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

落日映苍穹つ 2023-07-10 08:27 7阅读 0赞

> 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:**

![format_png][]

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:**

![format_png 1][]

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][Link 1]

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：**

![format_png][]

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：**

![format_png 1][]

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][GitHub]

### 参考 ###

[Jerry学长题解][Jerry]

[LeetCode]: https://leetcode.com/problems/linked-list-in-binary-tree/
[format_png]: https://imgconvert.csdnimg.cn/aHR0cHM6Ly9jZG4uanNkZWxpdnIubmV0L2doL1dvbno1MTMwL015LVByaXZhdGUtSW1nSG9zdC9pbWcvc2FtcGxlXzFfMTcyMC5wbmc?x-oss-process=image/format,png
[format_png 1]: https://imgconvert.csdnimg.cn/aHR0cHM6Ly9jZG4uanNkZWxpdnIubmV0L2doL1dvbno1MTMwL015LVByaXZhdGUtSW1nSG9zdC9pbWcvc2FtcGxlXzJfMTcyMC5wbmc?x-oss-process=image/format,png
[Link 1]: https://leetcode-cn.com/problems/linked-list-in-binary-tree/
[GitHub]: https://github.com/Wonz5130/LeetCode-Solutions/blob/master/solutions/1367-Linked-List-in-Binary-Tree/1367.py
[Jerry]: https://leetcode-cn.com/problems/linked-list-in-binary-tree/solution/zhe-ti-jiu-shi-subtreeyi-mao-yi-yang-by-jerry_nju/