# Maximum subtree of the same color

You are given a 2D integer array edges representing a tree with n nodes, numbered from 0 to n - 1, rooted at node 0, where edges[i] = [u_i, v_i] means there is an edge between the nodes v_i and u_i.

You are also given a 0-indexed integer array colors of size n, where colors[i] is the color assigned to node i.

We want to find a node v such that every node in the subtree of v has the same color.

Return the size of such a subtree with the maximum number of nodes possible.

```
class Solution {
private List<Integer>[] g;
private int[] colors;
private int[] size;
private int ans;
public int maximumSubtreeSize(int[][] edges, int[] colors) {//build adjacency list
int n = edges.length + 1;
g = new List[n];
size = new int[n];
this.colors = colors;
Arrays.fill(size, 1);
Arrays.setAll(g, i -> new ArrayList<>());
for (var e : edges) {
int a = e[0], b = e[1];
g[a].add(b);
g[b].add(a);
}
dfs(0, -1);
return ans;
}
private boolean dfs(int a, int fa) {
boolean ok = true;
for (int b : g[a]) {
if (b != fa) {
boolean t = dfs(b, a);
ok = ok && colors[a] == colors[b] && t;
size[a] += size[b];
}
}
if (ok) {
ans = Math.max(ans, size[a]);
}
return ok;
}
}
```

```
class Solution:
def maximumSubtreeSize(self, edges: List[List[int]], colors: List[int]) -> int:
def dfs(a: int, fa: int) -> bool:
ok = True
for b in g[a]:
if b != fa:
t = dfs(b, a)
ok = ok and colors[a] == colors[b] and t
size[a] += size[b]
if ok:
nonlocal ans
ans = max(ans, size[a])
return ok
n = len(edges) + 1
g = [[] for _ in range(n)]
size = [1] * n
for a, b in edges:
g[a].append(b)
g[b].append(a)
ans = 0
dfs(0, -1)
return ans
```

To solve this problem using Depth-First Search (DFS), traverse the tree starting from the root node. During the traversal, keep track of the colors of nodes in each subtree. For each node, recursively determine if all nodes in its subtree have the same color. If they do, compute the size of this subtree and update the maximum size found.

Specifically, implement a DFS function that returns both the size of the subtree and a boolean indicating whether all nodes in the subtree have the same color. If the subtree rooted at a node satisfies the condition, compare its size with the current maximum and update if larger. This approach ensures an efficient traversal and comparison, yielding the desired result.

## Related Problems

You are controlling a robot that is located somewhere in a room. The room is modeled as an m x n binary grid where 0 represents a wall and 1 represents an empty slot.

The robot starts at an unknown location in the room that is guaranteed to be empty, and you do not have access to the grid, but you can move the robot using the given API Robot.

You are tasked to use the robot to clean the entire room (i.e., clean every empty cell in the room). The robot with the four given APIs can move forward, turn left, or turn right. Each turn is 90 degrees.

When the robot tries to move into a wall cell, its bumper sensor detects the obstacle, and it stays on the current cell.

A **transformation sequence** from word beginWord to word endWord using a dictionary wordList is a sequence of words beginWord -> s1 -> s2 -> ... -> sk such that:

Every adjacent pair of words differs by a single letter.

Every si for 1 <= i <= k is in wordList. Note that beginWord does not need to be in wordList.

sk == endWord

Given two words, beginWord and endWord, and a dictionary wordList, return *the ***number of words*** in the ***shortest transformation sequence*** from* beginWord *to* endWord*, or *0* if no such sequence exists.*

Given an m x n 2D binary grid grid which represents a map of '1's (land) and '0's (water), return *the number of islands*.

An **island** is surrounded by water and is formed by connecting adjacent lands horizontally or vertically. You may assume all four edges of the grid are all surrounded by water.