var ni,k,i,j,max,m,n:longint;
    a,b:array [1..25001] of integer;
    q:array [1..1001] of byte;
    s:boolean;

begin
  readln (k);
  for j:=1 to k do
    begin
      readln (n);
      for i:=1 to n do
        begin
          if i<>n then
            read (a[i])
          else
            readln (a[i]);
          if a[i]<>0 then
            inc (b[a[i]]);
        end;
      max:=b[1];
      for i:=2 to n do
        if max<b[i] then
          max:=b[i];
      for i:=1 to n do
        if (b[i]=max) then
          inc (m);
      ni:=0;
      if m=1 then
        begin
          q[j]:=1;
          s:=true;
        end
      else
        for i:=1 to n do
          if (a[i]=0)and(b[i]<>0) then
            begin
              q[j]:=1;
              if ni=1 then
                begin
                  q[j]:=0;
                  break;
                end;
              inc (ni);
              s:=true;
            end;
      if not s then
        q[j]:=0;
      s:=false;
      fillchar(a,sizeof(a),0);
      fillchar(b,sizeof(b),0);
    end;
  for i:=1 to k do
    if q[i]=1 then
      writeln ('YES')
    else
      writeln ('NO');
end.

You need to use the O(n) algo. Here is my AC solution.
It checks if the graph is the tree. If is then YES.


program Collective_Guarantee;
const MaxN=25000;
label fin;
type List=array[1..MaxN] of Integer;
var  Tree,Sign:List;
     N,i,j,z,K,M,m1:integer;
     res:boolean;
begin
     readln(M);
for m1:=1 to M do
begin readln(N);
     z:=0;
     res:=true;
     for i:=1 to N do
     begin
          read(Tree[i]);
          if Tree[i]=0 then
            inc(z);
     end;
     if z<>1 then
     begin
          res:=false;
          goto fin;
     end else
     begin
           FillChar(Sign,SizeOf(Sign),0);
          K:=0;
          while true do
          begin
               i:=1;
               inc(K);
                 while (Sign[i]>0) and (i<=N) do
                  inc(i);
               if i<>N+1 then
               begin
                       while i<>0 do
                       begin
                         if Sign[i]=0 then
                            Sign[i]:=K else
                         if Sign[i]=K then
                         begin
                              res:=false;
                              goto fin;
                         end else
                         if Sign[i]<K then
                           i:=0;
                         if i<>0 then
                           i:=Tree[i];
                    end;
               end else
                goto fin;
           end;
     end;
fin: if res=false then
        writeln('NO') else
        writeln('YES');
end;
end.