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feat: add solutions to lc problem: No.3004 (doocs#2206)
No.3004.Maximum Subtree of the Same Color
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# [3004. Maximum Subtree of the Same Color](https://leetcode.cn/problems/maximum-subtree-of-the-same-color)
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[English Version](/solution/3000-3099/3004.Maximum%20Subtree%20of%20the%20Same%20Color/README_EN.md)
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## 题目描述
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<!-- 这里写题目描述 -->
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<p>You are given a 2D integer array <code>edges</code> representing a tree with <code>n</code> nodes, numbered from <code>0</code> to <code>n - 1</code>, rooted at node <code>0</code>, where <code>edges[i] = [u<sub>i</sub>, v<sub>i</sub>]</code> means there is an edge between the nodes <code>v<sub>i</sub></code> and <code>u<sub>i</sub></code>.</p>
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<p>You are also given a <strong>0-indexed</strong> integer array <code>colors</code> of size <code>n</code>, where <code>colors[i]</code> is the color assigned to node <code>i</code>.</p>
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<p>We want to find a node <code>v</code> such that every node in the <span data-keyword="subtree-of-node">subtree</span> of <code>v</code> has the <strong>same</strong> color.</p>
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<p>Return <em>the size of such subtree with the <strong>maximum</strong> number of nodes possible.</em></p>
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<p>&nbsp;</p>
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<p><strong><img alt="" src="https://assets.leetcode.com/static_assets/others/20231216-134026.png" style="padding: 10px; background: rgb(255, 255, 255); border-radius: 0.5rem; width: 221px; height: 132px;" /></strong></p>
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<p><strong class="example">Example 1:</strong></p>
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<pre>
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<strong>Input:</strong> edges = [[0,1],[0,2],[0,3]], colors = [1,1,2,3]
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<strong>Output:</strong> 1
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<strong>Explanation:</strong> Each color is represented as: 1 -&gt; Red, 2 -&gt; Green, 3 -&gt; Blue. We can see that the subtree rooted at node 0 has children with different colors. Any other subtree is of the same color and has a size of 1. Hence, we return 1.
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</pre>
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<p><strong class="example">Example 2:</strong></p>
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<pre>
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<strong>Input:</strong> edges = [[0,1],[0,2],[0,3]], colors = [1,1,1,1]
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<strong>Output:</strong> 4
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<strong>Explanation:</strong> The whole tree has the same color, and the subtree rooted at node 0 has the most number of nodes which is 4. Hence, we return 4.
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</pre>
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<p><strong><img alt="" src="https://assets.leetcode.com/static_assets/others/20231216-134017.png" style="padding: 10px; background: rgb(255, 255, 255); border-radius: 0.5rem; width: 221px; height: 221px;" /></strong></p>
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<p><strong class="example">Example 3:</strong></p>
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<pre>
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<strong>Input:</strong> edges = [[0,1],[0,2],[2,3],[2,4]], colors = [1,2,3,3,3]
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<strong>Output:</strong> 3
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<strong>Explanation:</strong> Each color is represented as: 1 -&gt; Red, 2 -&gt; Green, 3 -&gt; Blue. We can see that the subtree rooted at node 0 has children with different colors. Any other subtree is of the same color, but the subtree rooted at node 2 has a size of 3 which is the maximum. Hence, we return 3.
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</pre>
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<p>&nbsp;</p>
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<p><strong>Constraints:</strong></p>
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<ul>
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<li><code>n == edges.length + 1</code></li>
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<li><code>1 &lt;= n &lt;= 5 * 10<sup>4</sup></code></li>
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<li><code>edges[i] == [u<sub>i</sub>, v<sub>i</sub>]</code></li>
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<li><code>0 &lt;= u<sub>i</sub>, v<sub>i</sub> &lt; n</code></li>
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<li><code>colors.length == n</code></li>
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<li><code>1 &lt;= colors[i] &lt;= 10<sup>5</sup></code></li>
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<li>The input is generated such that the graph represented by <code>edges</code> is a tree.</li>
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</ul>
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## 解法
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<!-- 这里可写通用的实现逻辑 -->
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**方法一:DFS**
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我们先根据题目给定的边的信息,构建一个邻接表 $g,ドル其中 $g[a]$ 表示节点 $a$ 的所有相邻节点。然后我们创建一个长度为 $n$ 的数组 $size,ドル其中 $size[a]$ 表示以节点 $a$ 为根的子树的节点数。
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接下来,我们设计一个函数 $dfs(a, fa),ドル它将返回以节点 $a$ 为根的子树是否满足题目要求。函数 $dfs(a, fa)$ 的执行过程如下:
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- 首先,我们用一个变量 $ok$ 记录以节点 $a$ 为根的子树是否满足题目要求,初始时 $ok$ 为 $true$。
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- 接着,我们遍历节点 $a$ 的所有相邻节点 $b,ドル如果 $b$ 不是 $a$ 的父节点 $fa,ドル那么我们递归调用 $dfs(b, a),ドル并将返回值保存到变量 $t$ 中,并且更新 $ok$ 为 $ok$ 与 $colors[a] = colors[b] \land t$ 的值,其中 $\land$ 表示逻辑与运算。然后,我们更新 $size[a] = size[a] + size[b]$。
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- 然后,我们判断 $ok$ 的值,如果 $ok$ 为 $true,ドル那么我们更新答案 $ans = \max(ans, size[a])$。
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- 最后,我们返回 $ok$ 的值。
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我们调用 $dfs(0, -1),ドル其中 0ドル$ 表示根节点的编号,$-1$ 表示根节点没有父节点。最终的答案即为 $ans$。
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时间复杂度 $O(n),ドル空间复杂度 $O(n)$。其中 $n$ 是节点的数量。
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<!-- tabs:start -->
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### **Python3**
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<!-- 这里可写当前语言的特殊实现逻辑 -->
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```python
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class Solution:
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def maximumSubtreeSize(self, edges: List[List[int]], colors: List[int]) -> int:
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def dfs(a: int, fa: int) -> bool:
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ok = True
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for b in g[a]:
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if b != fa:
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t = dfs(b, a)
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ok = ok and colors[a] == colors[b] and t
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size[a] += size[b]
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if ok:
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nonlocal ans
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ans = max(ans, size[a])
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return ok
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n = len(edges) + 1
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g = [[] for _ in range(n)]
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size = [1] * n
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for a, b in edges:
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g[a].append(b)
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g[b].append(a)
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ans = 0
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dfs(0, -1)
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return ans
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```
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### **Java**
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<!-- 这里可写当前语言的特殊实现逻辑 -->
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```java
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class Solution {
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private List<Integer>[] g;
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private int[] colors;
