找到一个节点，使得从该节点到叶节点的所有路径都具有相同的颜色

给定一个类型为{ X, Y }的二维数组edge[][]表示树中节点X和Y之间存在一条边，以及一个数组color[]表示第i^个节点的颜色值，任务是找到树的根节点，使得同一路径上根节点的所有子节点具有相同的颜色值。如果存在多个解决方案，则打印其中任何一个。否则，打印-1 。

例子：

Input:

Output: 2
Explanation:

All the child nodes on the path from the root node(= 2) to the leaf node(= 1) have the same value of the color(= 1).
All the child nodes on the path from the root node(= 2) to the leaf node(= 4) have the same value of the color(= 1).
Therefore, the required output is 2.

Input:

Output: 2
Explanation:

All the child nodes on the path from the root node(=2) to the leaf node(=9) have the same value of the color(= 4).
All the child nodes on the path from the root node(=2) to the leaf node(=1) have the same value of the color(= 1).
All the child nodes on the path from the root node(=2) to the leaf node(=5) have the same value of the color(= 2).
All the child nodes on the path from the root node(=2) to the leaf node(=6) have the same value of the color(= 3).

编程需要懂一点英语

方法：这个想法是迭代树的所有可能节点。对于每个^第i^个节点，检查它是否满足根节点的条件或不使用 DFS。如果发现为真，则打印节点。否则，打印-1 。请按照以下步骤解决问题：

初始化一个变量，比如root ，来存储满足条件的树的根节点。
迭代树的所有可能节点。将树的每个^第i^个节点视为根节点，并使用 DFS 检查从根节点到叶节点的路径上的所有子节点是否具有相同的颜色。如果发现为真，则打印节点。
否则，打印-1 。

下面是上述方法的实现：

C++

// C++ program to implement
// the above approach
 
#include 
using namespace std;
 
// Function to perform dfs on the tree
bool dfs(int node, int c, vector adj[],
         int color[], int visited[])
{
 
    // Mark visited node as true
    visited[node] = true;
 
    // If color does not match with
    // previous node on the same path
    if (color[node] != c) {
        return false;
    }
 
    // Check if current subtree
    // has all same colored nodes
    int f = 1;
 
    // Traverse all unvisited neighbors
    // node of the tree
    for (int j = 0; j < adj[node].size(); j++) {
 
        // Stores neighbors node
        // of the tree
        int neighbor = adj[node][j];
 
        // If current node is not
        // already visited
        if (!visited[neighbor]) {
 
            if (dfs(neighbor, c, adj,
                    color, visited)
                == false) {
 
                // Update f
                f = 0;
                break;
            }
        }
    }
    return f;
}
 
// Function to find the root node of
// the tree such that all child nodes
// on the same path have the same color
void findNode(int edges[][2],
              int color[], int n)
{
 
    // Store the adjacency list
    vector adj[n + 1];
 
    // Traverse all edges and form
    // the adjacency list
    for (int i = 0; i < n - 1; i++) {
        int a = edges[i][0];
        int b = edges[i][1];
        adj[a].push_back(b);
        adj[b].push_back(a);
    }
 
    // Store the root node such that all
    // child nodes on the same path have
    // the same color
    int ans = -1;
 
    // Iterate over all possible
    // nodes of the tree
    for (int i = 1; i <= n; i++) {
 
        // Check if node i satisfies
        // the condition of root node
        int f = 1;
 
        // Check if a node has been
        // visited or not
        int visited[n + 1] = { false };
 
        // Mark visited[i] as true
        visited[i] = true;
 
        // Traverse all the neighbors
        // of node i
        for (int j = 0; j < adj[i].size(); j++) {
 
            // Stores the current neighbor
            int neighbor = adj[i][j];
 
            // Perform DFS for current neighbor
            if (dfs(neighbor, color[neighbor],
                    adj, color, visited)
                == false) {
 
                // Update f
                f = 0;
                break;
            }
        }
 
        if (f == 1) {
            ans = i;
            break;
        }
    }
 
    // Print the answer
    cout << ans;
}
 
// Driver Code
int main()
{
 
    int n = 9;
    int color[n + 1] = { -1, 1, 1, 2, 2,
                         2, 3, 3, 4, 4 };
 
    int edges[][2] = { { 1, 2 }, { 2, 3 },
                       { 3, 4 }, { 4, 5 },
                       { 2, 7 }, { 7, 6 },
                       { 2, 8 }, { 8, 9 } };
 
    findNode(edges, color, n);
 
    return 0;
}

Java

// Java program to implement
// the above approach
import java.util.*;
class GFG
{
 
