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:
y
of x
in increasing order of their numbers, and call dfs(y)
.s[x]
to the end of the string dfsStr
.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:
dfsStr
and call dfs(i)
.dfsStr
is a palindrome, then set answer[i]
to true
. Otherwise, set answer[i]
to false
.Return the array answer
.
Example 1:
Input: parent = [-1,0,0,1,1,2], s = "aababa"
Output: [true,true,false,true,true,true]
Explanation:
dfs(0)
results in the string dfsStr = "abaaba"
, which is a palindrome.dfs(1)
results in the string dfsStr = "aba"
, which is a palindrome.dfs(2)
results in the string dfsStr = "ab"
, which is not a palindrome.dfs(3)
results in the string dfsStr = "a"
, which is a palindrome.dfs(4)
results in the string dfsStr = "b"
, which is a palindrome.dfs(5)
results in the string dfsStr = "a"
, which is a palindrome.Example 2:
Input: parent = [-1,0,0,0,0], s = "aabcb"
Output: [true,true,true,true,true]
Explanation:
Every call on dfs(x)
results in a palindrome string.
Constraints:
n == parent.length == s.length
1 <= n <= 105
0 <= parent[i] <= n - 1
for all i >= 1
.parent[0] == -1
parent
represents a valid tree.s
consists only of lowercase English letters.