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
qq_513970446 分钟前
力扣 hot100 Day37
算法·leetcode
不見星空27 分钟前
leetcode 每日一题 1865. 找出和为指定值的下标对
算法·leetcode
我爱Jack37 分钟前
时间与空间复杂度详解:算法效率的度量衡
java·开发语言·算法
DoraBigHead2 小时前
小哆啦解题记——映射的背叛
算法
Heartoxx2 小时前
c语言-指针与一维数组
c语言·开发语言·算法
孤狼warrior3 小时前
灰色预测模型
人工智能·python·算法·数学建模
京东云开发者3 小时前
京东零售基于国产芯片的AI引擎技术
算法
chao_7894 小时前
回溯题解——子集【LeetCode】二进制枚举法
开发语言·数据结构·python·算法·leetcode
十盒半价4 小时前
从递归到动态规划:手把手教你玩转算法三剑客
javascript·算法·trae
GEEK零零七4 小时前
Leetcode 1070. 产品销售分析 III
sql·算法·leetcode