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English Version

题目描述

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。

解法

Python3

# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, val=0, next=None):
#         self.val = val
#         self.next = next
# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution:
    def isSubPath(self, head: ListNode, root: TreeNode) -> bool:
        def dfs(head, root):
            if head is None:
                return True
            if root is None:
                return False
            if root.val != head.val:
                return False
            return dfs(head.next, root.left) or dfs(head.next, root.right)

        if root is None:
            return False
        return dfs(head, root) or self.isSubPath(head, root.left) or self.isSubPath(head, root.right)

Java

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode() {}
 *     ListNode(int val) { this.val = val; }
 *     ListNode(int val, ListNode next) { this.val = val; this.next = next; }
 * }
 */
/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
class Solution {
    public boolean isSubPath(ListNode head, TreeNode root) {
        if (root == null) {
            return false;
        }
        return dfs(head, root) || isSubPath(head, root.left) || isSubPath(head, root.right);
    }

    private boolean dfs(ListNode head, TreeNode root) {
        if (head == null) {
            return true;
        }
        if (root == null) {
            return false;
        }
        if (root.val != head.val) {
            return false;
        }
        return dfs(head.next, root.left) || dfs(head.next, root.right);
    }
}

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