• 欢迎光临~

# tree reconstruct 834

834. Sum of Distances in Tree
Hard

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.
```

Example 2:

```Input: n = 1, edges = []
Output: [0]
```

Example 3:

```Input: n = 2, edges = [[1,0]]
Output: [1,1]
```

Constraints:

• `1 <= n <= 3 * 104`
• `edges.length == n - 1`
• `edges[i].length == 2`
• `0 <= ai, bi < n`
• `ai != bi`
• The given input represents a valid tree.
```class Solution {
public int[] sumOfDistancesInTree(int n, int[][] edges) {
//0.建图，这么做可以加速每个节点访问其子节点的速度
List<Integer>[] graph = new List[n];
for(int i = 0; i < n; i++) graph[i] = new ArrayList();
for(int[] edge : edges) {
int a = edge[0], b = edge[1];
}
//将0作为root节点，计算所有子树的size
//1.define a root node  0
//2.dfs calculate all the subtree size
int[] size = new int[n];
subtreeSize(0, graph, -1, size);
//计算root节点到所有节点的距离和
//3.dfs calculate the root node's distance to all the other nodes
int[] result = new int[n];
result[0] = rootDistance(0, graph, -1, size);
//最后递归根据父节点计算每个节点到其他点的距离和
//4.dfs calculate all the node's distance to the other nodes base on the formula
calculateSubTree(0, graph, -1, size, result);
return result;
}
//计算子书size
private int subtreeSize(int curr, List<Integer>[] graph, int parent, int[] size){
int sum = 0;
for(int child : graph[curr]){
if(child == parent) continue;
sum += subtreeSize(child, graph, curr, size);
}
size[curr] = sum + 1;
return size[curr];
}
//计算子树距离和
private int rootDistance(int curr, List<Integer>[] graph, int parent, int[] size){
int result = 0;
for(int child : graph[curr]){
if(child == parent) continue;
//离所有child的距离，是child的所有距离和 + 1 + child的所有child个数（因为每增加一层，你离所有孙子的距离都得加一层）
result += rootDistance(child, graph, curr, size) + size[child];
}
return result;
}
//基于父节点计算子树距离和
private void calculateSubTree(int curr, List<Integer>[] graph, int parent, int[] size, int[] result){
if(parent != -1) result[curr] = result[parent] + graph.length - 2 * size[curr];
for(int child : graph[curr]){
if(child == parent) continue;
calculateSubTree(child, graph, curr, size, result);
}
}
}```