// Construct one Huffman tree and assigns the code bit strings and lengths.
// Update the total bit length for the current block.
// IN assertion: the field freq is set for all tree elements.
// OUT assertions: the fields len and code are set to the optimal bit length
// and corresponding code. The length opt_len is updated; static_len is
// also updated if stree is not null. The field max_code is set.
internal void build_tree(Deflate s)
{
short[] tree=dyn_tree;
short[] stree=stat_desc.static_tree;
int elems=stat_desc.elems;
int n, m; // iterate over heap elements
int max_code=-1; // largest code with non zero frequency
int node; // new node being created
// Construct the initial heap, with least frequent element in
// heap[1]. The sons of heap[n] are heap[2*n] and heap[2*n+1].
// heap[0] is not used.
s.heap_len = 0;
s.heap_max = HEAP_SIZE;
for(n=0; n<elems; n++) {
if(tree[n*2] != 0) {
s.heap[++s.heap_len] = max_code = n;
s.depth[n] = 0;
}
else{
tree[n*2+1] = 0;
}
}
// The pkzip format requires that at least one distance code exists,
// and that at least one bit should be sent even if there is only one
// possible code. So to avoid special checks later on we force at least
// two codes of non zero frequency.
while (s.heap_len < 2) {
node = s.heap[++s.heap_len] = (max_code < 2 ? ++max_code : 0);
tree[node*2] = 1;
s.depth[node] = 0;
s.opt_len--; if (stree!=null) s.static_len -= stree[node*2+1];
// node is 0 or 1 so it does not have extra bits
}
this.max_code = max_code;
// The elements heap[heap_len/2+1 .. heap_len] are leaves of the tree,
// establish sub-heaps of increasing lengths:
for(n=s.heap_len/2;n>=1; n--)
s.pqdownheap(tree, n);
// Construct the Huffman tree by repeatedly combining the least two
// frequent nodes.
node=elems; // next internal node of the tree
do{
// n = node of least frequency
n=s.heap[1];
s.heap[1]=s.heap[s.heap_len--];
s.pqdownheap(tree, 1);
m=s.heap[1]; // m = node of next least frequency
s.heap[--s.heap_max] = n; // keep the nodes sorted by frequency
s.heap[--s.heap_max] = m;
// Create a new node father of n and m
tree[node*2] = (short)(tree[n*2] + tree[m*2]);
s.depth[node] = (byte)(System.Math.Max(s.depth[n],s.depth[m])+1);
tree[n*2+1] = tree[m*2+1] = (short)node;
// and insert the new node in the heap
s.heap[1] = node++;
s.pqdownheap(tree, 1);
}
while(s.heap_len>=2);
s.heap[--s.heap_max] = s.heap[1];
// At this point, the fields freq and dad are set. We can now
// generate the bit lengths.
gen_bitlen(s);
// The field len is now set, we can generate the bit codes
gen_codes(tree, max_code, s.bl_count);
}