Problem Statement:

You are given a tree (i.e. a connected, undirected graph that has no cycles) consisting of n nodes numbered from 0 to n - 1 and exactly n - 1 edges. The root of the tree is the node 0, and each node of the tree has a label which is a lower-case character given in the string labels (i.e. The node with the number i has the label labels[i]).

The edges array is given on the form edges[i] = [ai, bi], which means there is an edge between nodes ai and bi in the tree.

Return an array of size n where ans[i] is the number of nodes in the subtree of the ith node which have the same label as node i.

A subtree of a tree T is the tree consisting of a node in T and all of its descendant nodes.

Example 1:
Input: n = 7, edges = [[0,1],[0,2],[1,4],[1,5],[2,3],[2,6]], labels = "abaedcd"
Output: [2,1,1,1,1,1,1]
Explanation: Node 0 has label 'a' and its sub-tree has node 2 with label 'a' as well, thus the answer is 2. Notice that any node is part of its sub-tree.
Node 1 has a label 'b'. The sub-tree of node 1 contains nodes 1,4 and 5, as nodes 4 and 5 have different labels than node 1, the answer is just 1 (the node itself).

Example 2:
Input: n = 4, edges = [[0,1],[1,2],[0,3]], labels = "bbbb"
Output: [4,2,1,1]
Explanation: The sub-tree of node 2 contains only node 2, so the answer is 1.
The sub-tree of node 3 contains only node 3, so the answer is 1.
The sub-tree of node 1 contains nodes 1 and 2, both have label 'b', thus the answer is 2.
The sub-tree of node 0 contains nodes 0, 1, 2 and 3, all with label 'b', thus the answer is 4.

Example 3:
Input: n = 5, edges = [[0,1],[0,2],[1,3],[0,4]], labels = "aabab"
Output: [3,2,1,1,1]

Constraints:

• 1 <= n <= 105
• edges.length == n - 1
• edges[i].length == 2
• 0 <= ai, bi < n
• ai != bi
• labels.length == n
• labels is consisting of only of lowercase English letters.

Solution:
Code:

class Solution {
    private int[] ans;
    private List<List<Integer>> adjlist;
    private Set<Integer> visited;
    public int[] countSubTrees(int n, int[][] edges, String labels) {
        ans= new int[n];
        adjlist=new ArrayList<>(n);
        for(int i=0;i<n;i++){
            adjlist.add(new ArrayList<>());
        }
        for(int[] e: edges){
            adjlist.get(e[0]).add(e[1]);
            adjlist.get(e[1]).add(e[0]);
        }
        visited=new HashSet<Integer>(n);
        dfs(0,labels);
        return ans;

    }
    private int[] dfs(int node, String labels){
        visited.add(node);
        int[] count=new int[26];
        for(int adj:adjlist.get(node)){
            if(!visited.contains(adj)){
                int[] adjcount=dfs(adj,labels);
                for(int i=0;i<26;i++){
                    count[i]+=adjcount[i];
                }
            }
        }
        char ch=labels.charAt(node);
        count[ch-'a']++;
        ans[node]=count[ch-'a'];
        return count;

    }
}