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  • 【leetcode】1367. Linked List in Binary Tree

    题目如下:

    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.

    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
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  • 原文地址:https://www.cnblogs.com/seyjs/p/12416400.html
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