LeetCode | List 1367. Linked List in Binary Tree Binary Tree in the [Python]

List LeetCode 1367. Linked List in Binary Tree binary tree [Medium] [Python] [DFS]

Problem

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.

problem

Power button

To give you a rootroot of a binary tree and headfor the first node of the list.

If the binary tree, there has been a downward path, and the value of each point exactly correspond to headheaded the list the value of each node, and then you return Trueit, otherwise return False.

Has been downward path means: start from a node in the tree, it has been a continuous downward path.

Example 1:

输入：head = [4,2,8], root = [1,4,4,null,2,2,null,1,null,6,8,null,null,null,null,1,3]
输出：true
解释：树中蓝色的节点构成了与链表对应的子路径。

Example 2:

输入：head = [1,4,2,6], root = [1,4,4,null,2,2,null,1,null,6,8,null,null,null,null,1,3]
输出：true

Example 3:

输入：head = [1,4,2,6,8], root = [1,4,4,null,2,2,null,1,null,6,8,null,null,null,null,1,3]
输出：false
解释：二叉树中不存在一一对应链表的路径。

prompt:

And a binary value for each node in the linked list are satisfied 1 <= node.val <= 100.
The number of nodes included in the list 1to 100between.
The number of nodes in the binary tree 1to 2500between.

Thinking

DFS double

第一重：找到起点。先判断当前节点，如果不对就判断左子树和右子树。
第二重：从找到的起点开始判断剩下的点。

Python3 Code

# 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:
    def isSubPath(self, head: ListNode, root: TreeNode) -> bool:
        if head == None:
            return True
        if root == None:
            return False
        # judge root, then judge root.left and root.right
        return self.isSub(head, root) or self.isSubPath(head, root.left) or self.isSubPath(head, root.right)
    
    def isSub(self, head, node):
        # list is over
        if head == None:
            return True
        # list is not over and tree is over
        if node == None:
            return False
        # not equal
        if not head.val == node.val:
            return False
        # equal, then left and right
        return self.isSub(head.next, node.left) or self.isSub(head.next, node.right)

Code address

GitHub link

reference

Jerry seniors problem solution