// Function to perform dfs on the tree
static boolean dfs(int node, int c, Vector adj[],
         int color[], boolean visited[])
{
 
    // Mark visited node as true
    visited[node] = true;
 
    // If color does not match with
    // previous node on the same path
    if (color[node] != c)
    {
        return false;
    }
 
    // Check if current subtree
    // has all same colored nodes
    boolean f = true;
 
    // Traverse all unvisited neighbors
    // node of the tree
    for (int j = 0; j < adj[node].size(); j++)
    {
 
        // Stores neighbors node
        // of the tree
        int neighbor = adj[node].get(j);
 
        // If current node is not
        // already visited
        if (!visited[neighbor])
        {
 
            if (dfs(neighbor, c, adj,
                    color, visited) == false)
            {
 
                // Update f
                f = false;
                break;
            }
        }
    }
    return f;
}
 
// Function to find the root node of
// the tree such that all child nodes
// on the same path have the same color
static void findNode(int edges[][],
              int color[], int n)
{
 
    // Store the adjacency list
    Vector []adj = new Vector[n + 1];
    for(int i = 0; i < n + 1; i++)
        adj[i] = new Vector();
 
    // Traverse all edges and form
    // the adjacency list
    for (int i = 0; i < n - 1; i++)
    {
        int a = edges[i][0];
        int b = edges[i][1];
        adj[a].add(b);
        adj[b].add(a);
    }
 
    // Store the root node such that all
    // child nodes on the same path have
    // the same color
    int ans = -1;
 
    // Iterate over all possible
    // nodes of the tree
    for (int i = 1; i <= n; i++)
    {
 
        // Check if node i satisfies
        // the condition of root node
        int f = 1;
 
        // Check if a node has been
        // visited or not
        boolean []visited = new boolean[n + 1];
 
        // Mark visited[i] as true
        visited[i] = true;
 
        // Traverse all the neighbors
        // of node i
        for (int j = 0; j < adj[i].size(); j++)
        {
 
            // Stores the current neighbor
            int neighbor = adj[i].get(j);
 
            // Perform DFS for current neighbor
            if (dfs(neighbor, color[neighbor],
                    adj, color, visited) == false)
            {
 
                // Update f
                f = 0;
                break;
            }
        }
        if (f == 1)
        {
            ans = i;
            break;
        }
    }
 
    // Print the answer
    System.out.print(ans);
}
 
// Driver Code
public static void main(String[] args)
{
 
    int n = 9;
    int color[] = { -1, 1, 1, 2, 2,
                         2, 3, 3, 4, 4 };
    int edges[][] = { { 1, 2 }, { 2, 3 },
                       { 3, 4 }, { 4, 5 },
                       { 2, 7 }, { 7, 6 },
                       { 2, 8 }, { 8, 9 } };
    findNode(edges, color, n);
}
}
 
// This code is contributed by 29AjayKumar

Python3

# Python program to implement
# the above approach
from typing import List
 
# Function to perform dfs on the tree
def dfs(node: int, c: int, adj: List[List[int]],
        color: List[int],
        visited: List[int]) -> bool:
 
    # Mark visited node as true
    visited[node] = True
 
    # If color does not match with
    # previous node on the same path
    if (color[node] != c):
        return False
 
    # Check if current subtree
    # has all same colored nodes
    f = 1
 
    # Traverse all unvisited neighbors
    # node of the tree
    for j in range(len(adj[node])):
 
        # Stores neighbors node
        # of the tree
        neighbor = adj[node][j]
 
        # If current node is not
        # already visited
        if (not visited[neighbor]):
            if not dfs(neighbor, c, adj, color, visited):
 
                # Update f
                f = 0
                break
    return f
 
# Function to find the root node of
# the tree such that all child nodes
# on the same path have the same color
def findNode(edges: List[List[int]], color: List[int], n: int) -> None:
 
    # Store the adjacency list
    adj = [[] for _ in range(n + 1)]
 
    # Traverse all edges and form
    # the adjacency list
    for i in range(n - 1):
        a = edges[i][0]
        b = edges[i][1]
        adj[a].append(b)
        adj[b].append(a)
 
