给定一棵树，其中N个顶点的编号从0到N – 1，并且Q查询包含树中的节点，任务是为多个查询找到给定节点的父节点。将第0个节点视为根节点，并将根节点的父级作为根本身。
例子：
Tree:
0
/ \
1 2
| / \
3 4 5
Input: N = 2
Output: 0
Explanation:
Parent of node 2 is node 0 i.e root node
Input: N = 3
Output: 1
Explanation:
Parent of node 3 is node 1
方法：
默认情况下，我们将根节点的父级分配为根本身。然后，我们使用广度优先遍历(BFS)遍历树。当我们将节点s的子节点标记为已访问时，我们还将这些子节点的父节点分配为节点s。最后，对于不同的查询，将打印节点的parent []的值。
下面是上述方法的实现：
C++
// C++ implementation for
// the above approach
#include
using namespace std;
const int sz = 1e5;
// Adjacency list representation
// of the tree
vector tree[sz + 1];
// Boolean array to mark all the
// vertices which are visited
bool vis[sz + 1];
// Array of vector where ith index
// stores the path from the root
// node to the ith node
int ans[sz + 1];
// Function to create an
// edge between two vertices
void addEdge(int a, int b)
{
// Add a to b's list
tree[a].push_back(b);
// Add b to a's list
tree[b].push_back(a);
}
// Modified Breadth-First Function
void bfs(int node)
{
// Create a queue of {child, parent}
queue > qu;
// Push root node in the front of
qu.push({ node, 0 });
while (!qu.empty()) {
pair p = qu.front();
// Dequeue a vertex from queue
qu.pop();
ans[p.first] = p.second;
vis[p.first] = true;
// Get all adjacent vertices of the dequeued
// vertex s. If any adjacent has not
// been visited then enqueue it
for (int child : tree[p.first]) {
if (!vis[child]) {
qu.push({ child, p.first });
}
}
}
}
// Driver code
int main()
{
// Number of vertices
int n = 6;
addEdge(0, 1);
addEdge(0, 2);
addEdge(1, 3);
addEdge(2, 4);
addEdge(2, 5);
// Calling modified bfs function
bfs(0);
int q[] = { 2, 3 };
for (int i = 0; i < 2; i++) {
cout << ans[q[i]] << '\n';
}
return 0;
} Java
// Java implementation for
// the above approach
import java.util.*;
class GFG
{
static int sz = (int) 1e5;
// Adjacency list representation
// of the tree
static Vector []tree = new Vector[sz + 1];
// Boolean array to mark all the
// vertices which are visited
static boolean []vis = new boolean[sz + 1];
// Array of vector where ith index
// stores the path from the root
// node to the ith node
static int []ans = new int[sz + 1];
static class pair
{
int first, second;
public pair(int first, int second)
{
this.first = first;
this.second = second;
}
}
// Function to create an
// edge between two vertices
static void addEdge(int a, int b)
{
// Add a to b's list
tree[a].add(b);
// Add b to a's list
tree[b].add(a);
}
// Modified Breadth-First Function
static void bfs(int node)
{
// Create a queue of {child, parent}
Queue qu = new LinkedList<>();
// Push root node in the front of
qu.add(new pair(node, 0 ));
while (!qu.isEmpty())
{
pair p = qu.peek();
// Dequeue a vertex from queue
qu.remove();
ans[p.first] = p.second;
vis[p.first] = true;
// Get all adjacent vertices of the dequeued
// vertex s. If any adjacent has not
// been visited then enqueue it
for (int child : tree[p.first])
{
if (!vis[child])
{
qu.add(new pair(child, p.first ));
}
}
}
}
// Driver code
public static void main(String[] args)
{
// Number of vertices
int n = 6;
for (int i = 0; i < sz + 1; i++)
tree[i] = new Vector();
addEdge(0, 1);
addEdge(0, 2);
addEdge(1, 3);
addEdge(2, 4);
addEdge(2, 5);
// Calling modified bfs function
bfs(0);
int q[] = { 2, 3 };
for (int i = 0; i < 2; i++)
{
System.out.println(ans[q[i]]);
}
}
}
// This code is contributed by 29AjayKumar Python
# Python implementation for
# the above approach
sz = 10**5
# Adjacency list representation
# of the tree
tree = [[] for _ in range(sz + 1)]
# Boolean array to mark all the
# vertices which are visited
vis = [0] * (sz + 1)
# Array of vector where ith index
# stores the path from the root
# node to the ith node
ans = [0] * (sz + 1)
# Function to create an
# edge between two vertices
def addEdge(a, b):
# Add a to b's list
tree[a].append(b)
# Add b to a's list
tree[b].append(a)
# Modified Breadth-First Function
def bfs(node):
# Create a queue of child, parent
qu = []
# Push root node in the front of
qu.append([node, 0])
while (len(qu)):
p = qu[0]
# Dequeue a vertex from queue
qu.pop(0)
ans[p[0]] = p[1]
vis[p[0]] = True
# Get all adjacent vertices of the dequeued
# vertex s. If any adjacent has not
# been visited then enqueue it
for child in tree[p[0]]:
if (not vis[child]):
qu.append([child, p[0]])
# Driver code
# Number of vertices
n = 6
addEdge(0, 1)
addEdge(0, 2)
addEdge(1, 3)
addEdge(2, 4)
addEdge(2, 5)
# Calling modified bfs function
bfs(0)
q = [2, 3]
for i in range(2):
print(ans[q[i]])
# This code is contributed by SHUBHAMSINGH10C#
// C# implementation of the approach
using System;
using System.Collections.Generic;
class GFG
{
static int sz = (int) 1e5;
// Adjacency list representation
// of the tree
static List []tree = new List[sz + 1];
// Boolean array to mark all the
// vertices which are visited
static Boolean []vis = new Boolean[sz + 1];
// Array of vector where ith index
// stores the path from the root
// node to the ith node
static int []ans = new int[sz + 1];
public class pair
{
public int first, second;
public pair(int first, int second)
{
this.first = first;
this.second = second;
}
}
// Function to create an
// edge between two vertices
static void addEdge(int a, int b)
{
// Add a to b's list
tree[a].Add(b);
// Add b to a's list
tree[b].Add(a);
}
// Modified Breadth-First Function
static void bfs(int node)
{
// Create a queue of {child, parent}
Queue qu = new Queue();
// Push root node in the front of
qu.Enqueue(new pair(node, 0 ));
while (qu.Count != 0)
{
pair p = qu.Peek();
// Dequeue a vertex from queue
qu.Dequeue();
ans[p.first] = p.second;
vis[p.first] = true;
// Get all adjacent vertices of the dequeued
// vertex s. If any adjacent has not
// been visited then enqueue it
foreach (int child in tree[p.first])
{
if (!vis[child])
{
qu.Enqueue(new pair(child, p.first ));
}
}
}
}
// Driver code
public static void Main(String[] args)
{
// Number of vertices
for (int i = 0; i < sz + 1; i++)
tree[i] = new List();
addEdge(0, 1);
addEdge(0, 2);
addEdge(1, 3);
addEdge(2, 4);
addEdge(2, 5);
// Calling modified bfs function
bfs(0);
int []q = { 2, 3 };
for (int i = 0; i < 2; i++)
{
Console.WriteLine(ans[q[i]]);
}
}
}
// This code is contributed by 29AjayKumar 输出：
0
1