1367. Linked List in Binary Tree

Hint

Solution - Recursion

/**
 * 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 (!root) return false;
        return dfs(head, root) || isSubPath(head, root->left) || isSubPath(head, root->right);
    }
    bool dfs(ListNode* head, TreeNode* node)
    {
        if (!head) return true;
        if (!node) return false;
        if (node->val != head->val) return false;
        return dfs(head->next, node->left) || dfs(head->next, node->right);
    }
};

• T:
  $O(n\cdot m)$
• S:
  $O(n+m)$