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:
Linked List, Dynamic Programming, Tree

Solution 1.

/**
 * 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);
    }
}

// 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);
    }
};

# 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)
        )

/**
 * 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)
}

/**
 * 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)
    );
}

// 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)
    }
}

1367 - Linked List in Binary Tree

1367. Linked List in Binary Tree (Medium)

Solution 1.

All Problems

All Solutions

Welcome to Subscribe On Youtube

1367. Linked List in Binary Tree (Medium)

Solution 1.

All Problems

All Solutions