# 1519. Number of Nodes in the Sub-Tree With the Same Label
###### tags: `leetcode`
## Description
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
- This is a DFS problem that should accumulate the count from the leaf to the root
- Because the given information about the tree is the `edges`, transform them into a adjancent list

adjList.resize(n);
for (auto &i : edges)
{
    adjList[i[0]].push_back(i[1]);
    adjList[i[1]].push_back(i[0]);
}

- Then for the `DFS`, calculate the current count for all the alphabet and let the one that finish iteration with the answer of the current accumulated result of the alphabet the index has

for (auto &i : adjList[idx])
{
    if (i != prev)
    {
        temp = dfs(i, idx, edges, ans, labels);
        transform(cur.begin(), cur.end(), temp.begin(), cur.begin(), plus<int>());
    }

}
cur[labels[idx] - 'a']++;
ans[idx] = cur[labels[idx] - 'a'];

- When passing the vector and the string, remember to pass them with the reference in order to speed up
- In this case, I did not use reference with string and got TLE several time