    # Store the root node such that all
    # child nodes on the same path have
    # the same color
    ans = -1
 
    # Iterate over all possible
    # nodes of the tree
    for i in range(1, n + 1):
 
        # Check if node i satisfies
        # the condition of root node
        f = 1
 
        # Check if a node has been
        # visited or not
        visited = [False for _ in range(n + 1)]
 
        # Mark visited[i] as true
        visited[i] = True
 
        # Traverse all the neighbors
        # of node i
        for j in range(len(adj[i])):
 
            # Stores the current neighbor
            neighbor = adj[i][j]
 
            # Perform DFS for current neighbor
            if not dfs(neighbor, color[neighbor],
                       adj, color, visited):
 
                # Update f
                f = 0
                break
 
        if (f == 1):
            ans = i
            break
 
    # Print the answer
    print(ans)
 
# Driver Code
if __name__ == "__main__":
 
    n = 9
    color = [-1, 1, 1, 2, 2, 2, 3, 3, 4, 4]
    edges = [[1, 2], [2, 3], [3, 4], [4, 5], [2, 7], [7, 6], [2, 8], [8, 9]]
    findNode(edges, color, n)
 
# This code is contributed by sanjeev2552

C#

// C# program to implement
// the above approach
using System;
using System.Collections.Generic;
class GFG
{
 
// Function to perform dfs on the tree
static bool dfs(int node, int c, List []adj,
         int []color, bool []visited)
{
 
    // Mark visited node as true
    visited[node] = true;
 
    // If color does not match with
    // previous node on the same path
    if (color[node] != c)
    {
        return false;
    }
 
    // Check if current subtree
    // has all same colored nodes
    bool f = true;
 
    // Traverse all unvisited neighbors
    // node of the tree
    for (int j = 0; j < adj[node].Count; j++)
    {
 
        // Stores neighbors node
        // of the tree
        int neighbor = adj[node][j];
 
        // If current node is not
        // already visited
        if (!visited[neighbor])
        {
 
            if (dfs(neighbor, c, adj,
                    color, visited) == false)
            {
 
                // Update f
                f = false;
                break;
            }
        }
    }
    return f;
}
 
// Function to find the root node of
// the tree such that all child nodes
// on the same path have the same color
static void findNode(int [,]edges,
              int []color, int n)
{
 
    // Store the adjacency list
    List []adj = new List[n + 1];
    for(int i = 0; i < n + 1; i++)
        adj[i] = new List();
 
    // Traverse all edges and form
    // the adjacency list
    for (int i = 0; i < n - 1; i++)
    {
        int a = edges[i, 0];
        int b = edges[i, 1];
        adj[a].Add(b);
        adj[b].Add(a);
    }
 
    // Store the root node such that all
    // child nodes on the same path have
    // the same color
    int ans = -1;
 
    // Iterate over all possible
    // nodes of the tree
    for (int i = 1; i <= n; i++)
    {
 
        // Check if node i satisfies
        // the condition of root node
        int f = 1;
 
        // Check if a node has been
        // visited or not
        bool []visited = new bool[n + 1];
 
        // Mark visited[i] as true
        visited[i] = true;
 
        // Traverse all the neighbors
        // of node i
        for (int j = 0; j < adj[i].Count; j++)
        {
 
            // Stores the current neighbor
            int neighbor = adj[i][j];
 
            // Perform DFS for current neighbor
            if (dfs(neighbor, color[neighbor],
                    adj, color, visited) == false)
            {
 
                // Update f
                f = 0;
                break;
            }
        }
        if (f == 1)
        {
            ans = i;
            break;
        }
    }
 
    // Print the answer
    Console.Write(ans);
}
 
// Driver Code
public static void Main(String[] args)
{
    int n = 9;
    int []color = { -1, 1, 1, 2, 2,
                         2, 3, 3, 4, 4 };
    int [,]edges = { { 1, 2 }, { 2, 3 },
                       { 3, 4 }, { 4, 5 },
                       { 2, 7 }, { 7, 6 },
                       { 2, 8 }, { 8, 9 } };
    findNode(edges, color, n);
}
}
 
// This code is contributed by 29AjayKumar

Javascript

输出：

时间复杂度： O(N ² )
辅助空间： O(N)

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