Binary search tree

Binary search requires an order relation by which every element (item) can be compared with every other element in the sense of a total preorder. The part of the element which effectively takes place in the comparison is called its key. Whether duplicates, i. e. different elements with the same key, shall be allowed in the tree or not, does not depend on the order relation, but on the application only. For a search function supporting and handling duplicates in a tree, see section Searching with duplicates allowed.

In the context of binary search trees, a total preorder is realized most flexibly by means of a three-way comparison subroutine.

Searching in a binary search tree for a specific key can be programmed recursively or iteratively.

We begin by examining the root node. If the tree is null, the key we are searching for does not exist in the tree. Otherwise, if the key equals that of the root, the search is successful and we return the node. If the key is less than that of the root, we search the left subtree. Similarly, if the key is greater than that of the root, we search the right subtree. This process is repeated until the key is found or the remaining subtree is null. If the searched key is not found after a null subtree is reached, then the key is not present in the tree. This is easily expressed as a recursive algorithm (implemented in Python):

def search_recursively(key, node):
    if node == None or node.key == key:
        return node
    if key < node.key:
        return search_recursively(key, node.left)
    if key > node.key:
        return search_recursively(key, node.right)

The same algorithm can be implemented iteratively:

def search_iteratively(key, node):
    current_node = node
    while current_node != None:
        if key == current_node.key:
            return current_node
        if key < current_node.key:
            current_node = current_node.left
        else:  # key > current_node.key:
            current_node = current_node.right
    return current_node

These two examples do not support duplicates, that is, they rely on the order relation being a total order.

One can note that the recursive algorithm is tail recursive. In a language supporting tail call optimization, the recursive and iterative examples will compile to equivalent programs.

Because in the worst case this algorithm must search from the root of the tree to the leaf farthest from the root, the search operation takes time proportional to the tree's height (see tree terminology). On average, binary search trees with Nodes keys have O(log |Nodes|) height.[note 1] However, in the worst case, binary search trees can have O(|Nodes|) height, when the unbalanced tree resembles a linked list (degenerate tree).

If the order relation is only a total preorder, a reasonable extension of the functionality is the following: also in case of equality search down to the leaves. Thereby allowing to specify (or hard-wire) a direction, where to insert a duplicate, either to the right or left of all duplicates in the tree so far. If the direction is hard-wired, both choices, right and left, support a stack with insert duplicate as push operation and delete as the pop operation.[2]^:155

def search_duplicatesAllowed(key, node):
    current_node = node
    while current_node != None:
        if key < current_node.key:
            current_node = current_node.left
        else:  # key >= current_node.key:
            current_node = current_node.right
    return current_node

A binary tree sort equipped with such a search function becomes stable.

Insertion begins as a search would begin; if the key is not equal to that of the root, we search the left or right subtrees as before. Eventually, we will reach an external node and add the new key-value pair (here encoded as a record 'newNode') as its right or left child, depending on the node's key. In other words, we examine the root and recursively insert the new node to the left subtree if its key is less than that of the root, or the right subtree if its key is greater than or equal to the root.

Here's how a typical binary search tree insertion might be performed in a binary tree in C++:

void insert(Node*& root, int key, int value) {
  if (!root) 
    root = new Node(key, value);
  else if (key == root->key)
    root->value = value;
  else if (key < root->key)
    insert(root->left, key, value);
  else  // key > root->key
    insert(root->right, key, value);
}

Alternatively, a non-recursive version might be implemented like this. Using a pointer-to-pointer to keep track of where we came from lets the code avoid explicit checking for and handling of the case where it needs to insert a node at the tree root:[3]

void insert(Node** root, int key, int value) {
  Node **walk = root;
  while (*walk) { 
    int curkey = (*walk)->key;
    if (curkey == key) {
      (*walk)->value = value;
      return;
    }
    if (key > curkey) 
      walk = &(*walk)->right;
    else 
      walk = &(*walk)->left;
  }
  *walk = new Node(key, value);
}

The above destructive procedural variant modifies the tree in place. It uses only constant heap space (and the iterative version uses constant stack space as well), but the prior version of the tree is lost. Alternatively, as in the following Python example, we can reconstruct all ancestors of the inserted node; any reference to the original tree root remains valid, making the tree a persistent data structure:

def binary_tree_insert(node, key, value):
    if node == None:
        return NodeTree(None, key, value, None)
    if key == node.key:
        return NodeTree(node.left, key, value, node.right)
    if key < node.key:
        return NodeTree(binary_tree_insert(node.left, key, value), node.key, node.value, node.right)
    return NodeTree(node.left, node.key, node.value, binary_tree_insert(node.right, key, value))

The part that is rebuilt uses O(log n) space in the average case and O(n) in the worst case.

In either version, this operation requires time proportional to the height of the tree in the worst case, which is O(log n) time in the average case over all trees, but O(n) time in the worst case.

