<p>Given a binary tree <code>root</code> and a&nbsp;linked list with&nbsp;<code>head</code>&nbsp;as the first node.&nbsp;</p>

<p>Return True if all the elements in the linked list starting from the <code>head</code> correspond to some <em>downward path</em> connected in the binary tree&nbsp;otherwise return False.</p>

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

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<p><strong><img alt="" src="https://assets.leetcode.com/uploads/2020/02/12/sample_1_1720.png" style="width: 220px; height: 280px;" /></strong></p>

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

<p><strong class="example">Example 2:</strong></p>

<p><strong><img alt="" src="https://assets.leetcode.com/uploads/2020/02/12/sample_2_1720.png" style="width: 220px; height: 280px;" /></strong></p>

<pre>
<strong>Input:</strong> head = [1,4,2,6], root = [1,4,4,null,2,2,null,1,null,6,8,null,null,null,null,1,3]
<strong>Output:</strong> true
</pre>

<p><strong class="example">Example 3:</strong></p>

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

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li>The number of nodes in the tree will be in the range <code>[1, 2500]</code>.</li>
	<li>The number of nodes in the list will be in the range <code>[1, 100]</code>.</li>
	<li><code>1 &lt;= Node.val&nbsp;&lt;= 100</code>&nbsp;for each node in the linked list and binary tree.</li>
</ul>

# 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 head.next is None:
            # special case, only need to look for one value, DFS
            q = [root]
            while q:
                curr = q.pop()
                if curr.val == head.val:
                    return True
                if curr.right:
                    q.append(curr.right)
                if curr.left:
                    q.append(curr.left)
            return False
        # case where len of list is larger than 1
        head = self.add_backtracking(head)
        dummy = head.backtrack
        # DFS
        q = [(root, head)]
        while q:
            curr, test_node = q.pop()
            while test_node is not dummy and not curr.val == test_node.val:
                # try to match other backtrack locations
                test_node = test_node.backtrack
            # either matched or reset
            test_node = test_node.next
            if test_node is None:
                return True
            if curr.right:
                q.append((curr.right, test_node))
            if curr.left:
                q.append((curr.left, test_node))
        return False

    def add_backtracking(self, head: ListNode) -> ListNode:
        dummy = ListNode(val=-1, next=head)
        head.backtrack = dummy
        node = head.next
        node.backtrack = head
        border_end = head
        while node.next is not None:
            while (
                border_end is not dummy 
                and not border_end.val == node.val
            ):
                border_end = border_end.backtrack
            border_end = border_end.next
            node.next.backtrack = border_end
            node = node.next
        return head
59