Formatted question description: https://leetcode.ca/all/1367.html
1367. 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:
1 <= node.val <= 100
for each node in the linked list and binary tree.- The given linked list will contain between
1
and100
nodes. - The given binary tree will contain between
1
and2500
nodes.
Related Topics:
Linked List, Dynamic Programming, Tree
Solution 1.
// OJ: https://leetcode.com/problems/linked-list-in-binary-tree/
// Time: O(N * min(L, H))
// Space: O(H)
class Solution {
bool dfs(ListNode *head, TreeNode *root) {
if (!head) return true;
if (!root) return false;
return head->val == root->val && (dfs(head->next, root->left) || dfs(head->next, root->right));
}
public:
bool isSubPath(ListNode* head, TreeNode* root) {
if (!head) return true;
if (!root) return false;
return dfs(head, root) || isSubPath(head, root->left) || isSubPath(head, root->right);
}
};
Java
/**
* Definition for singly-linked list.
* public class ListNode {
* int val;
* ListNode next;
* ListNode(int x) { val = x; }
* }
*/
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
class Solution {
public boolean isSubPath(ListNode head, TreeNode root) {
int headVal = head.val;
Queue<TreeNode> listQueue = new LinkedList<TreeNode>();
Queue<TreeNode> searchQueue = new LinkedList<TreeNode>();
searchQueue.offer(root);
while (!searchQueue.isEmpty()) {
TreeNode node = searchQueue.poll();
if (node.val == headVal)
listQueue.offer(node);
TreeNode left = node.left, right = node.right;
if (left != null)
searchQueue.offer(left);
if (right != null)
searchQueue.offer(right);
}
ListNode temp = head.next;
while (temp != null && !listQueue.isEmpty()) {
int nextVal = temp.val;
int size = listQueue.size();
for (int i = 0; i < size; i++) {
TreeNode node = listQueue.poll();
TreeNode left = node.left, right = node.right;
if (left != null && left.val == nextVal)
listQueue.offer(left);
if (right != null && right.val == nextVal)
listQueue.offer(right);
}
temp = temp.next;
}
return !listQueue.isEmpty();
}
}