/*
* mm_remove_children - Remove the node from the Binary Tree as long as it has two children
*/
void *mm_remove_children(void *root, void *bp)
{
void *parent_node = mm_parent(root, bp);
void *replacement = mm_replace(LEFT(bp));
void *new_bp;
/* Remove the replacement and store the new bp */
new_bp = mm_remove(LEFT(bp), replacement);
SETLEFT(replacement, new_bp);
SETRIGHT(replacement, RIGHT(bp));
if(parent_node != NULL)
{
if(LEFT(parent_node) == bp)
SETLEFT(parent_node, replacement);
else
SETRIGHT(parent_node, replacement);
return root;
}
else
{
return replacement;
}
}
/*
* mm_insert - Insert a free block into the Binary Tree and return bp
*/
void *mm_insert(void *root, void* bp)
{
/* Determine if the tree is empty and initiate all nodes to NULL */
if(root == NULL)
{
SETLEFT(bp, NULL);
SETRIGHT(bp, NULL);
return bp;
}
else if(GETSIZE(bp) <= GETSIZE(root))
{
SETLEFT(root, mm_insert(LEFT(root),bp));
return root;
}
else if(GETSIZE(bp) > GETSIZE(root))
{
SETRIGHT(root, mm_insert(RIGHT(root),bp));
return root;
}
/* If there's an error, return -1 */
return -1;
}
/*
* mm_remove_child - Remove the node from the Binary Tree as long as it has one child
*/
void *mm_remove_child(void *root, void* bp)
{
void *parent_node = mm_parent(root, bp);
void *child;
/* Who is our child? Where is it? */
if(LEFT(bp) != NULL)
child = LEFT(bp);
else
child = RIGHT(bp);
if(parent_node != NULL)
{
if(LEFT(parent_node) == bp)
SETLEFT(parent_node, child);
else
SETRIGHT(parent_node, child);
return root;
}
else
{
return child;
}
}
Plane * CMRouter::initPlane(bool rotate90) {
Tile * bufferTile = TiAlloc();
TiSetType(bufferTile, Tile::BUFFER);
TiSetBody(bufferTile, NULL);
QRectF bufferRect(rotate90 ? m_maxRect90 : m_maxRect);
TileRect br;
qrectToTile(bufferRect, br);
bufferRect.adjust(-bufferRect.width(), -bufferRect.height(), bufferRect.width(), bufferRect.height());
//DebugDialog::debug("max rect", m_maxRect);
//DebugDialog::debug("max rect 90", m_maxRect90);
int l = fasterRealToTile(bufferRect.left());
int t = fasterRealToTile(bufferRect.top());
int r = fasterRealToTile(bufferRect.right());
int b = fasterRealToTile(bufferRect.bottom());
SETLEFT(bufferTile, l);
SETYMIN(bufferTile, t); // TILE is Math Y-axis not computer-graphic Y-axis
Plane * thePlane = TiNewPlane(bufferTile, br.xmini, br.ymini, br.xmaxi, br.ymaxi);
SETRIGHT(bufferTile, r);
SETYMAX(bufferTile, b); // TILE is Math Y-axis not computer-graphic Y-axis
// do not use InsertTile here
TiInsertTile(thePlane, &thePlane->maxRect, NULL, Tile::SPACE);
//infoTileRect("insert", thePlane->maxRect);
return thePlane;
}
/*
* mm_remove_node - If a given node has no children, lets remove it.
*/
void *mm_remove_node(void *root, void *bp)
{
void *parent_node = mm_parent(root, bp);
if(parent_node != NULL)
{
if(LEFT(parent_node) == bp)
SETLEFT(parent_node, NULL);
else
SETRIGHT(parent_node, NULL);
return root;
}
else
{
return NULL;
}
}
/* Delete node with given key.
