C :

#include<stdio.h>
#include<string.h>

int a[105][105]; 
int du[105];
int n,e,start,df;
void init();
void dfs(int);

int main()
{
	init();
	dfs(start);
	return 0;
}

void dfs(int x)
{
		for(int i=1;i<=n;i++)
	{
		if(a[x][i]==1)
		{
			a[x][i]=a[i][x]=0;
			dfs(i); 
		}
	}
	printf("%d ",x);
}

void init()
{
	memset(a,0,sizeof(a));
	memset(du,0,sizeof(du));
	scanf("%d%d",&n,&e);
	int x,y;
	for(int i=1;i<=e;i++)
	{
		scanf("%d%d",&x,&y);
		a[x][y]=a[y][x]=1;
		du[x]++; 
		du[y]++;
	}
	start=1;
	for(int i=1;i<=n;i++)
	  if(du[i]%2==1)  start=i;
	 }

C++ :

#include<iostream>
#include<cstring>
#include<cstdio> 
using namespace std;
int a[100][100];//邻接矩阵存储节点关系 
int du[100];//记录每个节点的度数 
int n,e,start;
void init();
void dfs(int);
int main()
{
	//freopen("euler1.in","r",stdin);
	//freopen("euler1.out","w",stdout);
	init();
	//cout<<"start="<<start<<endl;
	dfs(start);
	return 0;
}
void init()
{
	memset(a,0,sizeof(a));
	memset(du,0,sizeof(du));
	cin>>n>>e;
	int x,y;
	for(int i=1;i<=e;i++)
	{
		cin>>x>>y;
		a[x][y]=a[y][x]=1;//无向图 
		du[x]++;//x节点度数++ 
		du[y]++;//y节点度数++		
	}
	start=1;//遍历所有节点找到第一个奇点，没奇点则为1 
	for(int i=1;i<=n;i++)
	  if(du[i]%2==1)
	  {
	  	start=i;
	  	//break;
	  }
}
void dfs(int x)
{	 	
	for(int i=1;i<=n;i++)
	{
		if(a[x][i]==1)//找到下一个邻接点 
		{
			a[x][i]=a[i][x]=0;//两条边均设为0，不会重复访问 
			dfs(i);//以邻接点开始遍历 
		}
	}
	cout<<x<<' ';//输出当前节点	
}

Pascal :

program acm21184;
const maxn=100;
type node=record
          u,v,next:longint;
          end;
var
  d:array[1..maxn] of longint;
  num,i,j,k,n,e,u,v,t:longint;
  graph:array[1..maxn,1..maxn] of longint;
  head:array[1..maxn] of longint;
  vis:array[1..maxn] of boolean;
procedure dfs(x:longint);
var
  i:longint;
begin

   for i:=1 to n do
     if (graph[x,i]>0) then
       begin
         graph[x,i]:=0;
         graph[i,x]:=0;
         dfs(i);
       end;
     if k=0 then
   begin
     write(x); k:=k+1;
   end
  else write(' ',x);
end;
procedure init;
var i:longint;
begin
  fillchar(d,sizeof(d),0);
  fillchar(graph,sizeof(graph),0);
  readln(n,e);
  for i:=1 to e do
   begin
     readln(u,v);
     graph[u,v]:=1;
     graph[v,u]:=1;
     inc(d[u]);
     inc(d[v]);
   end;
  num:=1;
  for i:=1 to n do
   if d[i] mod 2<>0 then
      num:=i;
end;
begin
  fillchar(head,sizeof(head),0);
  t:=0;
  init;
  k:=0;
  dfs(num);
end.