原文:
给定一个 和一个元素 x ，任务是用值 x 打印节点的兄弟节点。
如果两个节点位于同一级别并且具有相同的父节点，则它们被视为兄弟节点。
示例:
输入: x = 100
输出:90 110 t3】说明:值为 90、100、110 的节点有相同的父节点，即值为 40 的节点。因此，节点 90 和 110 是给定节点 x( = 100)的兄弟节点。
输入: x = 30
输出:20 40 t3】说明:值为 20、30、40 的节点有相同的父节点，即值为 10 的节点。因此，节点 20 和 40 是给定节点 x( = 30)的兄弟节点。
方法:按照下面给出的步骤解决问题:
- 在给定的 n 元树上执行。
- 初始化一个 q 进行等级顺序遍历。
- 对于遇到的每个节点，。
- 将当前节点的子节点推入队列时，检查:这些子节点是否等于给定值 x 。如果发现为真，则打印除 x 之外的当前子节点作为所需答案。
- 否则，继续。
下面是上述方法的实现:
c
// c program for the above approach
#include
using namespace std;
// structure of a node of n-ary tree
struct node {
int key;
vector child;
};
// function to create a new node
node* newnode(int key)
{
node* temp = new node;
temp->key = key;
return temp;
}
// function to find the siblings
// of the node value
void siblings(node* root, int value)
{
int flag = 0;
if (root == null)
return;
// stores nodes level wise
queue q;
// push the root
q.push(root);
// continue until all levels
// are traversed
while (!q.empty()) {
// stores current node
node* temp = q.front();
q.pop();
// enqueue all children of the current node
for (int i = 0; i < temp->child.size(); i ) {
// if node value is found
if (temp->child[i]->key == value) {
flag = 1;
// print all children of current node
// except value as the answer
for (int j = 0; j < temp->child.size();
j ) {
if (value
!= temp->child[j]->key)
cout << temp->child[j]->key
<< " ";
}
break;
}
// push the child nodes
// of temp into the queue
q.push(temp->child[i]);
}
}
if (flag == 0)
cout << "no siblings!!";
}
node* constructtree()
{
node* root = newnode(10);
(root->child).push_back(newnode(20));
(root->child).push_back(newnode(30));
(root->child).push_back(newnode(40));
(root->child[0]->child).push_back(newnode(50));
(root->child[0]->child).push_back(newnode(60));
(root->child[1]->child).push_back(newnode(70));
(root->child[1]->child).push_back(newnode(80));
(root->child[2]->child).push_back(newnode(90));
(root->child[2]->child).push_back(newnode(100));
(root->child[2]->child).push_back(newnode(110));
return root;
}
// driver code
int main()
{
// stores root of the
// constructed tree
node* root = constructtree();
int x = 30;
// print siblings of node x
siblings(root, x);
return 0;
}
java 语言(一种计算机语言，尤用于创建网站)
// java program for the
// above approach
import java.util.*;
class gfg{
// structure of a node
// of n-ary tree
static class node
{
int key;
vector child =
new vector<>();
};
// function to create a
// new node
static node newnode(int key)
{
node temp = new node();
temp.key = key;
return temp;
}
// function to find the
// siblings of the node
/// value
static void siblings(node root,
int value)
{
int flag = 0;
if (root == null)
return;
// stores nodes level wise
queue q =
new linkedlist<>();
// push the root
q.add(root);
// continue until all
// levels are traversed
while (!q.isempty())
{
// stores current node
node temp = q.peek();
q.remove();
// enqueue all children
// of the current node
for (int i = 0;
i < temp.child.size();
i )
{
// if node value is found
if (temp.child.get(i).key ==
value)
{
flag = 1;
// print all children of current
// node
// except value as the answer
for (int j = 0;
j < temp.child.size();
j )
{
if (value !=
temp.child.get(j).key)
system.out.print(
temp.child.get(j).key " ");
}
break;
}
// push the child nodes
// of temp into the queue
q.add(temp.child.get(i));
}
}
if (flag == 0)
system.out.print("no siblings!!");
}
static node constructtree()
{
node root = newnode(10);
(root.child).add(newnode(20));
(root.child).add(newnode(30));
(root.child).add(newnode(40));
(root.child.get(0).child).add(newnode(50));
(root.child.get(0).child).add(newnode(60));
(root.child.get(1).child).add(newnode(70));
(root.child.get(1).child).add(newnode(80));
(root.child.get(2).child).add(newnode(90));
(root.child.get(2).child).add(newnode(100));
(root.child.get(2).child).add(newnode(110));
return root;
}
// driver code
public static void main(string[] args)
{
// stores root of the
// constructed tree
node root = constructtree();
int x = 30;
// print siblings of node x
siblings(root, x);
}
}
// this code is contributed by rajput-ji
c
// c# program for the
// above approach
using system;
using system.collections.generic;
class gfg{
// structure of a node
// of n-ary tree
public class node
{
public int key;
public list child =
new list();
};
// function to create a
// new node
static node newnode(int key)
{
node temp = new node();
temp.key = key;
return temp;
}
// function to find the
// siblings of the node
/// value
static void siblings(node root,
int value)
{
int flag = 0;
if (root == null)
return;
// stores nodes level
// wise
queue q =
new queue();
// push the root
q.enqueue(root);
// continue until all
// levels are traversed
while (q.count != 0)
{
// stores current node
node temp = q.peek();
q.dequeue();
// enqueue all children
// of the current node
for (int i = 0;
i < temp.child.count; i )
{
// if node value is found
if (temp.child[i].key == value)
{
flag = 1;
// print all children of
// current node except value
// as the answer
for (int j = 0;
j < temp.child.count; j )
{
if (value != temp.child[j].key)
console.write(temp.child[j].key " ");
}
break;
}
// push the child nodes
// of temp into the queue
q.enqueue(temp.child[i]);
}
}
if (flag == 0)
console.write("no siblings!!");
}
static node constructtree()
{
node root = newnode(10);
(root.child).add(newnode(20));
(root.child).add(newnode(30));
(root.child).add(newnode(40));
(root.child[0].child).add(newnode(50));
(root.child[0].child).add(newnode(60));
(root.child[1].child).add(newnode(70));
(root.child[1].child).add(newnode(80));
(root.child[2].child).add(newnode(90));
(root.child[2].child).add(newnode(100));
(root.child[2].child).add(newnode(110));
return root;
}
// driver code
public static void main(string[] args)
{
// stores root of the
// constructed tree
node root = constructtree();
int x = 30;
// print siblings of node x
siblings(root, x);
}
}
// this code is contributed by rajput-ji
java 描述语言
output:
20 40
时间复杂度:o(n2) 辅助空间: o(n)
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