LeetCode 1367. Linked List in Binary Tree二叉树中的列表【Medium】【Python】【DFS】

Problem

LeetCode

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.

问题

力扣

给你一棵以 root 为根的二叉树和一个 head 为第一个节点的链表。

如果在二叉树中，存在一条一直向下的路径，且每个点的数值恰好一一对应以 head 为首的链表中每个节点的值，那么请你返回 True ，否则返回 False 。

一直向下的路径的意思是：从树中某个节点开始，一直连续向下的路径。

示例 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。

示例 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

示例 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。

提示：

  • 二叉树和链表中的每个节点的值都满足 1 <= node.val <= 100 。
  • 链表包含的节点数目在 1 到 100 之间。
  • 二叉树包含的节点数目在 1 到 2500 之间。

思路

两重DFS

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代码
# 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)

代码地址

GitHub链接

参考

Jerry学长题解