A 
 is a special kind of binary tree in which each node, 
,
also stores a data value, 
, from some total order.  The data values in
a binary search tree obey the binary search tree property:  For
a node, 
, every data value stored in the subtree rooted at 
is less than 
 and every data value stored in the subtree rooted at
 is greater than 
.  An example of a 
 is shown in Figure 6.5.
The binary search tree property is extremely useful because it allows
us to quickly locate a value, 
, in a binary search tree.  To do this we start
searching for 
 at the root, 
.  When examining a node, 
, there
are three cases:
  T findEQ(T x) {
    Node *w = r;
    while (w != nil) {
      int comp = compare(x, w->x);
      if (comp < 0) {
        w = w->left;
      } else if (comp > 0) {
        w = w->right;
      } else {
        return w->x;
      }
    }
    return null;
  }
Two examples of searches in a binary search tree are shown in
Figure 6.6.  As the second example shows, even if we don't find 
in the tree, we still gain some valuable information.  If we look at
the last node, 
, at which Case 1 occurred, we see that 
 is the smallest
value in the tree that is greater than 
.  Similarly, the last node
at which Case 2 occurred contains the largest value in the tree that is
less than 
.  Therefore, by keeping track of the last node, 
,
at which Case 1 occurs, a 
 can implement the 
operation that returns the smallest value stored in the tree that is
greater than or equal to 
:
  T find(T x) {
    Node *w = r, *z = nil;
    while (w != nil) {
      int comp = compare(x, w->x);
      if (comp < 0) {
        z = w;
        w = w->left;
      } else if (comp > 0) {
        w = w->right;
      } else {
        return w->x;
      }
    }
    return z == nil ? null : z->x;
  }
  | 
To add a new value, 
, to a 
, we first search for
. If we find it, then there is no need to insert it.  Otherwise,
we store 
 at a leaf child of the last node, 
, encountered during the
search for 
. Whether the new node is the left or right child of 
 depends on the result of comparing 
 and 
.
  bool add(T x) {
    Node *p = findLast(x);
    Node *u = new Node;
    u->x = x;
    return addChild(p, u);
  }
  Node* findLast(T x) {
    Node *w = r, *prev = nil;
    while (w != nil) {
      prev = w;
      int comp = compare(x, w->x);
      if (comp < 0) {
        w = w->left;
      } else if (comp > 0) {
        w = w->right;
      } else {
        return w;
      }
    }
    return prev;
  }
  bool addChild(Node *p, Node *u) {
      if (p == nil) {
        r = u;              // inserting into empty tree
      } else {
        int comp = compare(u->x, p->x);
        if (comp < 0) {
          p->left = u;
        } else if (comp > 0) {
          p->right = u;
        } else {
          return false;   // u.x is already in the tree
        }
        u->parent = p;
      }
      n++;
      return true;
    }
An example is shown in Figure 6.7. The most time-consuming
part of this process is the initial search for 
Deleting a value stored in a node, 
, of a 
 is a
little more difficult.  If 
 is a leaf, then we can just detach 
from its parent.  Even better: If 
 has only one child, then we can
splice 
 from the tree by having 
 adopt 
's child:
  void splice(Node *u) {
    Node *s, *p;
    if (u->left != nil) {
      s = u->left;
    } else {
      s = u->right;
    }
    if (u == r) {
      r = s;
      p = nil;
    } else {
      p = u->parent;
      if (p->left == u) {
        p->left = s;
      } else {
        p->right = s;
      }
    }
    if (s != nil) {
      s->parent = p;
    }
    n--;
  }
Things get tricky, though, when 
 has two children.  In this case,
the simplest thing to do is to find a node, 
, that has less than
two children such that we can replace 
 with 
.  To maintain
the binary search tree property, the value 
 should be close to the
value of 
.  For example, picking 
 such that 
 is the smallest
value greater than 
 will do.  Finding the node 
 is easy; it is
the smallest value in the subtree rooted at 
.  This node can
be easily removed because it has no left child.  (See Figure 6.9)
  void remove(Node *u) {
    if (u->left == nil || u->right == nil) {
      splice(u);
      delete u;
    } else {
      Node *w = u->right;
      while (w->left != nil)
        w = w->left;
      u->x = w->x;
      splice(w);
      delete w;
    }
  }
  | 
The 
, 
, and 
 operations in a
 each involve following a path from the root of the
tree to some node in the tree. Without knowing more about the shape of
the tree it is difficult to say much about the length of this path,
except that it is less than 
, the number of nodes in the tree.
The following (unimpressive) theorem summarizes the performance of the
 data structure:
Theorem 6.1 compares poorly with Theorem 4.1, which shows
that the 
 structure can implement the 
 interface
with 
 expected time per operation.  The problem with the
 structure is that it can become unbalanced.
Instead of looking like the tree in Figure 6.5 it can look like a long
chain of 
 nodes, all but the last having exactly one child.
There are a number of ways of avoiding unbalanced binary search
trees, all of which lead to data structures that have 
 time operations. In Chapter 7 we show how 
expected time operations can be achieved with randomization.
In Chapter 8 we show how 
 amortized
time operations can be achieved with partial rebuilding operations.
In Chapter 9 we show how 
 worst-case
time operations can be achieved by simulating a tree that is not binary:
a tree in which nodes can have up to four children.
opendatastructures.org