题目如下：
Given a binary tree
rootand a linked list withheadas the first node.Return True if all the elements in the linked list starting from the
headcorrespond 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: trueExample 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 fromhead.Constraints:
1 <= node.val <= 100for each node in the linked list and binary tree.- The given linked list will contain between
1and100nodes.- The given binary tree will contain between
1and2500nodes.
解题思路：依次判断树中每个节点是否能作为链表的头结点即可。
代码如下：
# 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
【leetcode】1367. Linked List in Binary Tree
原文：https://www.cnblogs.com/seyjs/p/12416400.html