Another way to explain insertion is that in order to insert a new node in the tree, its key is first compared with that of the root. If its key is less than the root's, it is then compared with the key of the root's left child. If its key is greater, it is compared with the root's right child. This process continues, until the new node is compared with a leaf node, and then it is added as this node's right or left child, depending on its key: if the key is less than the leaf's key, then it is inserted as the leaf's left child, otherwise as the leaf's right child.

There are other ways of inserting nodes into a binary tree, but this is the only way of inserting nodes at the leaves and at the same time preserving the BST structure.

When removing a node from a binary search tree it is mandatory to maintain the in-order sequence of the nodes. There are many possibilities to do this. However, the following method which has been proposed by T. Hibbard in 1962[4] guarantees that the heights of the subject subtrees are changed by at most one. There are three possible cases to consider:

• Deleting a node with no children: simply remove the node from the tree.
• Deleting a node with one child: remove the node and replace it with its child.
• Deleting a node D with two children: choose either D's in-order predecessor or in-order successor E (see figure). Instead of deleting D, overwrite its key and value with E's.[note 2] If E does not have a child, remove E from its previous parent G. If E has a child F, it is a right child, so that it is to replace E at E's parent.

Deleting a node with two children from a binary search tree. First the leftmost node in the right subtree, the in-order successor E, is identified. Its value is copied into the node D being deleted. The in-order successor can then be easily deleted because it has at most one child. The same method works symmetrically using the in-order predecessor C.

In all cases, when D happens to be the root, make the replacement node root again.

Nodes with two children are harder to delete. A node's in-order successor is its right subtree's left-most child, and a node's in-order predecessor is the left subtree's right-most child. In either case, this node will have only one or no child at all. Delete it according to one of the two simpler cases above.

Consistently using the in-order successor or the in-order predecessor for every instance of the two-child case can lead to an unbalanced tree, so some implementations select one or the other at different times.

Runtime analysis: Although this operation does not always traverse the tree down to a leaf, this is always a possibility; thus in the worst case it requires time proportional to the height of the tree. It does not require more even when the node has two children, since it still follows a single path and does not visit any node twice.

def find_min(self):
    """Get minimum node in a subtree."""
    current_node = self
    while current_node.left_child:
        current_node = current_node.left_child
    return current_node

def replace_node_in_parent(self, new_value=None) -> None:
    if self.parent:
        if self == self.parent.left_child:
            self.parent.left_child = new_value
        else:
            self.parent.right_child = new_value
    if new_value:
        new_value.parent = self.parent

def binary_tree_delete(self, key) -> None:
    if key < self.key:
        self.left_child.binary_tree_delete(key)
        return
    if key > self.key:
        self.right_child.binary_tree_delete(key)
        return
    # Delete the key here
    if self.left_child and self.right_child:  # If both children are present
        successor = self.right_child.find_min()
        self.key = successor.key
        successor.binary_tree_delete(successor.key)
    elif self.left_child:  # If the node has only a *left* child
        self.replace_node_in_parent(self.left_child)
    elif self.right_child:  # If the node has only a *right* child
        self.replace_node_in_parent(self.right_child)
    else:
        self.replace_node_in_parent(None)  # This node has no children

Once the binary search tree has been created, its elements can be retrieved in-order by recursively traversing the left subtree of the root node, accessing the node itself, then recursively traversing the right subtree of the node, continuing this pattern with each node in the tree as it's recursively accessed. As with all binary trees, one may conduct a pre-order traversal or a post-order traversal, but neither are likely to be useful for binary search trees. An in-order traversal of a binary search tree will always result in a sorted list of node items (numbers, strings or other comparable items).

The code for in-order traversal in Python is given below. It will call callback (some function the programmer wishes to call on the node's value, such as printing to the screen) for every node in the tree.

def traverse_binary_tree(node, callback):
    if node == None:
        return
    traverse_binary_tree(node.leftChild, callback)
    callback(node.value)
    traverse_binary_tree(node.rightChild, callback)

Traversal requires O(n) time, since it must visit every node. This algorithm is also O(n), so it is asymptotically optimal.

Traversal can also be implemented iteratively. For certain applications, e.g. greater equal search, approximative search, an operation for single step (iterative) traversal can be very useful. This is, of course, implemented without the callback construct and takes O(1) on average and O(log n) in the worst case.

Sometimes we already have a binary tree, and we need to determine whether it is a BST. This problem has a simple recursive solution.

The BST property—every node on the right subtree has to be larger than the current node and every node on the left subtree has to be smaller than the current node—is the key to figuring out whether a tree is a BST or not. The greedy algorithm—simply traverse the tree, at every node check whether the node contains a value larger than the value at the left child and smaller than the value on the right child—does not work for all cases. Consider the following tree:

     20
    /  \
  10    30
       /  \
      5    40

In the tree above, each node meets the condition that the node contains a value larger than its left child and smaller than its right child hold, and yet it is not a BST: the value 5 is on the right subtree of the node containing 20, a violation of the BST property.

