> For the complete documentation index, see [llms.txt](https://hao-fu-1.gitbook.io/oj/llms.txt). Markdown versions of documentation pages are available by appending `.md` to page URLs; this page is available as [Markdown](https://hao-fu-1.gitbook.io/oj/binary_tree_and_divide_conquer/1519.-number-of-nodes-in-the-sub-tree-with-the-same-label.md).

# 1519. Number of Nodes in the Sub-Tree With the Same Label

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:**![](https://assets.leetcode.com/uploads/2020/07/01/q3e1.jpg)

```
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:**![](https://assets.leetcode.com/uploads/2020/07/01/q3e2.jpg)

```
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:**![](https://assets.leetcode.com/uploads/2020/07/01/q3e3.jpg)

```
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`
* `labels` is consisting of only of lower-case English letters.

对树上每个结点统计它和它子树上与它的值相同的结点数目，值范围为\[a, z]。树非按层统计结点=>DFS/分治。假设每个子节点对应的树对每个值在该子树的频次已经统计完毕，将频率汇总并对该结点的值频次加一，返回。

```python
class Solution:
    def countSubTrees(self, n: int, edges: List[List[int]], labels: str) -> List[int]:
        g, res = collections.defaultdict(list), [0] * n
        for (a, b) in edges:
            g[a].append(b)
            g[b].append(a)
        def dfs(i, last, g, labels, res):
            cnt = collections.Counter()
            for nei in g[i]:
                if nei == last: continue
                cnt += dfs(nei, i, g, labels, res)
            cnt[labels[i]] += 1
            res[i] = cnt[labels[i]]
            return cnt

        dfs(0, -1, g, labels, res)
        return res
            
```