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3 changes: 2 additions & 1 deletion Contents/00.Introduction/04.Solutions-List.md
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# LeetCode 题解(已完成 798 道)
# LeetCode 题解(已完成 799 道)

| 题号 | 标题 | 题解 | 标签 | 难度 |
| :------ | :------ | :------ | :------ | :------ |
Expand Down Expand Up @@ -429,6 +429,7 @@
| 0820 | [单词的压缩编码](https://leetcode.cn/problems/short-encoding-of-words/) | [Python](https://github.com/itcharge/LeetCode-Py/blob/main/Solutions/0820.%20%E5%8D%95%E8%AF%8D%E7%9A%84%E5%8E%8B%E7%BC%A9%E7%BC%96%E7%A0%81.md) | 字典树、数组、哈希表、字符串 | 中等 |
| 0830 | [较大分组的位置](https://leetcode.cn/problems/positions-of-large-groups/) | [Python](https://github.com/itcharge/LeetCode-Py/blob/main/Solutions/0830.%20%E8%BE%83%E5%A4%A7%E5%88%86%E7%BB%84%E7%9A%84%E4%BD%8D%E7%BD%AE.md) | 字符串 | 简单 |
| 0832 | [翻转图像](https://leetcode.cn/problems/flipping-an-image/) | [Python](https://github.com/itcharge/LeetCode-Py/blob/main/Solutions/0832.%20%E7%BF%BB%E8%BD%AC%E5%9B%BE%E5%83%8F.md) | 数组、双指针、矩阵、模拟 | 简单 |
| 0834 | [树中距离之和](https://leetcode.cn/problems/sum-of-distances-in-tree/) | [Python](https://github.com/itcharge/LeetCode-Py/blob/main/Solutions/0834.%20%E6%A0%91%E4%B8%AD%E8%B7%9D%E7%A6%BB%E4%B9%8B%E5%92%8C.md) | 树、深度优先搜索、图、动态规划 | 困难 |
| 0836 | [矩形重叠](https://leetcode.cn/problems/rectangle-overlap/) | [Python](https://github.com/itcharge/LeetCode-Py/blob/main/Solutions/0836.%20%E7%9F%A9%E5%BD%A2%E9%87%8D%E5%8F%A0.md) | 几何、数学 | 简单 |
| 0841 | [钥匙和房间](https://leetcode.cn/problems/keys-and-rooms/) | [Python](https://github.com/itcharge/LeetCode-Py/blob/main/Solutions/0841.%20%E9%92%A5%E5%8C%99%E5%92%8C%E6%88%BF%E9%97%B4.md) | 深度优先搜索、广度优先搜索、图 | 中等 |
| 0844 | [比较含退格的字符串](https://leetcode.cn/problems/backspace-string-compare/) | [Python](https://github.com/itcharge/LeetCode-Py/blob/main/Solutions/0844.%20%E6%AF%94%E8%BE%83%E5%90%AB%E9%80%80%E6%A0%BC%E7%9A%84%E5%AD%97%E7%AC%A6%E4%B8%B2.md) | 栈、双指针、字符串、模拟 | 简单 |
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| 题号 | 标题 | 题解 | 标签 | 难度 |
| :------ | :------ | :------ | :------ | :------ |
| 0310 | [最小高度树](https://leetcode.cn/problems/minimum-height-trees/) | [Python](https://github.com/itcharge/LeetCode-Py/blob/main/Solutions/0310.%20%E6%9C%80%E5%B0%8F%E9%AB%98%E5%BA%A6%E6%A0%91.md) | 深度优先搜索、广度优先搜索、图、拓扑排序 | 中等 |
| 0834 | [树中距离之和](https://leetcode.cn/problems/sum-of-distances-in-tree/) | | 树、深度优先搜索、图、动态规划 | 困难 |
| 0834 | [树中距离之和](https://leetcode.cn/problems/sum-of-distances-in-tree/) | [Python](https://github.com/itcharge/LeetCode-Py/blob/main/Solutions/0834.%20%E6%A0%91%E4%B8%AD%E8%B7%9D%E7%A6%BB%E4%B9%8B%E5%92%8C.md) | 树、深度优先搜索、图、动态规划 | 困难 |
| 2581 | [统计可能的树根数目](https://leetcode.cn/problems/count-number-of-possible-root-nodes/) | | 树、深度优先搜索、哈希表、动态规划 | 困难 |

### 状态压缩 DP 题目
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| 题号 | 标题 | 题解 | 标签 | 难度 |
| :------ | :------ | :------ | :------ | :------ |
| 0310 | [最小高度树](https://leetcode.cn/problems/minimum-height-trees/) | [Python](https://github.com/itcharge/LeetCode-Py/blob/main/Solutions/0310.%20%E6%9C%80%E5%B0%8F%E9%AB%98%E5%BA%A6%E6%A0%91.md) | 深度优先搜索、广度优先搜索、图、拓扑排序 | 中等 |
| 0834 | [树中距离之和](https://leetcode.cn/problems/sum-of-distances-in-tree/) | | 树、深度优先搜索、图、动态规划 | 困难 |
| 0834 | [树中距离之和](https://leetcode.cn/problems/sum-of-distances-in-tree/) | [Python](https://github.com/itcharge/LeetCode-Py/blob/main/Solutions/0834.%20%E6%A0%91%E4%B8%AD%E8%B7%9D%E7%A6%BB%E4%B9%8B%E5%92%8C.md) | 树、深度优先搜索、图、动态规划 | 困难 |
| 2581 | [统计可能的树根数目](https://leetcode.cn/problems/count-number-of-possible-root-nodes/) | | 树、深度优先搜索、哈希表、动态规划 | 困难 |

