\datethis @*Intro. This program tests an amazingly simple algorithm that generates all $n$-node trees with the property that the $k$th node in preorder has $t_k$ link fields. (A link field is either zero or it points to a subtree.) In the special case that $t_k=2$ for $1\le k\le n$, we get Skarbek's algorithm for binary trees, Algorithm 7.2.1.6B. In the special case that $t_k=t$ for $1\le k\le n$, we get an algorithm that was sent to me by James Korsh in December 2004. I~happened to notice that Korsh's idea works in the general case considered here; but I'll let him have the fun of constructing a formal proof, because the basic insights are essentially his. The number of trees generated does not appear to have a simple formula in general. But one can show bijectively that such trees are equivalent to integer sequences $a_1\ldots a_{n-1}$ with the property that $a_1\ge\cdots\ge a_{n-1}\ge0$ and $a_k\le\sum_{j=1}^{n-k}(t_j-1)$. The numbers $t_1$, \dots, $t_n$ are input on the command line. @d maxn 100 /* $n$ should be less than this */ @c #include int h[maxn]; /* the table of degrees; $h_k=t_k-1$ */ int l[maxn][maxn]; /* the links (right to left) */ int count; main(int argc, char*argv[]) { register int j,k,n,r,x,y; @; for (j=1;j; for (j=1,x=l[1][0]; x==j+1; j=x,x=l[j][0]) l[j][0]=0, l[j][h[j]]=x; if (j==n) break; for (r=1;l[j][r]==0;r++); for (k=0,y=l[j][r]; l[y][0]; k=y,y=l[y][0]); if (k) l[k][0]=0; @+ else l[j][r]=0; l[j][0]=0, l[j][r-1]=y, l[y][0]=x; } } @ @d abort(m) {@+fprintf(stderr,"%s!\n",m); exit(j);@+} @= n=argc-1; if (n==0) { fprintf(stderr,"Usage: %s t1 t2 ... tn\n",argv[0]); exit(0); } if (n>=maxn) abort("I can't handle that many degrees"); for (j=1;j<=n;j++) { if (sscanf(argv[j],"%d",&h[j])!=1) abort("unreadable degree"); h[j]--; if (h[j]<0) abort("Each degree must be positive"); if (h[j]>=maxn) abort("Degree is too large"); } @ For each tree, we print out the array of links, with link~0 last. @= count++; printf("%d:",count); for (j=1;j<=n;j++) { for (k=h[j];k>=0;k--) printf(" %d",l[j][k]); if (j