Check if DFS Strings Are Palindromes
You are given a tree rooted at node 0, consisting of n nodes numbered from 0 to n - 1. The tree is represented by an array parent of size n, where parent[i] is the parent of node i. Since node 0 is the root, parent[0] == -1.
You are also given a string s of length n, where s[i] is the character assigned to node i.
Consider an empty string dfsStr, and define a recursive function dfs(int x) that takes a node x as a parameter and performs the following steps in order:
- Iterate over each child
yofxin increasing order of their numbers, and calldfs(y). - Add the character
s[x]to the end of the stringdfsStr.
Note that dfsStr is shared across all recursive calls of dfs.
You need to find a boolean array answer of size n, where for each index i from 0 to n - 1, you do the following:
- Empty the string
dfsStrand calldfs(i). - If the resulting string
dfsStris a palindrome, then setanswer[i]totrue. Otherwise, setanswer[i]tofalse.
Return the array answer.
Example 1
Input
parent = [-1,0,0,1,1,2], s = "aababa"Output
[true,true,false,true,true,true]Calling
dfs(0) and dfs(1) produces palindromes, dfs(2) produces "ab" which is not a palindrome, and each leaf produces a one-character palindrome.Example 2
Input
parent = [-1,0,0,0,0], s = "aabcb"Output
[true,true,true,true,true]Every call on
dfs(x) results in a palindrome string.Constraints
- n == parent.length == s.length
- 1 <= n <= 10^5
- 0 <= parent[i] <= n - 1 for all i >= 1.
- parent[0] == -1
- parent represents a valid tree.
- s consists only of lowercase English letters.