KEY is the key to be deleted, ROOTP is the address of the root of tree,
COMPAR the comparison function. */
void *
__tdelete (const void *key, void **vrootp, __compar_fn_t compar)
{
node p, q, r, retval;
int cmp;
node *rootp = (node *) vrootp;
node root, unchained;
/* Stack of nodes so we remember the parents without recursion. It's
_very_ unlikely that there are paths longer than 40 nodes. The tree
would need to have around 250.000 nodes. */
int stacksize = 40;
int sp = 0;
node **nodestack = alloca (sizeof (node *) * stacksize);
if (rootp == NULL)
return NULL;
p = DEREFNODEPTR(rootp);
if (p == NULL)
return NULL;
CHECK_TREE (p);
root = DEREFNODEPTR(rootp);
while ((cmp = (*compar) (key, root->key)) != 0)
{
if (sp == stacksize)
{
node **newstack;
stacksize += 20;
newstack = alloca (sizeof (node *) * stacksize);
nodestack = memcpy (newstack, nodestack, sp * sizeof (node *));
}
nodestack[sp++] = rootp;
p = DEREFNODEPTR(rootp);
if (cmp < 0)
{
rootp = LEFTPTR(p);
root = LEFT(p);
}
else
{
rootp = RIGHTPTR(p);
root = RIGHT(p);
}
if (root == NULL)
return NULL;
}
/* This is bogus if the node to be deleted is the root... this routine
really should return an integer with 0 for success, -1 for failure
and errno = ESRCH or something. */
retval = p;
/* We don't unchain the node we want to delete. Instead, we overwrite
it with its successor and unchain the successor. If there is no
successor, we really unchain the node to be deleted. */
root = DEREFNODEPTR(rootp);
r = RIGHT(root);
q = LEFT(root);
if (q == NULL || r == NULL)
unchained = root;
else
{
node *parentp = rootp, *up = RIGHTPTR(root);
node upn;
for (;;)
{
if (sp == stacksize)
{
node **newstack;
stacksize += 20;
newstack = alloca (sizeof (node *) * stacksize);
nodestack = memcpy (newstack, nodestack, sp * sizeof (node *));
}
nodestack[sp++] = parentp;
parentp = up;
upn = DEREFNODEPTR(up);
if (LEFT(upn) == NULL)
break;
up = LEFTPTR(upn);
}
unchained = DEREFNODEPTR(up);
}
/* We know that either the left or right successor of UNCHAINED is NULL.
R becomes the other one, it is chained into the parent of UNCHAINED. */
r = LEFT(unchained);
if (r == NULL)
r = RIGHT(unchained);
if (sp == 0)
SETNODEPTR(rootp,r);
else
{
q = DEREFNODEPTR(nodestack[sp-1]);
if (unchained == RIGHT(q))
SETRIGHT(q,r);
else
SETLEFT(q,r);
}
if (unchained != root)
root->key = unchained->key;
if (!RED(unchained))
{
/* Now we lost a black edge, which means that the number of black
edges on every path is no longer constant. We must balance the
tree. */
/* NODESTACK now contains all parents of R. R is likely to be NULL
in the first iteration. */
/* NULL nodes are considered black throughout - this is necessary for
correctness. */
while (sp > 0 && (r == NULL || !RED(r)))
{
node *pp = nodestack[sp - 1];
p = DEREFNODEPTR(pp);
/* Two symmetric cases. */
if (r == LEFT(p))
{
/* Q is R's brother, P is R's parent. The subtree with root
R has one black edge less than the subtree with root Q. */
q = RIGHT(p);
if (RED(q))
{
/* If Q is red, we know that P is black. We rotate P left
so that Q becomes the top node in the tree, with P below
it. P is colored red, Q is colored black.
This action does not change the black edge count for any
leaf in the tree, but we will be able to recognize one
of the following situations, which all require that Q
is black. */
SETBLACK(q);
SETRED(p);
/* Left rotate p. */
SETRIGHT(p,LEFT(q));
SETLEFT(q,p);
SETNODEPTR(pp,q);
/* Make sure pp is right if the case below tries to use
it. */
nodestack[sp++] = pp = LEFTPTR(q);
q = RIGHT(p);
}
/* We know that Q can't be NULL here. We also know that Q is
black. */
if ((LEFT(q) == NULL || !RED(LEFT(q)))
&& (RIGHT(q) == NULL || !RED(RIGHT(q))))
{
/* Q has two black successors. We can simply color Q red.
