Question

1367. Linked List in Binary Tree

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Description

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:

• The number of nodes in the tree will be in the range [1, 2500].
• The number of nodes in the list will be in the range [1, 100].
• 1 <= Node.val <= 100 for each node in the linked list and binary tree.

Solutions

Solution 1: Recursion

We design a recursive function $dfs(head, root)$, which indicates whether the linked list $head$ corresponds to a subpath on the path starting with $root$ in the binary tree. The logic of the function $dfs(head, root)$ is as follows:

• If the linked list $head$ is empty, it means that the linked list has been traversed, return true;
• If the binary tree $root$ is empty, it means that the binary tree has been traversed, but the linked list has not been traversed yet, return false;
• If the value of the binary tree $root$ is not equal to the value of the linked list $head$, return false;
• Otherwise, return $dfs(head.next, root.left)$ or $dfs(head.next, root.right)$.

In the main function, we call $dfs(head, root)$ for each node of the binary tree. As long as one returns true, it means that the linked list is a subpath of the binary tree, return true; if all nodes return false, it means that the linked list is not a subpath of the binary tree, return false.

The time complexity is $O(n^2)$, and the space complexity is $O(n)$. Where $n$ is the number of nodes in the 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)
    )

/**
 * 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);
  }
}

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *   int val;
 *   ListNode *next;
 *   ListNode() : val(0), next(nullptr) {}
 *   ListNode(int x) : val(x), next(nullptr) {}
 *   ListNode(int x, ListNode *next) : val(x), next(next) {}
 * };
 */
/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *   int val;
 *   TreeNode *left;
 *   TreeNode *right;
 *   TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *   TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *   TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
class Solution {
public:
  bool isSubPath(ListNode* head, TreeNode* root) {
    if (!root) {
      return false;
    }
    return dfs(head, root) || isSubPath(head, root->left) || isSubPath(head, root->right);
  }

  bool dfs(ListNode* head, TreeNode* root) {
    if (!head) {
      return true;
    }
    if (!root || head->val != root->val) {
      return false;
    }
    return dfs(head->next, root->left) || dfs(head->next, 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)
  }
}