【leetcode】1367. Linked List in Binary Tree

Topics are as follows:

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:

• 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.

Outline of Solution : sequentially determining whether each node in the tree can be used as the first node to the list.

code show as below:

# Definition for singly-linked list.
# class ListNode(object):
#     def __init__(self, x):
#         self.val = x
#         self.next = None

# Definition for a binary tree node.
# class TreeNode(object):
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution(object):
    res = False
    def compare(self,tree_node,list_node):
        if self.res == True:
            return
        elif list_node.next == None and list_node.val == tree_node.val:
            self.res = True
            return
        elif list_node.val != tree_node.val:
            return
        elif list_node.next == None:
            return
        if tree_node.left != None:
            self.compare(tree_node.left,list_node.next)
        if tree_node.right != None:
            self.compare(tree_node.right,list_node.next)

    def isSubPath(self, head, root):
        """
        :type head: ListNode
        :type root: TreeNode
        :rtype: bool
        """
        self.res = False

        def recursive(tree_node):
            self.compare(tree_node,head)
            if self.res == True:
                return
            if tree_node.left != None:
                recursive(tree_node.left)
            if tree_node.right != None:
                recursive(tree_node.right)

        recursive(root)
        return self.res