The whole subtree with root P is now missing one black
edge. Note that this action can temporarily make the
tree invalid (if P is red). But we will exit the loop
in that case and set P black, which both makes the tree
valid and also makes the black edge count come out
right. If P is black, we are at least one step closer
to the root and we'll try again the next iteration. */
SETRED(q);
r = p;
}
else
{
/* Q is black, one of Q's successors is red. We can
repair the tree with one operation and will exit the
loop afterwards. */
if (RIGHT(q) == NULL || !RED(RIGHT(q)))
{
/* The left one is red. We perform the same action as
in maybe_split_for_insert where two red edges are
adjacent but point in different directions:
Q's left successor (let's call it Q2) becomes the
top of the subtree we are looking at, its parent (Q)
and grandparent (P) become its successors. The former
successors of Q2 are placed below P and Q.
P becomes black, and Q2 gets the color that P had.
This changes the black edge count only for node R and
its successors. */
node q2 = LEFT(q);
if (RED(p))
SETRED(q2);
else
SETBLACK(q2);
SETRIGHT(p,LEFT(q2));
SETLEFT(q,RIGHT(q2));
SETRIGHT(q2,q);
SETLEFT(q2,p);
SETNODEPTR(pp,q2);
SETBLACK(p);
}
else
{
/* It's the right one. Rotate P left. P becomes black,
and Q gets the color that P had. Q's right successor
also becomes black. This changes the black edge
count only for node R and its successors. */
if (RED(p))
SETRED(q);
else
SETBLACK(q);
SETBLACK(p);
SETBLACK(RIGHT(q));
/* left rotate p */
SETRIGHT(p,LEFT(q));
SETLEFT(q,p);
SETNODEPTR(pp,q);
}
/* We're done. */
sp = 1;
r = NULL;
}
}
else
{
/* Comments: see above. */
q = LEFT(p);
if (RED(q))
{
SETBLACK(q);
SETRED(p);
SETLEFT(p,RIGHT(q));
SETRIGHT(q,p);
SETNODEPTR(pp,q);
nodestack[sp++] = pp = RIGHTPTR(q);
q = LEFT(p);
}
if ((RIGHT(q) == NULL || !RED(RIGHT(q)))
&& (LEFT(q) == NULL || !RED(LEFT(q))))
{
SETRED(q);
r = p;
}
else
{
if (LEFT(q) == NULL || !RED(LEFT(q)))
{
node q2 = RIGHT(q);
if (RED(p))
SETRED(q2);
else
SETBLACK(q2);
SETLEFT(p,RIGHT(q2));
SETRIGHT(q,LEFT(q2));
SETLEFT(q2,q);
SETRIGHT(q2,p);
SETNODEPTR(pp,q2);
SETBLACK(p);
}
else
{
if (RED(p))
SETRED(q);
else
SETBLACK(q);
SETBLACK(p);
SETBLACK(LEFT(q));
SETLEFT(p,RIGHT(q));
SETRIGHT(q,p);
SETNODEPTR(pp,q);
}
sp = 1;
r = NULL;
}
}
--sp;
}
if (r != NULL)
SETBLACK(r);
}
free (unchained);
return retval;
}
/* Find or insert datum into search tree.
KEY is the key to be located, ROOTP is the address of tree root,
COMPAR the ordering function. */
void *
__tsearch (const void *key, void **vrootp, __compar_fn_t compar)
{
node q, root;
node *parentp = NULL, *gparentp = NULL;
node *rootp = (node *) vrootp;
node *nextp;
int r = 0, p_r = 0, gp_r = 0; /* No they might not, Mr Compiler. */
#ifdef USE_MALLOC_LOW_BIT
static_assert (alignof (max_align_t) > 1, "malloc must return aligned ptrs");
#endif
if (rootp == NULL)
return NULL;
/* This saves some additional tests below. */
root = DEREFNODEPTR(rootp);
if (root != NULL)
SETBLACK(root);
CHECK_TREE (root);
nextp = rootp;
while (DEREFNODEPTR(nextp) != NULL)
{
root = DEREFNODEPTR(rootp);
r = (*compar) (key, root->key);
if (r == 0)
return root;
maybe_split_for_insert (rootp, parentp, gparentp, p_r, gp_r, 0);
/* If that did any rotations, parentp and gparentp are now garbage.
