• Time:O(n)
• Space:O(n)

## C++

``````class Solution {
public:
vector<int> sumOfDistancesInTree(int N, vector<vector<int>>& edges) {
vector<int> ans(N);
vector<int> count(N, 1);
vector<unordered_set<int>> tree(N);

for (const auto& e : edges) {
const int u = e[0];
const int v = e[1];
tree[u].insert(v);
tree[v].insert(u);
}

postorder(tree, 0, -1, count, ans);
preorder(tree, 0, -1, count, ans);
return ans;
}

private:
void postorder(const vector<unordered_set<int>>& tree, int node, int parent,
vector<int>& count, vector<int>& ans) {
for (const int child : tree[node]) {
if (child == parent)
continue;
postorder(tree, child, node, count, ans);
count[node] += count[child];
ans[node] += ans[child] + count[child];
}
}

void preorder(const vector<unordered_set<int>>& tree, int node, int parent,
vector<int>& count, vector<int>& ans) {
for (const int child : tree[node]) {
if (child == parent)
continue;
// count[child] nodes are 1 step closer from child than parent
// (N - count[child]) nodes are 1 step farther from child than parent
ans[child] = ans[node] - count[child] + (tree.size() - count[child]);
preorder(tree, child, node, count, ans);
}
}
};
``````

## JAVA

``````class Solution {
public int[] sumOfDistancesInTree(int N, int[][] edges) {
int[] ans = new int[N];
int[] count = new int[N];
Set<Integer>[] tree = new Set[N];

Arrays.fill(count, 1);

for (int i = 0; i < N; ++i)
tree[i] = new HashSet<>();

for (int[] e : edges) {
final int u = e[0];
final int v = e[1];
}

postorder(tree, 0, -1, count, ans);
preorder(tree, 0, -1, count, ans);
return ans;
}

private void postorder(Set<Integer>[] tree, int node, int parent, int[] count, int[] ans) {
for (final int child : tree[node]) {
if (child == parent)
continue;
postorder(tree, child, node, count, ans);
count[node] += count[child];
ans[node] += ans[child] + count[child];
}
}

private void preorder(Set<Integer>[] tree, int node, int parent, int[] count, int[] ans) {
for (final int child : tree[node]) {
if (child == parent)
continue;
// count[child] nodes are 1 step closer from child than parent
// (N - count[child]) nodes are 1 step farther from child than parent
ans[child] = ans[node] - count[child] + (tree.length - count[child]);
preorder(tree, child, node, count, ans);
}
}
}
``````

## Python

``````class Solution:
def sumOfDistancesInTree(self, N: int, edges: List[List[int]]) -> List[int]:
ans = [0] * N
count = [1] * N
tree = defaultdict(set)

for u, v in edges:

def postorder(node, parent=None):
for child in tree[node]:
if child == parent:
continue
postorder(child, node)
count[node] += count[child]
ans[node] += ans[child] + count[child]

def preorder(node, parent=None):
for child in tree[node]:
if child == parent:
continue
# count[child] nodes are 1 step closer from child than parent
# (N - count[child]) nodes are 1 step farther from child than parent
ans[child] = ans[node] - count[child] + (N - count[child])
preorder(child, node)

postorder(0)
preorder(0)
return ans
``````