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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
and100
nodes. - The given binary tree will contain between
1
and2500
nodes.
Related Topics:
Linked List, Dynamic Programming, Tree
Solution 1.
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/** * 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(); } } ############ /** * Definition for singly-linked list. * 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; } return dfs(head, root) || isSubPath(head, root.left) || isSubPath(head, root.right); } private boolean dfs(ListNode head, TreeNode root) { if (head == null) { return true; } if (root == null || head.val != root.val) { return false; } return dfs(head.next, root.left) || dfs(head.next, root.right); } }
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// 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 (!head) return true; if (!root) return false; if (head->val != root->val) return false; return match(head->next, root->left) || match(head->next, root->right); } public: bool isSubPath(ListNode* head, TreeNode* root) { if (!head) return true; if (!root) return false; if (match(head, root)) return true; return isSubPath(head, root->left) || isSubPath(head, root->right); } };
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# 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: def dfs(head, root): if head is None: return True if root is None: return False if root.val != head.val: return False return dfs(head.next, root.left) or dfs(head.next, root.right) if root is None: return False return ( dfs(head, root) or self.isSubPath(head, root.left) or self.isSubPath(head, root.right) )
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/** * Definition for singly-linked list. * 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 } return dfs(head, root) || isSubPath(head, root.Left) || isSubPath(head, root.Right) } func dfs(head *ListNode, root *TreeNode) bool { if head == nil { return true } if root == nil || head.Val != root.Val { return false } return dfs(head.Next, root.Left) || dfs(head.Next, root.Right) }
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/** * Definition for singly-linked list. * 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) => { if (head == null) { return true; } if (root == null || head.val !== root.val) { return false; } return dfs(head.next, root.left) || dfs(head.next, root.right); }; function isSubPath(head: ListNode | null, root: TreeNode | null): boolean { if (root == null) { return false; } return ( dfs(head, root) || isSubPath(head, root.left) || isSubPath(head, root.right) ); }
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// 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 { if head.is_none() { return true; } if root.is_none() { return false; } let node = head.as_ref().unwrap(); 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(); Self::dfs(head, root) || Self::my_is_sub_path(head, &node.left) || Self::my_is_sub_path(head, &node.right) } pub fn is_sub_path(head: Option<Box<ListNode>>, root: Option<Rc<RefCell<TreeNode>>>) -> bool { Self::my_is_sub_path(&head, &root) } }