Register

# 12.5. 2-3 Trees¶

## 12.5.1. 2-3 Trees¶

This section presents a data structure called the 2-3 tree. The 2-3 tree is not a binary tree, but instead its shape obeys the following definition:

1. A node contains one or two keys.

2. Every internal node has either two children (if it contains one key) or three children (if it contains two keys). Hence the name.

3. All leaves are at the same level in the tree, so the tree is always height balanced.

In addition to these shape properties, the 2-3 tree has a search tree property analogous to that of a BST. For every node, the values of all descendants in the left subtree are less than the value of the first key, while values in the center subtree are greater than or equal to the value of the first key. If there is a right subtree (equivalently, if the node stores two keys), then the values of all descendants in the center subtree are less than the value of the second key, while values in the right subtree are greater than or equal to the value of the second key. To maintain these shape and search properties requires that special action be taken when nodes are inserted and deleted. The 2-3 tree has the advantage over the BST in that the 2-3 tree can be kept height balanced at relatively low cost. Here is an example 2-3 tree.

Figure 12.5.1: An example of a 2-3 tree.

Nodes are shown as rectangular boxes with two key fields. (These nodes actually would contain complete records or pointers to complete records, but the figures will show only the keys.) Internal nodes with only two children have an empty right key field. Leaf nodes might contain either one or two keys. Here is an implementation for the 2-3 tree node class.

// 2-3 tree node implementation
class TTNode<Key extends Comparable<? super Key>,E> {
private E lval;        // The left record
private Key lkey;        // The node's left key
private E rval;        // The right record
private Key rkey;        // The node's right key
private TTNode<Key,E> left;   // Pointer to left child
private TTNode<Key,E> center; // Pointer to middle child
private TTNode<Key,E> right;  // Pointer to right child

public TTNode() { center = left = right = null; }
public TTNode(Key lk, E lv, Key rk, E rv,
TTNode<Key,E> p1, TTNode<Key,E> p2,
TTNode<Key,E> p3) {
lkey = lk; rkey = rk;
lval = lv; rval = rv;
left = p1; center = p2; right = p3;
}

public boolean isLeaf() { return left == null; }
public TTNode<Key,E> lchild() { return left; }
public TTNode<Key,E> rchild() { return right; }
public TTNode<Key,E> cchild() { return center; }
public Key lkey() { return lkey; }  // Left key
public E lval() { return lval; }  // Left value
public Key rkey() { return rkey; }  // Right key
public E rval() { return rval; }  // Right value
public void setLeft(Key k, E e) { lkey = k; lval = e; }
public void setRight(Key k, E e) { rkey = k; rval = e; }
public void setLeftChild(TTNode<Key,E> it) { left = it; }
public void setCenterChild(TTNode<Key,E> it)
{ center = it; }
public void setRightChild(TTNode<Key,E> it)
{ right = it; }
}


Note that this sample declaration does not distinguish between leaf and internal nodes and so is space inefficient, because leaf nodes store three pointers each. We can use a class hierarcy to implement separate internal and leaf node types.

From the defining rules for 2-3 trees we can derive relationships between the number of nodes in the tree and the depth of the tree. A 2-3 tree of height $k$ has at least $2^{k-1}$ leaves, because if every internal node has two children it degenerates to the shape of a complete binary tree. A 2-3 tree of height $k$ has at most $3^{k-1}$ leaves, because each internal node can have at most three children.

Searching for a value in a 2-3 tree is similar to searching in a BST. Search begins at the root. If the root does not contain the search key $K$, then the search progresses to the only subtree that can possibly contain $K$. The value(s) stored in the root node determine which is the correct subtree. For example, if searching for the value 30 in the tree of Figure 12.5.1, we begin with the root node. Because 30 is between 18 and 33, it can only be in the middle subtree. Searching the middle child of the root node yields the desired record. If searching for 15, then the first step is again to search the root node. Because 15 is less than 18, the first (left) branch is taken. At the next level, we take the second branch to the leaf node containing 15. If the search key were 16, then upon encountering the leaf containing 15 we would find that the search key is not in the tree. Here is an implementation for the 2-3 tree search method.

private E findhelp(TTNode<Key,E> root, Key k) {
if (root == null) { return null; }          // val not found
if (k.compareTo(root.lkey()) == 0) { return root.lval(); }
if ((root.rkey() != null) && (k.compareTo(root.rkey())
== 0))
{ return root.rval(); }
if (k.compareTo(root.lkey()) < 0) {       // Search left
return findhelp(root.lchild(), k);
}
else if (root.rkey() == null) {           // Search center
return findhelp(root.cchild(), k);
}
else if (k.compareTo(root.rkey()) < 0) {  // Search center
return findhelp(root.cchild(), k);
}
else { return findhelp(root.rchild(), k); } // Search right
}


Insertion into a 2-3 tree is similar to insertion into a BST to the extent that the new record is placed in the appropriate leaf node. Unlike BST insertion, a new child is not created to hold the record being inserted, that is, the 2-3 tree does not grow downward. The first step is to find the leaf node that would contain the record if it were in the tree. If this leaf node contains only one value, then the new record can be added to that node with no further modification to the tree, as illustrated in the following visualization.