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private int[] size;
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private int ans;
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public int maximumSubtreeSize(int[][] edges, int[] colors) {
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int n = edges.length + 1;
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g = new List[n];
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size = new int[n];
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this.colors = colors;
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Arrays.fill(size, 1);
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Arrays.setAll(g, i -> new ArrayList<>());
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for (var e : edges) {
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int a = e[0], b = e[1];
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g[a].add(b);
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g[b].add(a);
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}
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dfs(0, -1);
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return ans;
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}
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private boolean dfs(int a, int fa) {
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boolean ok = true;
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for (int b : g[a]) {
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if (b != fa) {
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boolean t = dfs(b, a);
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ok = ok && colors[a] == colors[b] && t;
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size[a] += size[b];
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}
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}
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if (ok) {
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ans = Math.max(ans, size[a]);
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}
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return ok;
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}
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}
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```
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### **C++**
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```cpp
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class Solution {
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public:
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int maximumSubtreeSize(vector<vector<int>>& edges, vector<int>& colors) {
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int n = edges.size() + 1;
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vector<int> g[n];
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vector<int> size(n, 1);
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for (auto& e : edges) {
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int a = e[0], b = e[1];
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g[a].push_back(b);
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g[b].push_back(a);
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}
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int ans = 0;
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function<bool(int, int)> dfs = [&](int a, int fa) {
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bool ok = true;
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for (int b : g[a]) {
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if (b != fa) {
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bool t = dfs(b, a);
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ok = ok && colors[a] == colors[b] && t;
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size[a] += size[b];
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}
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}
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if (ok) {
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ans = max(ans, size[a]);
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}
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return ok;
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};
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dfs(0, -1);
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return ans;
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}
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};
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```
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### **Go**
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```go
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func maximumSubtreeSize(edges [][]int, colors []int) (ans int) {
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n := len(edges) + 1
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g := make([][]int, n)
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for _, e := range edges {
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a, b := e[0], e[1]
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g[a] = append(g[a], b)
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g[b] = append(g[b], a)
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}
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size := make([]int, n)
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var dfs func(int, int) bool
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dfs = func(a, fa int) bool {
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size[a] = 1
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ok := true
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for _, b := range g[a] {
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if b != fa {
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t := dfs(b, a)
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ok = ok && t && colors[a] == colors[b]
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size[a] += size[b]
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}
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}
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if ok {
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ans = max(ans, size[a])
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}
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return ok
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}
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dfs(0, -1)
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return
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}
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```
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### **TypeScript**
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```ts
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function maximumSubtreeSize(edges: number[][], colors: number[]): number {
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const n = edges.length + 1;
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const g: number[][] = Array.from({ length: n }, () => []);
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for (const [a, b] of edges) {
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g[a].push(b);
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g[b].push(a);
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}
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const size: number[] = Array(n).fill(1);
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let ans = 0;
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const dfs = (a: number, fa: number): boolean => {
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let ok = true;
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for (const b of g[a]) {
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if (b !== fa) {
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const t = dfs(b, a);
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ok = ok && t && colors[a] === colors[b];
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size[a] += size[b];
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}
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}
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if (ok) {
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ans = Math.max(ans, size[a]);
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}
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return ok;
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};
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dfs(0, -1);
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return ans;
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}
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```
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### **...**
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```
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```
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<!-- tabs:end -->

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