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#834 - Sum of Distances in Tree
Problem Description
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.
Solution
/**
* @param {number} n
* @param {number[][]} edges
* @return {number[]}
*/
var sumOfDistancesInTree = function(n, edges) {
const graph = Array.from({ length: n }, () => []);
const count = new Array(n).fill(1);
const result = new Array(n).fill(0);
for (const [a, b] of edges) {
graph[a].push(b);
graph[b].push(a);
}
function dfs1(node, parent) {
for (const child of graph[node]) {
if (child !== parent) {
dfs1(child, node);
count[node] += count[child];
result[node] += result[child] + count[child];
}
}
}
function dfs2(node, parent) {
for (const child of graph[node]) {
if (child !== parent) {
result[child] = result[node] - count[child] + (n - count[child]);
dfs2(child, node);
}
}
}
dfs1(0, -1);
dfs2(0, -1);
return result;
};