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
山登绝顶我为峰 3(^v^)327 分钟前
如何录制带备注的演示文稿(LaTex Beamer + Pympress)
c++·线性代数·算法·计算机·密码学·音视频·latex
Two_brushes.2 小时前
【算法】宽度优先遍历BFS
算法·leetcode·哈希算法·宽度优先
森焱森4 小时前
水下航行器外形分类详解
c语言·单片机·算法·架构·无人机
QuantumStack5 小时前
【C++ 真题】P1104 生日
开发语言·c++·算法
写个博客6 小时前
暑假算法日记第一天
算法
绿皮的猪猪侠6 小时前
算法笔记上机训练实战指南刷题
笔记·算法·pta·上机·浙大
hie988947 小时前
MATLAB锂离子电池伪二维(P2D)模型实现
人工智能·算法·matlab
杰克尼7 小时前
BM5 合并k个已排序的链表
数据结构·算法·链表
.30-06Springfield8 小时前
决策树(Decision tree)算法详解(ID3、C4.5、CART)
人工智能·python·算法·决策树·机器学习
我不是哆啦A梦8 小时前
破解风电运维“百模大战”困局,机械版ChatGPT诞生?
运维·人工智能·python·算法·chatgpt