本文共 22608 字,大约阅读时间需要 75 分钟。
from:
| ||
---|---|---|
in | ||
Average | Worst case | |
Space | O(n) | O(n) |
Search | O(log n) | O(n) |
Insert | O(log n) | O(n) |
Delete | O(log n) | O(n) |
In , a binary search tree (BST), which may sometimes also be called an ordered or sorted binary tree, is a which has the following properties:
Generally, the information represented by each node is a record rather than a single data element. However, for sequencing purposes, nodes are compared according to their keys rather than any part of their associated records.
The major advantage of binary search trees over other data structures is that the related and such as can be very efficient.
Binary search trees are a fundamental used to construct more abstract data structures such as , , and .
Contents[] |
Operations on a binary search tree require comparisons between nodes. These comparisons are made with calls to a comparator, which is a that computes the total order (linear order) on any two values. This comparator can be explicitly or implicitly defined, depending on the language in which the BST is implemented.
Searching a binary search tree for a specific value can be a or process. This explanation covers a recursive method.
We begin by examining the . If the tree is null, the value we are searching for does not exist in the tree. Otherwise, if the value equals the root, the search is successful. If the value is less than the root, search the left subtree. Similarly, if it is greater than the root, search the right subtree. This process is repeated until the value is found or the indicated subtree is null. If the searched value is not found before a null subtree is reached, then the item must not be present in the tree.
Here is the search algorithm in the :
# 'node' refers to the parent-node in this case def search_binary_tree(node, key): if node is None: return None # key not found if key < node.key: return search_binary_tree(node.leftChild, key) elif key > node.key: return search_binary_tree(node.rightChild, key) else: # key is equal to node key return node.value # found key
… or equivalent :
searchBinaryTree _ NullNode = Nothing searchBinaryTree key (Node nodeKey nodeValue (leftChild, rightChild)) = case compare key nodeKey of LT -> searchBinaryTree key leftChild GT -> searchBinaryTree key rightChild EQ -> Just nodeValue
This operation requires (log n) time in the average case, but needs (n) time in the worst case, when the unbalanced tree resembles a ().
Assuming that BinarySearchTree is a class with a member function "search(int)" and a pointer to the root node, the algorithm is also easily implemented in terms of an iterative approach. The algorithm enters a loop, and decides whether to branch left or right depending on the value of the node at each parent node.
bool BinarySearchTree::search(int val){ Node *next = this->root(); while (next != NULL) { if (val == next->value()) { return true; } else if (val < next->value()) { next = next->left(); } else { next = next->right(); } } //not found return false;}
Insertion begins as a search would begin; if the root is not equal to the value, we search the left or right subtrees as before. Eventually, we will reach an external node and add the value as its right or left child, depending on the node's value. In other words, we examine the root and recursively insert the new node to the left subtree if the new value is less than the root, or the right subtree if the new value is greater than or equal to the root.
Here's how a typical binary search tree insertion might be performed in :
/* Inserts the node pointed to by "newNode" into the subtree rooted at "treeNode" */ void InsertNode(Node* &treeNode, Node *newNode) { if (treeNode == NULL) treeNode = newNode; else if (newNode->key < treeNode->key) InsertNode(treeNode->left, newNode); else InsertNode(treeNode->right, newNode); }
The above "destructive" procedural variant modifies the tree in place. It uses only constant space, but the previous version of the tree is lost. Alternatively, as in the following example, we can reconstruct all ancestors of the inserted node; any reference to the original tree root remains valid, making the tree a :
def binary_tree_insert(node, key, value): if node is None: return TreeNode(None, key, value, None) if key == node.key: return TreeNode(node.left, key, value, node.right) if key < node.key: return TreeNode(binary_tree_insert(node.left, key, value), node.key, node.value, node.right) else: return TreeNode(node.left, node.key, node.value, binary_tree_insert(node.right, key, value))
The part that is rebuilt uses Θ(log n) space in the average case and O(n) in the worst case (see ).
In either version, this operation requires time proportional to the height of the tree in the worst case, which is (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 value is first compared with the value of the root. If its value is less than the root's, it is then compared with the value of the root's left child. If its value 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 value.
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.
Here is an iterative approach to inserting into a binary search tree in :
private Node m_root; public void insert(int data) { if (m_root == null) { m_root = new TreeNode(data, null, null); return; } Node root = m_root; while (root != null) { // Not the same value twice if (data == root.getData()) { return; } else if (data < root.getData()) { // insert left if (root.getLeft() == null) { root.setLeft(new TreeNode(data, null, null)); return; } else { root = root.getLeft(); } } else { // insert right if (root.getRight() == null) { root.setRight(new TreeNode(data, null, null)); return; } else { root = root.getRight(); } } }}
Below is a recursive approach to the insertion method.
private Node m_root; public void insert(int data){ if (m_root == null) { m_root = TreeNode(data, null, null); }else{ internalInsert(m_root, data); }} private static void internalInsert(Node node, int data){ // Not the same value twice if (data == node.getValue()) { return; } else if (data < node.getValue()) { if (node.getLeft() == null) { node.setLeft(new TreeNode(data, null, null)); }else{ internalInsert(node.getLeft(), data); } }else{ if (node.getRight() == null) { node.setRight(new TreeNode(data, null, null)); }else{ internalInsert(node.getRight(), data); } }}
There are three possible cases to consider:
As with all binary trees, a node's in-order successor is the left-most child of its right subtree, and a node's in-order predecessor is the right-most child of its left subtree. In either case, this node will have zero or one children. 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 tree, so good implementations add inconsistency to this selection.
