13.1

A `BinaryTrie` encodes a set of
bit integers in a binary tree.
All leaves in the tree have depth
and each integer is encoded as a
root-to-leaf path. The path for the integer
turns left at level
if the
th most significant bit of
is a 0 and turns right if it
is a 1. Figure 13.1 shows an example for the case
,
in which the trie stores the integers 3(0011), 9(1001), 12(1100),
and 13(1101).

Because the search path for a value depends on the bits of , it will be helpful to name the children of a node, , ( ) and ( ). These child pointers will actually serve double-duty. Since the leaves in a binary trie have no children, the pointers are used to string the leaves together into a doubly-linked list. For a leaf in the binary trie ( ) is the node that comes before in the list and ( ) is the node that follows in the list. A special node, , is used both before the first node and after the last node in the list (see Section 3.2).

Each node,
, also contains an additional pointer
. If
's
left child is missing, then
points to the smallest leaf in
's subtree. If
's right child is missing, then
points
to the largest leaf in
's subtree. An example of a `BinaryTrie`,
showing
pointers and the doubly-linked list at the leaves, is
shown in Figure 13.2.

The
operation in a `BinaryTrie` is fairly straightforward.
We try to follow the search path for
in the trie. If we reach a leaf,
then we have found
. If we reach a node
where we cannot proceed
(because
is missing a child), then we follow
, which takes us
either to the smallest leaf larger than
or the largest leaf smaller than
. Which of these two cases occurs depends on whether
is missing
its left or right child, respectively. In the former case (
is missing
its left child), we have found the node we want. In the latter case (
is missing its right child), we can use the linked list to reach the node
we want. Each of these cases is illustrated in Figure 13.3.

T find(T x) { int i, c = 0, ix = it.intValue(x); Node u = r; for (i = 0; i < w; i++) { c = (ix >>> w-i-1) & 1; if (u.child[c] == null) break; u = u.child[c]; } if (i == w) return u.x; // found it u = (c == 0) ? u.jump : u.jump.child[next]; return u == dummy ? null : u.x; }The running-time of the method is dominated by the time it takes to follow a root-to-leaf path, so it runs in time.

The
operation in a `BinaryTrie` is also fairly straightforward,
but has a lot of work to do:

- It follows the search path for until reaching a node where it can no longer proceed.
- It creates the remainder of the search path from to a leaf that contains .
- It adds the node, , containing to the linked list of leaves (it has access to the predecessor, , of in the linked list from the pointer of the last node, , encountered during step 1.)
- It walks back up the search path for adjusting pointers at the nodes whose pointer should now point to .

boolean add(T x) { int i, c = 0, ix = it.intValue(x); Node u = r; // 1 - search for ix until falling out of the trie for (i = 0; i < w; i++) { c = (ix >>> w-i-1) & 1; if (u.child[c] == null) break; u = u.child[c]; } if (i == w) return false; // already contains x - abort Node pred = (c == right) ? u.jump : u.jump.child[0]; u.jump = null; // u will have two children shortly // 2 - add path to ix for (; i < w; i++) { c = (ix >>> w-i-1) & 1; u.child[c] = newNode(); u.child[c].parent = u; u = u.child[c]; } u.x = x; // 3 - add u to linked list u.child[prev] = pred; u.child[next] = pred.child[next]; u.child[prev].child[next] = u; u.child[next].child[prev] = u; // 4 - walk back up, updating jump pointers Node v = u.parent; while (v != null) { if ((v.child[left] == null && (v.jump == null || it.intValue(v.jump.x) > ix)) || (v.child[right] == null && (v.jump == null || it.intValue(v.jump.x) < ix))) v.jump = u; v = v.parent; } n++; return true; }This method performs one walk down the search path for and one walk back up. Each step of these walks takes constant time, so the method runs in time.

The operation undoes the work of . Like , it has a lot of work to do:

- It follows the search path for until reaching the leaf, , containing .
- It removes from the doubly-linked list.
- It deletes and then walks back up the search path for deleting nodes until reaching a node that has a child that is not on the search path for .
- It walks upwards from to the root updating any pointers that point to .

boolean remove(T x) { // 1 - find leaf, u, containing x int i, c, ix = it.intValue(x); Node u = r; for (i = 0; i < w; i++) { c = (ix >>> w-i-1) & 1; if (u.child[c] == null) return false; u = u.child[c]; } // 2 - remove u from linked list u.child[prev].child[next] = u.child[next]; u.child[next].child[prev] = u.child[prev]; Node v = u; // 3 - delete nodes on path to u for (i = w-1; i >= 0; i--) { c = (ix >>> w-i-1) & 1; v = v.parent; v.child[c] = null; if (v.child[1-c] != null) break; } // 4 - update jump pointers c = (ix >>> w-i-1) & 1; v.jump = u.child[1-c]; v = v.parent; i--; for (; i >= 0; i--) { c = (ix >>> w-i-1) & 1; if (v.jump == u) v.jump = u.child[1-c]; v = v.parent; } n--; return true; }

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