Instead of making a decision based solely on the values of a node and its children, we also need information flowing down from the parent as well. In the case of the tree above, if we could remember about the node containing the value 20, we would see that the node with value 5 is violating the BST property contract.

So the condition we need to check at each node is:

• if the node is the left child of its parent, then it must be smaller than (or equal to) the parent and it must pass down the value from its parent to its right subtree to make sure none of the nodes in that subtree is greater than the parent
• if the node is the right child of its parent, then it must be larger than the parent and it must pass down the value from its parent to its left subtree to make sure none of the nodes in that subtree is lesser than the parent.

A recursive solution in C++ can explain this further:

struct TreeNode {
    int key;
    int value;
    struct TreeNode *left;
    struct TreeNode *right;
};

bool isBST(struct TreeNode *node, int minKey, int maxKey) {
    if (node == NULL) return true;
    if (node->key < minKey || node->key > maxKey) return false;
    
    return isBST(node->left, minKey, node->key-1) && isBST(node->right, node->key+1, maxKey);
}

node->key+1 and node->key-1 are done to allow only distinct elements in BST.

If we want the same elements to also be present, then we can use only node->key in both places.

The initial call to this function can be something like this:

if (isBST(root, INT_MIN, INT_MAX)) {
    puts("This is a BST.");
} else {
    puts("This is NOT a BST!");
}

Essentially we keep creating a valid range (starting from [MIN_VALUE, MAX_VALUE]) and keep shrinking it down for each node as we go down recursively.

As pointed out in section #Traversal, an in-order traversal of a binary search tree returns the nodes sorted. Thus we only need to keep the last visited node while traversing the tree and check whether its key is smaller (or smaller/equal, if duplicates are to be allowed in the tree) compared to the current key.

There are also parallel algorithms for binary search trees, including inserting/deleting multiple elements, construction from array, filter with certain predicator, flatten to an array, merging/substracting/intersecting two trees, etc. These algorithms can be implemented using join-based tree algorithms, which can also keep the tree balanced using several balancing schemes (including AVL tree, red-black tree, weight-balanced tree and treap) .

A binary search tree can be used to implement a simple sorting algorithm. Similar to heapsort, we insert all the values we wish to sort into a new ordered data structure—in this case a binary search tree—and then traverse it in order.

The worst-case time of build_binary_tree is O(n²)—if you feed it a sorted list of values, it chains them into a linked list with no left subtrees. For example, build_binary_tree([1, 2, 3, 4, 5]) yields the tree (1 (2 (3 (4 (5))))).

There are several schemes for overcoming this flaw with simple binary trees; the most common is the self-balancing binary search tree. If this same procedure is done using such a tree, the overall worst-case time is O(n log n), which is asymptotically optimal for a comparison sort. In practice, the added overhead in time and space for a tree-based sort (particularly for node allocation) make it inferior to other asymptotically optimal sorts such as heapsort for static list sorting. On the other hand, it is one of the most efficient methods of incremental sorting, adding items to a list over time while keeping the list sorted at all times.

Binary search trees can serve as priority queues: structures that allow insertion of arbitrary key as well as lookup and deletion of the minimum (or maximum) key. Insertion works as previously explained. Find-min walks the tree, following left pointers as far as it can without hitting a leaf:

// Precondition: T is not a leaf
function find-min(T):
    while hasLeft(T):
        T ? left(T)
    return key(T)

Find-max is analogous: follow right pointers as far as possible. Delete-min (max) can simply look up the minimum (maximum), then delete it. This way, insertion and deletion both take logarithmic time, just as they do in a binary heap, but unlike a binary heap and most other priority queue implementations, a single tree can support all of find-min, find-max, delete-min and delete-max at the same time, making binary search trees suitable as double-ended priority queues.[2]^:156

D. A. Heger (2004)[5] presented a performance comparison of binary search trees. Treap was found to have the best average performance, while red-black tree was found to have the smallest number of performance variations.

Tree rotations are very common internal operations in binary trees to keep perfect, or near-to-perfect, internal balance in the tree.

If a search tree is not intended to be modified, and it is known exactly how often each item will be accessed, it is possible to construct[6] an optimal binary search tree, which is a search tree where the average cost of looking up an item (the expected search cost) is minimized.

Even if we only have estimates of the search costs, such a system can considerably speed up lookups on average. For example, if we have a BST of English words used in a spell checker, we might balance the tree based on word frequency in text corpora, placing words like the near the root and words like agerasia near the leaves. Such a tree might be compared with Huffman trees, which similarly seek to place frequently used items near the root in order to produce a dense information encoding; however, Huffman trees store data elements only in leaves, and these elements need not be ordered.

If the sequence in which the elements in the tree will be accessed is unknown in advance, splay trees can be used which are asymptotically as good as any static search tree we can construct for any particular sequence of lookup operations.

Alphabetic trees are Huffman trees with the additional constraint on order, or, equivalently, search trees with the modification that all elements are stored in the leaves. Faster algorithms exist for optimal alphabetic binary trees (OABTs).