Leetcode 1367. Linked List in Binary Tree (二叉树好题)

  1. Linked List in Binary Tree
    Medium
    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:

The number of nodes in the tree will be in the range [1, 2500].

The number of nodes in the list will be in the range [1, 100].

1 <= Node.val <= 100 for each node in the linked list and binary tree.

解法1:

这题其实并不容易。要在整个二叉树里面,对每个节点调用helper()函数,用前中后序遍历应该都可以。helper()则是用的分解问题的方法。

cpp 复制代码
/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode() : val(0), next(nullptr) {}
 *     ListNode(int x) : val(x), next(nullptr) {}
 *     ListNode(int x, ListNode *next) : val(x), next(next) {}
 * };
 */
/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
class Solution {
public:
    bool isSubPath(ListNode* head, TreeNode* root) {
        if (!head) return true;
        if (!root) return false;
        if (helper(head, root)) return true;
        return isSubPath(head, root->left) || isSubPath(head, root->right);  
    }
private:
    bool helper(ListNode* head, TreeNode* root) {
        if (!head) return true;
        if (!root) return false;
        if (head->val != root->val) return false;
        return helper(head->next, root->left) || helper(head->next, root->right);
    }
};

我一开始写成下面这样,但是不对。因为它只调用了一次helper(),如果链表是{2,2,1},而树里面存在一个path{2,2,2,1},结果应该返回true。但这个解法里面,读到第3个2的时候,发现不对,已经没法走回头路了。
应该在整个树里面,对每个节点都调用helper(),用前/中/后序遍历都可以。

cpp 复制代码
class Solution {
public:
    bool isSubPath(ListNode* head, TreeNode* root) {
        origHead = head;
        origRoot = root;
        helper(head, root);
        return gPathExist;
    }
private:
    bool gPathExist = false;
    TreeNode *origRoot = NULL;
    ListNode *origHead = NULL;
    void helper(ListNode *head, TreeNode *root) {
        if (!head) {
            gPathExist = true;
            return;
        }
        if (!root || gPathExist) return;
        if (root->val == head->val) {
            helper(head->next, root->left);
            helper(head->next, root->right);
        } else {
            if (root == origRoot) {
                helper(origHead, root->left);
                helper(origHead, root->right);
            } else if (head != origHead) {
                helper(origHead, root);
                helper(origHead, root);
            }
        }
    }
};
相关推荐
sp_fyf_202418 分钟前
计算机前沿技术-人工智能算法-大语言模型-最新研究进展-2024-11-01
人工智能·深度学习·神经网络·算法·机器学习·语言模型·数据挖掘
香菜大丸39 分钟前
链表的归并排序
数据结构·算法·链表
jrrz082839 分钟前
LeetCode 热题100(七)【链表】(1)
数据结构·c++·算法·leetcode·链表
oliveira-time1 小时前
golang学习2
算法
面试鸭1 小时前
离谱!买个人信息买到网安公司头上???
java·开发语言·职场和发展
南宫生2 小时前
贪心算法习题其四【力扣】【算法学习day.21】
学习·算法·leetcode·链表·贪心算法
懒惰才能让科技进步2 小时前
从零学习大模型(十二)-----基于梯度的重要性剪枝(Gradient-based Pruning)
人工智能·深度学习·学习·算法·chatgpt·transformer·剪枝
Ni-Guvara3 小时前
函数对象笔记
c++·算法
测试19983 小时前
2024软件测试面试热点问题
自动化测试·软件测试·python·测试工具·面试·职场和发展·压力测试
泉崎3 小时前
11.7比赛总结
数据结构·算法