Losing head/tail while rewiring
Wrong move: Pointer updates overwrite references before they are saved.
Usually fails on: List becomes disconnected mid-operation.
Fix: Store next pointers first and use a dummy head for safer joins.
Move from brute-force thinking to an efficient approach using linked list strategy.
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, 2500].[1, 100].1 <= Node.val <= 100 for each node in the linked list and binary tree.Problem summary: 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.
Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.
Pattern signal: Linked List · Tree
[4,2,8] [1,4,4,null,2,2,null,1,null,6,8,null,null,null,null,1,3]
[1,4,2,6] [1,4,4,null,2,2,null,1,null,6,8,null,null,null,null,1,3]
[1,4,2,6,8] [1,4,4,null,2,2,null,1,null,6,8,null,null,null,null,1,3]
Source-backed implementations are provided below for direct study and interview prep.
// Accepted solution for LeetCode #1367: Linked List in Binary Tree
/**
* 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);
}
}
// Accepted solution for LeetCode #1367: Linked List in Binary Tree
/**
* 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)
}
# Accepted solution for LeetCode #1367: Linked List in Binary Tree
# 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: Optional[ListNode], root: Optional[TreeNode]) -> bool:
def dfs(head, root):
if head is None:
return True
if root is None or 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)
)
// Accepted solution for LeetCode #1367: Linked List in Binary Tree
// 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::cell::RefCell;
use std::rc::Rc;
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)
}
}
// Accepted solution for LeetCode #1367: Linked List in Binary Tree
/**
* 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);
}
Use this to step through a reusable interview workflow for this problem.
Copy all n nodes into an array (O(n) time and space), then use array indexing for random access. Operations like reversal or middle-finding become trivial with indices, but the O(n) extra space defeats the purpose of using a linked list.
Most linked list operations traverse the list once (O(n)) and re-wire pointers in-place (O(1) extra space). The brute force often copies nodes to an array to enable random access, costing O(n) space. In-place pointer manipulation eliminates that.
Review these before coding to avoid predictable interview regressions.
Wrong move: Pointer updates overwrite references before they are saved.
Usually fails on: List becomes disconnected mid-operation.
Fix: Store next pointers first and use a dummy head for safer joins.
Wrong move: Recursive traversal assumes children always exist.
Usually fails on: Leaf nodes throw errors or create wrong depth/path values.
Fix: Handle null/base cases before recursive transitions.