Running Time 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.
Here is the code in Python:
def findMin(self): ''' Finds the smallest element that is a child of *self* ''' 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): ''' Removes the reference to *self* from *self.parent* and replaces it with *new_value*. ''' 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): if key < self.key: self.left_child.binary_tree_delete(key) elif key > self.key: self.right_child.binary_tree_delete(key) else: # delete the key here if self.left_child and self.right_child: # if both children are present # get the smallest node that's bigger than *self* successor = self.right_child.findMin() self.key = successor.key # if *successor* has a child, replace it with that # at this point, it can only have a *right_child* # if it has no children, *right_child* will be "None" successor.replace_node_in_parent(successor.right_child) elif self.left_child or self.right_child: # if the node has only one child if self.left_child: self.replace_node_in_parent(self.left_child) else: self.replace_node_in_parent(self.right_child) else: # this node has no children self.replace_node_in_parent(None)
bool BinarySearchTree::remove(int value) { if (root == NULL) return false; else { if (root->getValue() == value) { BSTNode auxRoot(0); auxRoot.setLeftChild(root); BSTNode* removedNode = root->remove(value, &auxRoot); root = auxRoot.getLeft(); if (removedNode != NULL) { delete removedNode; return true; } else return false; } else { BSTNode* removedNode = root->remove(value, NULL); if (removedNode != NULL) { delete removedNode; return true; } else return false; } }} BSTNode* BSTNode::remove(int value, BSTNode *parent) { if (value < this->value) { if (left != NULL) return left->remove(value, this); else return NULL; } else if (value > this->value) { if (right != NULL) return right->remove(value, this); else return NULL; } else { if (left != NULL && right != NULL) { this->value = right->minValue(); return right->remove(this->value, this); } else if (parent->left == this) { parent->left = (left != NULL) ? left : right; return this; } else if (parent->right == this) { parent->right = (left != NULL) ? left : right; return this; } }} int BSTNode::minValue() { if (left == NULL) return value; else return left->minValue();}
Once the binary search tree has been created, its elements can be retrieved by 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 or a , but neither are likely to be useful for binary search trees.
The code for in-order traversal in Python is given below. It will call callback for every node in the tree.
def traverse_binary_tree(node, callback): if node is None: return traverse_binary_tree(node.leftChild, callback) callback(node.value) traverse_binary_tree(node.rightChild, callback)
Traversal requires time, since it must visit every node. This algorithm is also O(n), so it is .
The Code for in-order traversal in Language C is given below.
void InOrderTraversal(struct Node *n){ struct Node *Cur, *Pre; if(n==NULL) return; Cur = n; while(Cur != NULL) { if(Cur->lptr == NULL) { printf("\t%d",Cur->val); Cur= Cur->rptr; } else { Pre = Cur->lptr; while(Pre->rptr !=NULL && Pre->rptr != Cur) Pre = Pre->rptr; if (Pre->rptr == NULL) { Pre->rptr = Cur; Cur = Cur->lptr; } else { Pre->rptr = NULL; printf("\t%d",Cur->val); Cur = Cur->rptr; } } }}
A binary search tree can be used to implement a simple but efficient . Similar to , 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, building our result:
def build_binary_tree(values): tree = None for v in values: tree = binary_tree_insert(tree, v) return tree def get_inorder_traversal(root): ''' Returns a list containing all the values in the tree, starting at *root*. Traverses the tree in-order(leftChild, root, rightChild). ''' result = [] traverse_binary_tree(root, lambda element: result.append(element)) return result
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 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 . If this same procedure is done using such a tree, the overall worst-case time is(nlog n), which is for a . In practice, the poor performance and added overhead in time and space for a tree-based sort (particularly for node ) make it inferior to other asymptotically optimal sorts such as 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.
There are many types of binary search trees. and are both forms of . A is a binary search tree that automatically moves frequently accessed elements nearer to the root. In a ("tree "), each node also holds a (randomly chosen) priority and the parent node has higher priority than its children. are trees optimized for fast searches.
Two other titles describing binary search trees are that of a complete and degenerate tree.
A complete tree is a tree with n levels, where for each level d <= n - 1, the number of existing nodes at level d is equal to 2d. This means all possible nodes exist at these levels. An additional requirement for a complete binary tree is that for the nth level, while every node does not have to exist, the nodes that do exist must fill from left to right.
A degenerate tree is a tree where for each parent node, there is only one associated child node. What this means is that in a performance measurement, the tree will essentially behave like a linked list data structure.
