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:

 

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 

思路:这题其实就是个DFS的思路，每次往上传递char array的信息；但是build graph的时候，题目有个变态的case，[[0,2],[0,3],[1,2]] ，也就是边的信息不是那么的明确，谁是父节点，或者儿子节点，要用depth信息来判断；这个很恶心；这个题目出的水平不是很高，考察点很low，但是test cases很变态，所以不适合拿来面试；

class Solution {
    private class Node {
        public int index;
        public Character c;
        public int depth = 0;
        public int[] chars;
        public Node(int index, Character c, int depth) {
            this.index = index;
            this.c = c;
            this.depth = depth;
            this.chars = new int[26];
        }
    }
    
    public int[] countSubTrees(int n, int[][] edges, String labels) {
        if(edges == null || edges.length == 0 || edges[0].length == 0) {
            return new int[0];
        }
        HashMap<Integer, Node> nodes = new HashMap<>();
        for(int i = 0; i < n; i++) {
            Node node = null;
            char c = labels.charAt(i);
            if(i == 0) {
                node = new Node(i, c, 1);
                node.chars[c - 'a'] = 1;
            } else {
                node = new Node(i, c, 0);
                node.chars[c - 'a'] = 1;
            }
            nodes.put(i, node);
        }
        HashMap<Integer, List<Node>> graph = buildGraph(n, edges, labels, nodes);
        dfs(0, graph, nodes);
        int[] res = new int[n];
        for(int i = 0; i < n; i++) {
            Node node = nodes.get(i);
            res[i] = node.chars[node.c - 'a'];
        }
        return res;
    }
    
    private Node dfs(int index, HashMap<Integer, List<Node>> graph, HashMap<Integer, Node> nodes) {
        Node node = nodes.get(index);
        for(Node neighbor: graph.get(index)) {
            Node n = dfs(neighbor.index, graph, nodes);
            for(int i = 0; i < 26; i++) {
                node.chars[i] += n.chars[i];
            }
        }
        return node;
    }
    
    private HashMap<Integer, List<Node>> buildGraph(int n, int[][] edges, String labels, 
                                                    HashMap<Integer, Node> nodes ) {
        HashMap<Integer, List<Node>> graph = new HashMap<>();
        for(int i = 0; i < n; i++) {
            graph.put(i, new ArrayList<Node>());
        }
        for(int i = 0; i < edges.length; i++) {
            int a = edges[i][0];
            int b = edges[i][1];
            // a -> b;
            if(nodes.get(a).depth > 0) {
                nodes.get(b).depth = nodes.get(a).depth + 1;
                graph.get(a).add(nodes.get(b));
            } else {
                nodes.get(a).depth = nodes.get(b).depth + 1;
                graph.get(b).add(nodes.get(a));
            }
        }
        return graph;
    }
}