1367. Linked List in Binary Tree

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1367. Linked List in Binary Tree

Description

Given a binary tree root and a linked list withheadas 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:

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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:

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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:

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

Hints/Notes

  • N/A

Solution

Language: C++

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/**
 * 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;
        }
        bool ans = false;
        if (head->val == root->val) {
            ans = check(head, root);
        }
        if (ans)
            return true;
        return isSubPath(head, root->left) || isSubPath(head, root->right);
    }

    bool check(ListNode* head, TreeNode* root) {
        if (!head) {
            return true;
        }
        if (!root) {
            return false;
        }
        if (root->val != head->val) {
            return false;
        }

        return check(head->next, root->left) || check(head->next, root->right);
    }
};