本文整理汇总了C++中NodeRef::subtree方法的典型用法代码示例。如果您正苦于以下问题：C++ NodeRef::subtree方法的具体用法？C++ NodeRef::subtree怎么用？C++ NodeRef::subtree使用的例子？那么, 这里精选的方法代码示例或许可以为您提供帮助。您也可以进一步了解该方法所在类NodeRef的用法示例。
在下文中一共展示了NodeRef::subtree方法的4个代码示例，这些例子默认根据受欢迎程度排序。您可以为喜欢或者感觉有用的代码点赞，您的评价将有助于系统推荐出更棒的C++代码示例。
示例1: moveLeft
void Path::moveLeft(unsigned Level) {
assert(Level != 0 && "Cannot move the root node");
// Go up the tree until we can go left.
unsigned l = 0;
if (valid()) {
l = Level - 1;
while (path[l].offset == 0) {
assert(l != 0 && "Cannot move beyond begin()");
--l;
}
} else if (height() < Level)
// end() may have created a height=0 path.
path.resize(Level + 1, Entry(nullptr, 0, 0));
// NR is the subtree containing our left sibling.
--path[l].offset;
NodeRef NR = subtree(l);
// Get the rightmost node in the subtree.
for (++l; l != Level; ++l) {
path[l] = Entry(NR, NR.size() - 1);
NR = NR.subtree(NR.size() - 1);
}
path[l] = Entry(NR, NR.size() - 1);
}示例2: moveRight
void Path::moveRight(unsigned Level) {
assert(Level != 0 && "Cannot move the root node");
// Go up the tree until we can go right.
unsigned l = Level - 1;
while (l && atLastEntry(l))
--l;
// NR is the subtree containing our right sibling. If we hit end(), we have
// offset(0) == node(0).size().
if (++path[l].offset == path[l].size)
return;
NodeRef NR = subtree(l);
for (++l; l != Level; ++l) {
path[l] = Entry(NR, 0);
NR = NR.subtree(0);
}
path[l] = Entry(NR, 0);
}示例3: getRightSibling
NodeRef Path::getRightSibling(unsigned Level) const {
// The root has no siblings.
if (Level == 0)
return NodeRef();
// Go up the tree until we can go right.
unsigned l = Level - 1;
while (l && atLastEntry(l))
--l;
// We can't go right.
if (atLastEntry(l))
return NodeRef();
// NR is the subtree containing our right sibling.
NodeRef NR = path[l].subtree(path[l].offset + 1);
// Keep left all the way down.
for (++l; l != Level; ++l)
NR = NR.subtree(0);
return NR;
}示例4: getLeftSibling
NodeRef Path::getLeftSibling(unsigned Level) const {
// The root has no siblings.
if (Level == 0)
return NodeRef();
// Go up the tree until we can go left.
unsigned l = Level - 1;
while (l && path[l].offset == 0)
--l;
// We can't go left.
if (path[l].offset == 0)
return NodeRef();
// NR is the subtree containing our left sibling.
NodeRef NR = path[l].subtree(path[l].offset - 1);
// Keep right all the way down.
for (++l; l != Level; ++l)
NR = NR.subtree(NR.size() - 1);
return NR;
}