D. A. Heger (2004) presented a performance comparison of binary search trees. was found to have the best average performance, while was found to have the smallest amount of performance fluctuations.
If we don't plan on modifying a search tree, and we know exactly how often each item will be accessed, we can construct 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 you have a BST of English words used in a , you might balance the tree based on word frequency in , placing words like "the" near the root and words like "agerasia" near the leaves. Such a tree might be compared with , which similarly seek to place frequently-used items near the root in order to produce a dense information encoding; however, Huffman trees only store data elements in leaves and these elements need not be ordered.
If we do not know the sequence in which the elements in the tree will be accessed in advance, we can use 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).
Example:
procedure Optimum Search Tree(f, f´, c): for j = 0 to n do c[j, j] = 0, F[j, j] = f´j for d = 1 to n do for i = 0 to (n − d) do j = i + d F[i, j] = F[i, j − 1] + f´ + f´j c[i, j] = MIN(i<=j){c[i, k − 1] + c[k, j]} + F[i, j]
二叉查找树(Binary Search Tree),或者是一棵空树,或者是具有下列性质的:
二叉排序树的查找过程和类似,通常采取二叉作为二叉排序树的。中序遍历二叉排序树可得到一个关键字的有序序列,一个无序序列可以通过构造一棵二叉排序树变成一个有序序列,构造树的过程即为对无序序列进行排序的过程。每次插入的新的结点都是二叉排序树上新的叶子结点,在进行插入操作时,不必移动其它结点,只需改动某个结点的指针,由空变为非空即可。搜索,插入,删除的复杂度等于树高,期望O(logn),最坏O(n)(数列有序,树退化成线性表).
虽然二叉排序树的最坏效率是O(n),但它支持动态查询,且有很多改进版的二叉排序树可以使树高为O(logn),如,,等.故不失为一种好的动态排序方法.
目录[] |
在二元排序樹b中查找x的過程為:
/* 以下代码为C++写成, 下同 */Status SearchBST(BiTree T, KeyType key, BiTree f, BiTree &p){ //在根指针T所指二元排序樹中递归地查找其關键字等於key的數據元素,若查找成功, //則指针p指向該數據元素節點,并返回TRUE,否則指针指向查找路徑上訪問的最後 //一個節點并返回FALSE,指针f指向T的雙親,其初始调用值為NULL if(!T) { //查找不成功 p=f; return false; } else if (key == T->data.key) { //查找成功 p=T; return true; } else if (key < T->data.key) //在左子樹中繼續查找 return SearchBST(T->lchild, key, T, p); else //在右子樹中繼續查找 return SearchBST(T->rchild, key, T, p);}
向一个二叉排序树b中插入一个结点s的算法,过程为:
/*当二叉排序树T中不存在关键字等于e.key的数据元素时,插入e并返回TRUE,否则返回FALSE*/Status InsertBST(BiTree &T, ElemType e){ if(!SearchBST(T, e.key, NULL,p){ s = new BiTNode; s->data = e; s->lchild = s->rchild = NULL; if(!p) T=s; //被插结点*s为新的根结点 else if (e.key < p->data.key) p->lchld = s; //被子插结点*s为左孩子 else p->rchild = s; //被插结点*s为右孩子 return true; } else return false; //树中已有关键字相同的结点,不再插入 }}
在二叉排序树删去一个结点,分三种情况讨论:
Status DeleteBST(BiTree &T, KeyType key){ //若二叉排序树T中存在关键字等于key的数据元素时,则删除该数据元素,并返回 //TRUE;否则返回FALSE if(!T) return false; //不存在关键字等于key的数据元素 else{ if(key == T->data.key) { // 找到关键字等于key的数据元素 return Delete(T); } else if(key > T->data.key) return DeleteBST(T->lchild, key); else return DeleteBST(T->rchild, key); }} Status Delete(BiTree &p){ //从二叉排序树中删除结点p,并重接它的左或右子树 if(!p->rchild){ //右子树空则只需重接它的左子树 q=p; p=p->lchild; delete q; } else if(!p->lchild){ //左子树空只需重接它的右子树 q=p; p=p->rchild; delete q; } else{ //左右子树均不空 q=p; s=p->lchild; while(s->rchild){ q=s; s=s->rchild; } //转左,然后向右到尽头 p->data = s->data; //s指向被删结点的“前驱” if(q!=p) q->rchild = s->lchild; //重接*q的右子树 else q->lchild = s->lchild; //重接*q的左子树 delete s; } return true;}
每个结点的Ci为该结点的层次数。最坏情况下,当先后插入的关键字有序时,构成的二叉排序树蜕变为,树的深度为n,其平均查找长度为(和顺序查找相同),最好的情况是二叉排序树的形态和折半查找的判定树相同,其平均查找长度和log 2(n)成正比(O(log 2(n)))。
请参见主条目。
这些均可以使查找树的高度为O(log(n))
转载地址:http://ukbxi.baihongyu.com/