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:
1 <= node.val <= 100 for each node in the linked list and binary tree.
The given linked list will contain between 1 and 100 nodes.
The given binary tree will contain between 1 and 2500 nodes.
来源：力扣（LeetCode）
链接：https://leetcode-cn.com/problems/linked-list-in-binary-tree
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# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, x):
#         self.val = x
#         self.next = None
# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None
class Solution:
    # 比对是否存在和链表head相同的路径
    def isSame(self, head: ListNode, root: TreeNode) -> bool:
        if not head: return True # 没有剩余链表节点返回真
        if head and root and root.val == head.val: # 当前节点相同，比较左节点或右节点
            return self.isSame(head.next, root.left) or self.isSame(head.next, root.right)
        return False # 当前节点不同，返回假
    def isSubPath(self, head: ListNode, root: TreeNode) -> bool:
        if root: # 当前节点不为空
            # 返回当前节点是否和链表一致，或者左子树是否含有链表路径，或者右子树是否含有链表路径
            return self.isSame(head, root) or self.isSubPath(head, root.left) or self.isSubPath(head, root.right)
        return False