2 changes: 1 addition & 1 deletion README.md
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- [动态规划优化题目](./Contents/10.Dynamic-Programming/11.DP-Optimization/04.DP-Optimization-List.md)

## 11. 附加内容
## [12. LeetCode 题解(已完成 798 道)](./Contents/00.Introduction/04.Solutions-List.md)
## [12. LeetCode 题解(已完成 799 道)](./Contents/00.Introduction/04.Solutions-List.md)
108 changes: 108 additions & 0 deletions Solutions/0834. 树中距离之和.md
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# [0834. 树中距离之和](https://leetcode.cn/problems/sum-of-distances-in-tree/)

- 标签:树、深度优先搜索、图、动态规划
- 难度:困难

## 题目大意

**描述**:给定一个无向、连通的树。树中有 $n$ 个标记为 0ドル \sim n - 1$ 的节点以及 $n - 1$ 条边 。

给定整数 $n$ 和数组 $edges,ドル其中 $edges[i] = [ai, bi]$ 表示树中的节点 $ai$ 和 $bi$ 之间有一条边。

**要求**:返回长度为 $n$ 的数组 $answer,ドル其中 $answer[i]$ 是树中第 $i$ 个节点与所有其他节点之间的距离之和。

**说明**:

- 1ドル \le n \le 3 \times 10^4$。
- $edges.length == n - 1$。
- $edges[i].length == 2$。
- 0ドル \le ai, bi < n$。
- $ai \ne bi$。
- 给定的输入保证为有效的树。

**示例**:

- 示例 1:

![](https://assets.leetcode.com/uploads/2021/07/23/lc-sumdist1.jpg)

```Python
输入: n = 6, edges = [[0,1],[0,2],[2,3],[2,4],[2,5]]
输出: [8,12,6,10,10,10]
解释: 树如图所示。
我们可以计算出 dist(0,1) + dist(0,2) + dist(0,3) + dist(0,4) + dist(0,5)
也就是 1 +たす 1 +たす 2 +たす 2 +たす 2 =わ 8。 因此,answer[0] = 8,以此类推。
```

- 示例 2:

![](https://assets.leetcode.com/uploads/2021/07/23/lc-sumdist3.jpg)

```Python
输入: n = 2, edges = [[1,0]]
输出: [1,1]
```

## 解题思路

### 思路 1:树形 DP + 二次遍历换根法

最容易想到的做法是:枚举 $n$ 个节点,以每个节点为根节点进行树形 DP。

对于节点 $u,ドル定义 $dp[u]$ 为:以节点 $u$ 为根节点的树,它的所有子节点到它的距离之和。

然后进行一轮深度优先搜索,在搜索的过程中得到以节点 $v$ 为根节点的树,节点 $v$ 与所有其他子节点之间的距离之和 $dp[v]$。还能得到子树的节点个数 $sizes[v]$。

对于节点 $v$ 来说,其对 $dp[u]$ 的贡献为:节点 $v$ 与所有其他子节点之间的距离之和,再加上需要经过 $u \rightarrow v$ 这条边的节点个数,即 $dp[v] + sizes[v]$。

可得到状态转移方程为:$dp[u] = \sum_{v \in graph[u]}(dp[v] + sizes[v])$。

这样,对于 $n$ 个节点来说,需要进行 $n$ 次树形 DP,这种做法的时间复杂度为 $O(n^2),ドル而 $n$ 的范围为 $[1, 3 \times 10^4],ドル这样做会导致超时,因此需要进行优化。

我们可以使用「二次遍历换根法」进行优化,从而在 $O(n)$ 的时间复杂度内解决这道题。

以编号为 0ドル$ 的节点为根节点,进行两次深度优先搜索。

1. 第一次遍历:从编号为 0ドル$ 的根节点开始,自底向上地计算出节点 0ドル$ 到其他的距离之和,记录在 $ans[0]$ 中。并且统计出以子节点为根节点的子树节点个数 $sizes[v]$。
2. 第二次遍历:从编号为 0ドル$ 的根节点开始,自顶向下地枚举每个点,计算出将每个点作为新的根节点时,其他节点到根节点的距离之和。如果当前节点为 $v,ドル其父节点为 $u,ドル则自顶向下计算出 $ans[u]$ 之后,我们将根节点从 $u$ 换为节点 $v,ドル子树上的点到新根节点的距离比原来都小了 1ドル,ドル非子树上剩下所有点到新根节点的距离比原来都大了 1ドル$。则可以据此计算出节点 $v$ 与其他节点的距离和为:$ans[v] = ans[u] + n - 2 \times sizes[u]$。

### 思路 1:代码

```Python
class Solution:
def sumOfDistancesInTree(self, n: int, edges: List[List[int]]) -> List[int]:
graph = [[] for _ in range(n)]

for u, v in edges:
graph[u].append(v)
graph[v].append(u)


ans = [0 for _ in range(n)]

sizes = [1 for _ in range(n)]
def dfs(u, fa, depth):
ans[0] += depth
for v in graph[u]:
if v == fa:
continue
dfs(v, u, depth + 1)
sizes[u] += sizes[v]

def reroot(u, fa):
for v in graph[u]:
if v == fa:
continue
ans[v] = ans[u] + n - 2 * size[v]
reroot(v, u)

dfs(0, -1, 0)
reroot(0, -1)
return ans
```

### 思路 1:复杂度分析

- **时间复杂度**:$O(n),ドル其中 $n$ 为树的节点个数。
- **空间复杂度**:$O(n)$。

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