834. Sum of Distances in Tree —— weekly contest 84

Posted 2020-11-05 jinjin-2018

Sum of Distances in Tree

An undirected, connected tree with N nodes labelled 0...N-1 and N-1 edges are given.

The ith edge connects nodes edges[i][0] and edges[i][1] together.

Return a list ans, where ans[i] is the sum of the distances between node i and all other nodes.

Example 1:

Input: N = 6, edges = [[0,1],[0,2],[2,3],[2,4],[2,5]]
Output: [8,12,6,10,10,10]
Explanation: 
Here is a diagram of the given tree:
  0
 / 1  2
    /|  3 4 5
We can see that dist(0,1) + dist(0,2) + dist(0,3) + dist(0,4) + dist(0,5)
equals 1 + 1 + 2 + 2 + 2 = 8.  Hence, answer[0] = 8, and so on.

Note: 1 <= N <= 10000

 1 class Solution {
 2 public:
 3     //相当于图来存来处理
 4     vector<vector<int>> tree;
 5     vector<int> res;
 6     vector<int> subc;
 7     int n;
 8     vector<int> sumOfDistancesInTree(int N, vector<vector<int>>& edges) {
 9         //tree.rvec(N,vector<int>(N));
10         //init
11         n = N;
12         tree.resize(N);            //初始化的函数
13         res.assign(N,0);
14         subc.assign(N,0);
15         //build tree
16         for(auto e : edges){               //遍历的技巧
17             tree[e[0]].push_back(e[1]);
18             tree[e[1]].push_back(e[0]);
19         }
20         set<int> visited1;
21         set<int> visited2;
22         DFS_POST(0,visited1);         //初始root任何值都行
23         DFS_PRE(0,visited2);
24         return res;
25         
26     }
27     void DFS_POST(int root,set<int> &visited){            //传引用保存修改值
28         visited.insert(root);
29         for(auto i : tree[root]){
30             if(visited.find(i) == visited.end() ){
31                 DFS_POST(i,visited);
32                 subc[root] += subc[i];
33                 res[root] += res[i] + subc[i];    
34             }
35         }
36         subc[root]++;  //加上自身节点
37     }
38     void DFS_PRE(int root,set<int> &visited){
39         visited.insert(root);
40         for(auto i : tree[root]){
41             if(visited.find(i) == visited.end()){
42                 res[i] = res[root] - subc[i] + n - subc[i];         //算法核心
43                 DFS_PRE(i,visited);
44             }
45         }
46     }
47     
48 };

主要函数是初始化的函数。

主要算法思想是先序和后续的递归遍历（DFS）。

实现O（n²）的算法核心方程是：res[i] = res[root] - subc[i] + n - subc[i];