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

  • Hardness: \(\color{orange}\textsf{Medium}\)
  • Ralated Topics: Hash Table、Tree、Depth-First Search、Breadth-First Search、Counting

一、題目

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:
q3e1

  • 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: q3e2

  • 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: q3e3

  • Input: n = 5, edges = [[0,1],[0,2],[1,3],[0,4]], labels = “aabab”
  • Output: [3,2,1,1,1]

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 lowercase English letters.

二、分析

  • 這一題是後序遍歷，利用後序遍歷可以得到遍歷過後的訊息來解題。
  • 我們要的訊息是：當遍歷完子節點之後，每個字元已出現了幾遍，而根節點以下的所有節點出現過的字元次數，就是所有子節點統計數字的相加再加上根節點自己。
  • 這邊要注意的是，因為每次深入子節點，都需要新增一組陣列，會大量使用到記憶體，所以在處理上要小心空間複雜度的處理。技巧是，能用參考就用參考。
  • 由於這題的 n 的限制較大，只能運用陣列的加法，時間複雜度為\(O(26)\)，若場景有確認過，每個字符出現的次數限定在 10 以內，可以改用 int 來記錄，其加法的時間複雜度可以大大的下降。

三、解題

1. DFS

  • Time complexity: \(O(26\times n)\)
  • Space complexity: \(O(26\times k), k 為深度\)
vector<int> res;
vector<int> countSubTrees(int n, vector<vector<int>>& edges, string labels) {
    vector<vector<int>> graph(n);       // 先將 undirected graph 轉成每個節點有哪些鄰居
    for (const auto& e : edges) {
        graph[e[0]].push_back(e[1]);
        graph[e[1]].push_back(e[0]);
    }
    res.assign(n, 0);
    dfs(graph, labels, -1, 0);          // 起點為 0，而節點不為負，過 last 可假定為任意負數
    return res;
}
vector<int> dfs(vector<vector<int>>& graph, string& labels, int last, int curr) {
    vector<int> path(26, 0);            // 用一陣列記錄字符出現的次數
    path[labels[curr]-'a']++;
    for (const auto& next : graph[curr]) {
        if (last == next) continue;     // 進到上一輪的數字則跳過
        vector<int> tmp = dfs(graph, labels, curr, next);
        add(path, tmp);                 // 將遍歷完的結果加起來
    }
    res[curr] = path[labels[curr]-'a']; // 在後序的時間點，把統計完的結果記錄下來
    return path;
}
void add(vector<int>& a, vector<int>& b) {  // 用參考的方法做陣列的加法，也不回傳，可以省下空間
    for (int i = 0; i < 26; i++) 
        a[i] += b[i];
}

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