【leetcode】834. Sum of Distances in Tree（图算法）

　　There is an undirected connected tree with n nodes labeled from 0 to n - 1 and n - 1 edges. You are given the integer n and the array edges where edges[i] = [ai, bi] indicates that there is an edge between nodes ai and bi in the tree. Return an array answer of length n where answer[i] is the sum of the distances between the ith node in the tree 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: The tree is shown above. 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.

class Solution {

public:

    vector<int> sumOfDistancesInTree(int n, vector<vector<int>>& edges) {

        //好久没做和图论相关的题目了 图论的题目 dfs bfs？

        //链接图链表

        vector<int> res(n),count(n);

        vector<vector<int>> tree(n);

        for(auto &edge:edges)

        {

            tree[edge[0]].push_back(edge[1]);

            tree[edge[1]].push_back(edge[0]);

        }

        helper(tree,0,-1,count,res);

        helper2(tree,0,-1,count,res);

        return res;

    }

    void helper(vector<vector<int>> &tree,int cur,int pre,vector<int>&count,vector<int>& res)

    {

        for(int i:tree[cur])

        {

            if(i==pre) continue;

            helper(tree,i,cur,count,res);

            count[cur]+=count[i];//这个统计更新的逻辑 还是不懂。。

            res[cur]+=res[i]+count[i];

        }

        ++count[cur];

    }

    void helper2(vector<vector<int>> &tree,int cur,int pre,vector<int>&count,vector<int>& res)

    {

        for(int i:tree[cur])

        {

            if(i==pre) continue;

            res[i]=res[cur]-count[i]+count.size()-count[i];

            helper2(tree, i, cur, count, res);

        }

    }

};

巴特西

【leetcode】834. Sum of Distances in Tree（图算法）

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