Settings

Saving...
Server Error
Resubmit

If we insert the new record into a leaf node $L$ that already contains two records, then more space must be created. Consider the two records of node $L$ and the record to be inserted without further concern for which two were already in $L$ and which is the new record. The first step is to split $L$ into two nodes. Thus, a new node—call it $L'$—must be created from free store. $L$ receives the record with the least of the three key values. $L'$ receives the greatest of the three. The record with the middle of the three key value is passed up to the parent node along with a pointer to $L'$. This is called a promotion. The promoted key is then inserted into the parent. If the parent currently contains only one record (and thus has only two children), then the promoted record and the pointer to $L'$ are simply added to the parent node. If the parent is full, then the split-and-promote process is repeated. Here is an example of a a simple promotion.

Settings

Saving...
Server Error
Resubmit

Here is an illustration for what happens when promotions require the root to split, adding a new level to the tree. Note that all leaf nodes continue to have equal depth.

Settings

Saving...
Server Error
Resubmit

Here is an implementation for the insertion process.

private TTNode<Key,E> inserthelp(TTNode<Key,E> rt, Key k, E e) {
TTNode<Key,E> retval;
if (rt == null) {// Empty tree: create a leaf node for root
return new TTNode<Key,E>(k, e, null, null, null, null, null);
}
if (rt.isLeaf()) { // At leaf node: insert here
return rt.add(new TTNode<Key,E>(k, e, null, null, null, null, null));
}
if (k.compareTo(rt.lkey()) < 0) { // Insert left
retval = inserthelp(rt.lchild(), k, e);
if (retval == rt.lchild()) { return rt; }
}
else if((rt.rkey() == null) || (k.compareTo(rt.rkey()) < 0)) {
retval = inserthelp(rt.cchild(), k, e);
if (retval == rt.cchild()) { return rt; }
}
else { // Insert right
retval = inserthelp(rt.rchild(), k, e);
if (retval == rt.rchild()) { return rt; }
}
}

// Add a new key/value pair to the node. There might be a subtree
// associated with the record being added. This information comes
// in the form of a 2-3 tree node with one key and a (possibly null)
// subtree through the center pointer field.
if (rkey == null) { // Only one key, add here
if (lkey.compareTo(it.lkey()) < 0) {
rkey = it.lkey(); rval = it.lval();
center = it.lchild(); right = it.cchild();
}
else {
rkey = lkey; rval = lval; right = center;
lkey = it.lkey(); lval = it.lval();
center = it.cchild();
}
return this;
}
else if (lkey.compareTo(it.lkey()) >= 0) { // Add left
TTNode<Key,E> N1 = new TTNode<Key,E>(lkey, lval, null, null, it, this, null);
it.setLeftChild(left);
left = center; center = right; right = null;
lkey = rkey; lval = rval; rkey = null; rval = null;
return N1;
}
else if (rkey.compareTo(it.lkey()) >= 0) { // Add center
it.setCenterChild(new TTNode<Key,E>(rkey, rval, null, null, it.cchild(), right, null));
it.setLeftChild(this);
rkey = null; rval = null; right = null;
return it;
}
TTNode<Key,E> N1 = new TTNode<Key,E>(rkey, rval, null, null, this, it, null);
it.setLeftChild(right);
right = null; rkey = null; rval = null;
return N1;
}
}


Note that inserthelp takes three parameters. The first is a pointer to the root of the current subtree, named rt. The second is the key for the record to be inserted, and the third is the record itself. The return value for inserthelp is a pointer to a 2-3 tree node. If rt is unchanged, then a pointer to rt is returned. If rt is changed (due to the insertion causing the node to split), then a pointer to the new subtree root is returned, with the key value and record value in the leftmost fields, and a pointer to the (single) subtree in the center pointer field. This revised node will then be added to the parent as illustrated by the splitting visualization above.

When deleting a record from the 2-3 tree, there are three cases to consider. The simplest occurs when the record is to be removed from a leaf node containing two records. In this case, the record is simply removed, and no other nodes are affected. The second case occurs when the only record in a leaf node is to be removed. The third case occurs when a record is to be removed from an internal node. In both the second and the third cases, the deleted record is replaced with another that can take its place while maintaining the correct order, similar to removing a node from a BST. If the tree is sparse enough, there is no such record available that will allow all nodes to still maintain at least one record. In this situation, sibling nodes are merged together. The delete operation for the 2-3 tree is excessively complex and will not be described further. Instead, a complete discussion of deletion will be postponed until the next section, where it can be generalized for a particular variant of the B-tree.

The 2-3 tree insert and delete routines do not add new nodes at the bottom of the tree. Instead they cause leaf nodes to split or merge, possibly causing a ripple effect moving up the tree to the root. If necessary the root will split, causing a new root node to be created and making the tree one level deeper. On deletion, if the last two children of the root merge, then the root node is removed and the tree will lose a level. In either case, all leaf nodes are always at the same level. When all leaf nodes are at the same level, we say that a tree is height balanced. Because the 2-3 tree is height balanced, and every internal node has at least two children, we know that the maximum depth of the tree is $\log n$. Thus, all 2-3 tree insert, find, and delete operations require $\Theta(\log n)$ time.

Click here for another visualization that will let you construct and interact with a 2-3 tree. Actually, this visualization is for a data structure that is more general than just a 2-3 tree. To see how a 2-3 would behave, be sure to use the “Max Degree = 3” setting. This visualization was written by David Galles of the University of San Francisco as part of his Data Structure Visualizations package.