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

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文章讨论了一个编程问题，如何判断给定的单链表是否表示二叉树中的某个向下路径。作者提供了两种解决方案，一种是递归地在每个节点上使用辅助函数，另一种是迭代地遍历树并更新全局状态。

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。

/**
 * 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。

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);
            }
        }
    }
};