Given a tree (i.e. a connected, undirected graph that has no cycles) consisting ofnnodes numbered from0ton - 1and exactlyn - 1edges. Therootof the tree is the node0, and each node of the tree hasa labelwhich is a lower-case character given in the stringlabels(i.e. The node with the numberihas the labellabels[i]

Theedgesarray is given on the formedges[i] = [ai, bi], which means there is an edge between nodesaiandbiin the tree.

Returnan array of sizenwhereans[i]is the number of nodes in the subtree of theithnode which have the same label as nodei.

Asubtreeof a treeTis the tree consisting of a node inTand all of its descendantnodes.

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]

Example 4:

Input: n = 6, edges = [[0,1],[0,2],[1,3],[3,4],[4,5]], labels = "cbabaa"
Output: [1,2,1,1,2,1]

Example 5:

Input: n = 7, edges = [[0,1],[1,2],[2,3],[3,4],[4,5],[5,6]], labels = "aaabaaa"
Output: [6,5,4,1,3,2,1]

Constraints:

• 1 <= n <= 10^5
• edges.length == n - 1
• edges[i].length == 2
• 0 <= ai,bi< n
• ai!=bi
• labels.length == n
• labelsis consisting of only of lower-case English letters.

class Solution {
    public int[] countSubTrees(int n, int[][] edges, String labels) {
        int[] ans = new int[n];
        Set<Integer> used = new HashSet();
        Map<Integer, List<Integer>> map = new HashMap();
        for(int[] edge: edges) {
            map.computeIfAbsent(edge[0], l -> new ArrayList()).add(edge[1]);
            map.computeIfAbsent(edge[1], l -> new ArrayList()).add(edge[0]);
        }
        dfs(map, ans, labels, used, 0);
        return ans;
    }
    
    public int[] dfs(Map<Integer, List<Integer>> map, int[] ans, String labels, Set<Integer> used, int root) {
        int[] count = new int[26];
        if(used.add(root)) {
            char c = labels.charAt(root);
            for(int child: map.get(root)) {
                int[] cur = dfs(map, ans, labels, used, child);
                for(int i = 0; i < 26; i++) {
                    count[i] += cur[i];
                }
            }
            count[c - 'a']++;
            ans[root] = count[c-'a'];
        }
        return count;
    }
}

坑逼

用DFS的方法做，

因為這個edges裡面會出現edge[0]是child，反而edge[1]是parent的奇葩情況，所以我們只能先把樹做成圖（雙向連線），此外我們還需要一個used hashset來存放已經遍歷過的root

既然只有26個字母，我們就計算每一個subtree所包含的字母，把它們存到array中，這使我們要用dfs往更深層遞迴。

然後要把每個根的child上的字母數加起來，才等於我們要求的。