LC.P834[树中距离之和]

方法一:换根DP

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class Solution {

int n;
List<Integer>[] g;
int[] ans, size;

public int[] sumOfDistancesInTree(int n, int[][] edges) {
this.n = n;
g = new ArrayList[n];
Arrays.setAll(g, e -> new ArrayList<>());
for (int[] edge : edges) {
int x = edge[0], y = edge[1];
g[x].add(y);
g[y].add(x);
}
ans = new int[n];
size = new int[n];
dfs(0, -1, 0);
reroot(0, -1);
return ans;
}

/**
* @param x 当前遍历到的节点
* @param pa 父节点
* @param depth 0~当前节点的深度
*/
private void dfs(int x, int pa, int depth) {
ans[0] += depth;
size[x] = 1;
for (int y : g[x]) {
if (y != pa) {
dfs(y, x, depth + 1);
size[x] += size[y];
}
}
}

private void reroot(int x, int pa) {
for (int y : g[x]) {
if (y != pa) {
ans[y] = ans[x] + n - 2 * size[y];
reroot(y, x);
}
}
}
}

  • 时间复杂度:$O(n)$
  • 空间复杂度:$O(n)$