AVL树是一种自平衡二叉搜索树(BST)，其中左子树和右子树的高度之差对于所有节点都不能超过一个。
作为AVL树的示例树
_0.jpg)
上面的树是AVL，因为每个节点的左右子树的高度差小于或等于1。
不是AVL树的示例树
_1.jpg)
上面的树不是AVL，因为8和12的左右子树的高度差大于1。
为什么选择AVL树?
大多数BST操作(例如，搜索，最大，最小，插入，删除等)花费O(h)时间，其中h是BST的高度。对于偏斜的二叉树，这些操作的成本可能变为O(n)。如果确保每次插入和删除后树的高度都保持O(Logn)，则可以保证所有这些操作的O(Logn)上限。 AVL树的高度始终为O(Logn)，其中n是树中的节点数(有关证明，请参阅此视频讲座)。
插入
为了确保给定的树在每次插入后仍保持AVL，我们必须增强标准的BST插入操作以执行一些重新平衡。以下是可以执行的两个基本操作，以在不违反BST属性的情况下重新平衡BST(键(左)<键(根)<键(右))。
1)左旋
2)右旋
T1, T2 and T3 are subtrees of the tree
rooted with y (on the left side) or x (on
the right side)
y x
/ \ Right Rotation / \
x T3 - - - - - - - > T1 y
/ \ < - - - - - - - / \
T1 T2 Left Rotation T2 T3
Keys in both of the above trees follow the
following order
keys(T1) < key(x) < keys(T2) < key(y) < keys(T3)
So BST property is not violated anywhere.插入步骤
令新插入的节点为w
1)对w执行标准BST插入。
2)从w开始，向上移动并找到第一个不平衡节点。令z为第一个不平衡节点，y为从w到z的路径中z的子代，w为从w到z的路径中z的孙代。
3)通过对以z为根的子树进行适当的旋转来重新平衡树。可能有4种可能的情况需要处理，因为x，y和z可以以4种方式排列。以下是可能的4种安排：
a)y是z的左子代，而x是y的左子代(Left Left Case)
b)y是z的左子代，x是y的右子代(左右案例)
c)y是z的正确子代，x是y的正确子代(正确的大小写)
d)y是z的右子代，x是y的左子代(Right Left Case)
以下是上述4种情况下要执行的操作。在所有情况下，我们只需要重新平衡以z为根的子树，并且当以z为根的子树的高度(经过适当的旋转后)与插入之前的高度相同时，整个树就变得平衡了。 (有关证明，请参见此视频讲座)
a)左左盒
T1, T2, T3 and T4 are subtrees.
z y
/ \ / \
y T4 Right Rotate (z) x z
/ \ - - - - - - - - -> / \ / \
x T3 T1 T2 T3 T4
/ \
T1 T2b)左右案例
z z x
/ \ / \ / \
y T4 Left Rotate (y) x T4 Right Rotate(z) y z
/ \ - - - - - - - - -> / \ - - - - - - - -> / \ / \
T1 x y T3 T1 T2 T3 T4
/ \ / \
T2 T3 T1 T2c)对对的情况
z y
/ \ / \
T1 y Left Rotate(z) z x
/ \ - - - - - - - -> / \ / \
T2 x T1 T2 T3 T4
/ \
T3 T4d)左右案例
z z x
/ \ / \ / \
T1 y Right Rotate (y) T1 x Left Rotate(z) z y
/ \ - - - - - - - - -> / \ - - - - - - - -> / \ / \
x T4 T2 y T1 T2 T3 T4
/ \ / \
T2 T3 T3 T4插入示例：
_2.jpg)
_3.jpg)
_4.jpg)
_5.jpg)
_6.jpg)
执行
以下是AVL树插入的实现。以下实现使用递归BST插入来插入新节点。在递归BST插入中，插入后，我们以自下而上的方式逐一获得指向所有祖先的指针。因此，我们不需要父指针向上移动。递归代码本身向上移动并访问新插入节点的所有祖先。
1)执行正常的BST插入。
2)当前节点必须是新插入节点的祖先之一。更新当前节点的高度。
3)获取当前节点的平衡因子(左子树的高度–右子树的高度)。
4)如果平衡因子大于1，则当前节点不平衡，并且我们处于“左左”情况或“左右”情况。要检查是否为左左大小写，请将新插入的密钥与左子树根中的密钥进行比较。
5)如果平衡因子小于-1，则当前节点不平衡，我们处于“右对右”情况或“右向左”情况。要检查是否为Right Right，请将新插入的密钥与right子树根中的密钥进行比较。
C++
// C++ program to insert a node in AVL tree
#include
using namespace std;
// An AVL tree node
class Node
{
public:
int key;
Node *left;
Node *right;
int height;
};
// A utility function to get maximum
// of two integers
int max(int a, int b);
// A utility function to get the
// height of the tree
int height(Node *N)
{
if (N == NULL)
return 0;
return N->height;
}
// A utility function to get maximum
// of two integers
int max(int a, int b)
{
return (a > b)? a : b;
}
/* Helper function that allocates a
new node with the given key and
NULL left and right pointers. */
Node* newNode(int key)
{
Node* node = new Node();
node->key = key;
node->left = NULL;
node->right = NULL;
node->height = 1; // new node is initially
// added at leaf
return(node);
}
// A utility function to right
// rotate subtree rooted with y
// See the diagram given above.
