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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 and 100 nodes.
• The given binary tree will contain between 1 and 2500 nodes.

Related Topics:

## Solution 1.

• /**
* 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) {
searchQueue.offer(root);
while (!searchQueue.isEmpty()) {
TreeNode node = searchQueue.poll();
listQueue.offer(node);
TreeNode left = node.left, right = node.right;
if (left != null)
searchQueue.offer(left);
if (right != null)
searchQueue.offer(right);
}
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();
}
}

############

/**
* public class ListNode {
*     int val;
*     ListNode next;
*     ListNode() {}
*     ListNode(int val) { this.val = val; }
*     ListNode(int val, ListNode next) { this.val = val; this.next = next; }
* }
*/
/**
* Definition for a binary tree node.
* public class TreeNode {
*     int val;
*     TreeNode left;
*     TreeNode right;
*     TreeNode() {}
*     TreeNode(int val) { this.val = val; }
*     TreeNode(int val, TreeNode left, TreeNode right) {
*         this.val = val;
*         this.left = left;
*         this.right = right;
*     }
* }
*/
class Solution {
public boolean isSubPath(ListNode head, TreeNode root) {
if (root == null) {
return false;
}
}

private boolean dfs(ListNode head, TreeNode root) {
return true;
}
if (root == null || head.val != root.val) {
return false;
}
}
}

• // OJ: https://leetcode.com/problems/linked-list-in-binary-tree/
// Time: O(N * min(L, H))
// Space: O(H)
class Solution {
bool match(ListNode *head, TreeNode *root) {
if (!root) return false;
if (head->val != root->val) return false;
}
public:
bool isSubPath(ListNode* head, TreeNode* root) {
if (!root) return false;
}
};

• # Definition for singly-linked list.
# class ListNode:
#     def __init__(self, val=0, next=None):
#         self.val = val
#         self.next = next
# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution:
def isSubPath(self, head: ListNode, root: TreeNode) -> bool:
return True
if root is None:
return False
return False

if root is None:
return False
return (
)


• /**
* type ListNode struct {
*     Val int
*     Next *ListNode
* }
*/
/**
* Definition for a binary tree node.
* type TreeNode struct {
*     Val int
*     Left *TreeNode
*     Right *TreeNode
* }
*/
func isSubPath(head *ListNode, root *TreeNode) bool {
if root == nil {
return false
}
}

func dfs(head *ListNode, root *TreeNode) bool {
return true
}
if root == nil || head.Val != root.Val {
return false
}
}

• /**
* class ListNode {
*     val: number
*     next: ListNode | null
*     constructor(val?: number, next?: ListNode | null) {
*         this.val = (val===undefined ? 0 : val)
*         this.next = (next===undefined ? null : next)
*     }
* }
*/

/**
* Definition for a binary tree node.
* class TreeNode {
*     val: number
*     left: TreeNode | null
*     right: TreeNode | null
*     constructor(val?: number, left?: TreeNode | null, right?: TreeNode | null) {
*         this.val = (val===undefined ? 0 : val)
*         this.left = (left===undefined ? null : left)
*         this.right = (right===undefined ? null : right)
*     }
* }
*/

const dfs = (head: ListNode | null, root: TreeNode | null) => {
return true;
}
if (root == null || head.val !== root.val) {
return false;
}
};

function isSubPath(head: ListNode | null, root: TreeNode | null): boolean {
if (root == null) {
return false;
}
return (
);
}


• // Definition for singly-linked list.
// #[derive(PartialEq, Eq, Clone, Debug)]
// pub struct ListNode {
//   pub val: i32,
//   pub next: Option<Box<ListNode>>
// }
//
// impl ListNode {
//   #[inline]
//   fn new(val: i32) -> Self {
//     ListNode {
//       next: None,
//       val
//     }
//   }
// }
// Definition for a binary tree node.
// #[derive(Debug, PartialEq, Eq)]
// pub struct TreeNode {
//   pub val: i32,
//   pub left: Option<Rc<RefCell<TreeNode>>>,
//   pub right: Option<Rc<RefCell<TreeNode>>>,
// }
//
// impl TreeNode {
//   #[inline]
//   pub fn new(val: i32) -> Self {
//     TreeNode {
//       val,
//       left: None,
//       right: None
//     }
//   }
// }
use std::rc::Rc;
use std::cell::RefCell;
impl Solution {
fn dfs(head: &Option<Box<ListNode>>, root: &Option<Rc<RefCell<TreeNode>>>) -> bool {
return true;
}
if root.is_none() {
return false;
}
let root = root.as_ref().unwrap().borrow();
if node.val != root.val {
return false;
}
Self::dfs(&node.next, &root.left) || Self::dfs(&node.next, &root.right)
}

fn my_is_sub_path(head: &Option<Box<ListNode>>, root: &Option<Rc<RefCell<TreeNode>>>) -> bool {
if root.is_none() {
return false;
}
let node = root.as_ref().unwrap().borrow();