True

        if not root: return False

        def go(node, ll, curr):
            if not curr: return True

            if not node: return False

            if node.val == curr.val:
                curr = curr.next
            elif node.val == ll.val:
                curr = ll.next
            else:
                return False

            return go(node.left, ll, curr) or go(node.right, ll, curr)


        return go(root, head, head) or self.isSubPath(head, root.left) or self.isSubPath(head, root.right)