That doesn't matter, because the values they contain are never
used again in that case. */
nextp = r < 0 ? LEFTPTR(root) : RIGHTPTR(root);
if (DEREFNODEPTR(nextp) == NULL)
break;
gparentp = parentp;
parentp = rootp;
rootp = nextp;
gp_r = p_r;
p_r = r;
}
q = (struct node_t *) malloc (sizeof (struct node_t));
if (q != NULL)
{
/* Make sure the malloc implementation returns naturally aligned
memory blocks when expected. Or at least even pointers, so we
can use the low bit as red/black flag. Even though we have a
static_assert to make sure alignof (max_align_t) > 1 there could
be an interposed malloc implementation that might cause havoc by
not obeying the malloc contract. */
#ifdef USE_MALLOC_LOW_BIT
assert (((uintptr_t) q & (uintptr_t) 0x1) == 0);
#endif
SETNODEPTR(nextp,q); /* link new node to old */
q->key = key; /* initialize new node */
SETRED(q);
SETLEFT(q,NULL);
SETRIGHT(q,NULL);
if (nextp != rootp)
/* There may be two red edges in a row now, which we must avoid by
rotating the tree. */
maybe_split_for_insert (nextp, rootp, parentp, r, p_r, 1);
}
return q;
}
/* Possibly "split" a node with two red successors, and/or fix up two red
edges in a row. ROOTP is a pointer to the lowest node we visited, PARENTP
and GPARENTP pointers to its parent/grandparent. P_R and GP_R contain the
comparison values that determined which way was taken in the tree to reach
ROOTP. MODE is 1 if we need not do the split, but must check for two red
edges between GPARENTP and ROOTP. */
static void
maybe_split_for_insert (node *rootp, node *parentp, node *gparentp,
int p_r, int gp_r, int mode)
{
node root = DEREFNODEPTR(rootp);
node *rp, *lp;
node rpn, lpn;
rp = RIGHTPTR(root);
rpn = RIGHT(root);
lp = LEFTPTR(root);
lpn = LEFT(root);
/* See if we have to split this node (both successors red). */
if (mode == 1
|| ((rpn) != NULL && (lpn) != NULL && RED(rpn) && RED(lpn)))
{
/* This node becomes red, its successors black. */
SETRED(root);
if (rpn)
SETBLACK(rpn);
if (lpn)
SETBLACK(lpn);
/* If the parent of this node is also red, we have to do
rotations. */
if (parentp != NULL && RED(DEREFNODEPTR(parentp)))
{
node gp = DEREFNODEPTR(gparentp);
node p = DEREFNODEPTR(parentp);
/* There are two main cases:
1. The edge types (left or right) of the two red edges differ.
2. Both red edges are of the same type.
There exist two symmetries of each case, so there is a total of
4 cases. */
if ((p_r > 0) != (gp_r > 0))
{
/* Put the child at the top of the tree, with its parent
and grandparent as successors. */
SETRED(p);
SETRED(gp);
SETBLACK(root);
if (p_r < 0)
{
/* Child is left of parent. */
SETLEFT(p,rpn);
SETNODEPTR(rp,p);
SETRIGHT(gp,lpn);
SETNODEPTR(lp,gp);
}
else
{
/* Child is right of parent. */
SETRIGHT(p,lpn);
SETNODEPTR(lp,p);
SETLEFT(gp,rpn);
SETNODEPTR(rp,gp);
}
SETNODEPTR(gparentp,root);
}
else
{
SETNODEPTR(gparentp,p);
/* Parent becomes the top of the tree, grandparent and
child are its successors. */
SETBLACK(p);
SETRED(gp);
if (p_r < 0)
{
/* Left edges. */
SETLEFT(gp,RIGHT(p));
SETRIGHT(p,gp);
}
else
{
/* Right edges. */
SETRIGHT(gp,LEFT(p));
SETLEFT(p,gp);
}
}
}
}
}