Node *rightRotate(Node *y)
{
Node *x = y->left;
Node *T2 = x->right;
// Perform rotation
x->right = y;
y->left = T2;
// Update heights
y->height = max(height(y->left),
height(y->right)) + 1;
x->height = max(height(x->left),
height(x->right)) + 1;
// Return new root
return x;
}
// A utility function to left
// rotate subtree rooted with x
// See the diagram given above.
Node *leftRotate(Node *x)
{
Node *y = x->right;
Node *T2 = y->left;
// Perform rotation
y->left = x;
x->right = T2;
// Update heights
x->height = max(height(x->left),
height(x->right)) + 1;
y->height = max(height(y->left),
height(y->right)) + 1;
// Return new root
return y;
}
// Get Balance factor of node N
int getBalance(Node *N)
{
if (N == NULL)
return 0;
return height(N->left) - height(N->right);
}
// Recursive function to insert a key
// in the subtree rooted with node and
// returns the new root of the subtree.
Node* insert(Node* node, int key)
{
/* 1. Perform the normal BST insertion */
if (node == NULL)
return(newNode(key));
if (key < node->key)
node->left = insert(node->left, key);
else if (key > node->key)
node->right = insert(node->right, key);
else // Equal keys are not allowed in BST
return node;
/* 2. Update height of this ancestor node */
node->height = 1 + max(height(node->left),
height(node->right));
/* 3. Get the balance factor of this ancestor
node to check whether this node became
unbalanced */
int balance = getBalance(node);
// If this node becomes unbalanced, then
// there are 4 cases
// Left Left Case
if (balance > 1 && key < node->left->key)
return rightRotate(node);
// Right Right Case
if (balance < -1 && key > node->right->key)
return leftRotate(node);
// Left Right Case
if (balance > 1 && key > node->left->key)
{
node->left = leftRotate(node->left);
return rightRotate(node);
}
// Right Left Case
if (balance < -1 && key < node->right->key)
{
node->right = rightRotate(node->right);
return leftRotate(node);
}
/* return the (unchanged) node pointer */
return node;
}
// A utility function to print preorder
// traversal of the tree.
// The function also prints height
// of every node
void preOrder(Node *root)
{
if(root != NULL)
{
cout << root->key << " ";
preOrder(root->left);
preOrder(root->right);
}
}
// Driver Code
int main()
{
Node *root = NULL;
/* Constructing tree given in
the above figure */
root = insert(root, 10);
root = insert(root, 20);
root = insert(root, 30);
root = insert(root, 40);
root = insert(root, 50);
root = insert(root, 25);
/* The constructed AVL Tree would be
30
/ \
20 40
/ \ \
10 25 50
*/
cout << "Preorder traversal of the "
"constructed AVL tree is \n";
preOrder(root);
return 0;
}
// This code is contributed by
// rathbhupendra C
// C program to insert a node in AVL tree
#include
#include
// An AVL tree node
struct Node
{
int key;
struct Node *left;
struct Node *right;
int height;
};
// A utility function to get maximum of two integers
int max(int a, int b);
// A utility function to get the height of the tree
int height(struct Node *N)
{
if (N == NULL)
return 0;
return N->height;
}
// A utility function to get maximum of two integers
int max(int a, int b)
{
return (a > b)? a : b;
}
/* Helper function that allocates a new node with the given key and
NULL left and right pointers. */
struct Node* newNode(int key)
{
struct Node* node = (struct Node*)
malloc(sizeof(struct Node));
node->key = key;
node->left = NULL;
node->right = NULL;
node->height = 1; // new node is initially added at leaf
return(node);
}
// A utility function to right rotate subtree rooted with y
// See the diagram given above.
struct Node *rightRotate(struct Node *y)
{
struct Node *x = y->left;
struct Node *T2 = x->right;
// Perform rotation
x->right = y;
y->left = T2;
// Update heights
y->height = max(height(y->left), height(y->right))+1;
x->height = max(height(x->left), height(x->right))+1;
// Return new root
return x;
}
// A utility function to left rotate subtree rooted with x
// See the diagram given above.
