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. * type ListNode struct { * Val int * Next *ListNode * } *//** * Definition for a binary tree node. * type TreeNode struct { * Val int * Left *TreeNode * Right *TreeNode * } */funcisSubPath(head*ListNode,root*TreeNode)bool{ifroot==nil{returnfalse}returndfs(head,root)||isSubPath(head,root.Left)||isSubPath(head,root.Right)}funcdfs(head*ListNode,root*TreeNode)bool{ifhead==nil{returntrue}ifroot==nil||head.Val!=root.Val{returnfalse}returndfs(head.Next,root.Left)||dfs(head.Next,root.Right)}