struct Node *leftRotate(struct Node *x)
{
struct Node *y = x->right;
struct Node *T2 = y->left;
// Perform rotation
y->left = x;
x->right = T2;
// Update heights
x->height = max(height(x->left), height(x->right))+1;
y->height = max(height(y->left), height(y->right))+1;
// Return new root
return y;
}
// Get Balance factor of node N
int getBalance(struct Node *N)
{
if (N == NULL)
return 0;
return height(N->left) - height(N->right);
}
// Recursive function to insert a key in the subtree rooted
// with node and returns the new root of the subtree.
struct Node* insert(struct Node* node, int key)
{
/* 1. Perform the normal BST insertion */
if (node == NULL)
return(newNode(key));
if (key < node->key)
node->left = insert(node->left, key);
else if (key > node->key)
node->right = insert(node->right, key);
else // Equal keys are not allowed in BST
return node;
/* 2. Update height of this ancestor node */
node->height = 1 + max(height(node->left),
height(node->right));
/* 3. Get the balance factor of this ancestor
node to check whether this node became
unbalanced */
int balance = getBalance(node);
// If this node becomes unbalanced, then
// there are 4 cases
// Left Left Case
if (balance > 1 && key < node->left->key)
return rightRotate(node);
// Right Right Case
if (balance < -1 && key > node->right->key)
return leftRotate(node);
// Left Right Case
if (balance > 1 && key > node->left->key)
{
node->left = leftRotate(node->left);
return rightRotate(node);
}
// Right Left Case
if (balance < -1 && key < node->right->key)
{
node->right = rightRotate(node->right);
return leftRotate(node);
}
/* return the (unchanged) node pointer */
return node;
}
// A utility function to print preorder traversal
// of the tree.
// The function also prints height of every node
void preOrder(struct Node *root)
{
if(root != NULL)
{
printf("%d ", root->key);
preOrder(root->left);
preOrder(root->right);
}
}
/* Driver program to test above function*/
int main()
{
struct Node *root = NULL;
/* Constructing tree given in the above figure */
root = insert(root, 10);
root = insert(root, 20);
root = insert(root, 30);
root = insert(root, 40);
root = insert(root, 50);
root = insert(root, 25);
/* The constructed AVL Tree would be
30
/ \
20 40
/ \ \
10 25 50
*/
printf("Preorder traversal of the constructed AVL"
" tree is \n");
preOrder(root);
return 0;
} Java
// Java program for insertion in AVL Tree
class Node {
int key, height;
Node left, right;
Node(int d) {
key = d;
height = 1;
}
}
class AVLTree {
Node root;
// A utility function to get the height of the tree
int height(Node N) {
if (N == null)
return 0;
return N.height;
}
// A utility function to get maximum of two integers
int max(int a, int b) {
return (a > b) ? a : b;
}
// A utility function to right rotate subtree rooted with y
// See the diagram given above.
Node rightRotate(Node y) {
Node x = y.left;
Node T2 = x.right;
// Perform rotation
x.right = y;
y.left = T2;
// Update heights
y.height = max(height(y.left), height(y.right)) + 1;
x.height = max(height(x.left), height(x.right)) + 1;
// Return new root
return x;
}
// A utility function to left rotate subtree rooted with x
// See the diagram given above.
Node leftRotate(Node x) {
Node y = x.right;
Node T2 = y.left;
// Perform rotation
y.left = x;
x.right = T2;
// Update heights
x.height = max(height(x.left), height(x.right)) + 1;
y.height = max(height(y.left), height(y.right)) + 1;
// Return new root
return y;
}
// Get Balance factor of node N
int getBalance(Node N) {
if (N == null)
return 0;
return height(N.left) - height(N.right);
}
Node insert(Node node, int key) {
/* 1. Perform the normal BST insertion */
if (node == null)
return (new Node(key));
if (key < node.key)
node.left = insert(node.left, key);
else if (key > node.key)
node.right = insert(node.right, key);
else // Duplicate keys not allowed
return node;
/* 2. Update height of this ancestor node */
node.height = 1 + max(height(node.left),
height(node.right));
/* 3. Get the balance factor of this ancestor
node to check whether this node became
unbalanced */
int balance = getBalance(node);
// If this node becomes unbalanced, then there
// are 4 cases Left Left Case
if (balance > 1 && key < node.left.key)
return rightRotate(node);
// Right Right Case
if (balance < -1 && key > node.right.key)
return leftRotate(node);
// Left Right Case
if (balance > 1 && key > node.left.key) {
node.left = leftRotate(node.left);
return rightRotate(node);
}
// Right Left Case
if (balance < -1 && key < node.right.key) {
node.right = rightRotate(node.right);
return leftRotate(node);
}
/* return the (unchanged) node pointer */
return node;
}
// A utility function to print preorder traversal
// of the tree.
// The function also prints height of every node
void preOrder(Node node) {
if (node != null) {
System.out.print(node.key + " ");
preOrder(node.left);
preOrder(node.right);
}
}
public static void main(String[] args) {
AVLTree tree = new AVLTree();
/* Constructing tree given in the above figure */
tree.root = tree.insert(tree.root, 10);
tree.root = tree.insert(tree.root, 20);
tree.root = tree.insert(tree.root, 30);
tree.root = tree.insert(tree.root, 40);
tree.root = tree.insert(tree.root, 50);
tree.root = tree.insert(tree.root, 25);
/* The constructed AVL Tree would be
30
/ \
20 40
/ \ \
10 25 50
*/
System.out.println("Preorder traversal" +
" of constructed tree is : ");
tree.preOrder(tree.root);
}
}
// This code has been contributed by Mayank JaiswalPython3
# Python code to insert a node in AVL tree
# Generic tree node class
class TreeNode(object):
def __init__(self, val):
self.val = val
self.left = None
self.right = None
self.height = 1
# AVL tree class which supports the
# Insert operation
class AVL_Tree(object):
# Recursive function to insert key in
# subtree rooted with node and returns
# new root of subtree.
def insert(self, root, key):
# Step 1 - Perform normal BST
if not root:
return TreeNode(key)
elif key < root.val:
root.left = self.insert(root.left, key)
else:
root.right = self.insert(root.right, key)
# Step 2 - Update the height of the
# ancestor node
root.height = 1 + max(self.getHeight(root.left),
self.getHeight(root.right))
# Step 3 - Get the balance factor
balance = self.getBalance(root)
# Step 4 - If the node is unbalanced,
# then try out the 4 cases
# Case 1 - Left Left
if balance > 1 and key < root.left.val:
return self.rightRotate(root)
# Case 2 - Right Right
if balance < -1 and key > root.right.val:
return self.leftRotate(root)
# Case 3 - Left Right
if balance > 1 and key > root.left.val:
root.left = self.leftRotate(root.left)
return self.rightRotate(root)
# Case 4 - Right Left
if balance < -1 and key < root.right.val:
root.right = self.rightRotate(root.right)
return self.leftRotate(root)
return root
def leftRotate(self, z):
y = z.right
T2 = y.left
# Perform rotation
y.left = z
z.right = T2
# Update heights
z.height = 1 + max(self.getHeight(z.left),
self.getHeight(z.right))
y.height = 1 + max(self.getHeight(y.left),
self.getHeight(y.right))
# Return the new root
return y
def rightRotate(self, z):
y = z.left
T3 = y.right
# Perform rotation
y.right = z
z.left = T3
# Update heights
z.height = 1 + max(self.getHeight(z.left),
self.getHeight(z.right))
y.height = 1 + max(self.getHeight(y.left),
self.getHeight(y.right))
# Return the new root
return y
def getHeight(self, root):
if not root:
return 0
return root.height
def getBalance(self, root):
if not root:
return 0
return self.getHeight(root.left) - self.getHeight(root.right)
def preOrder(self, root):
if not root:
return
print("{0} ".format(root.val), end="")
self.preOrder(root.left)
self.preOrder(root.right)
# Driver program to test above function
myTree = AVL_Tree()
root = None
root = myTree.insert(root, 10)
root = myTree.insert(root, 20)
root = myTree.insert(root, 30)
root = myTree.insert(root, 40)
root = myTree.insert(root, 50)
root = myTree.insert(root, 25)
"""The constructed AVL Tree would be
30
/ \
20 40
/ \ \
10 25 50"""
# Preorder Traversal
print("Preorder traversal of the",
"constructed AVL tree is")
myTree.preOrder(root)
print()
# This code is contributed by Ajitesh PathakC#
// C# program for insertion in AVL Tree
using System;
class Node
{
public int key, height;
public Node left, right;
public Node(int d)
{
key = d;
height = 1;
}
}
public class AVLTree
{
Node root;
// A utility function to get
// the height of the tree
int height(Node N)
{
if (N == null)
return 0;
return N.height;
}
// A utility function to get
// maximum of two integers
int max(int a, int b)
{
return (a > b) ? a : b;
}
// A utility function to right
// rotate subtree rooted with y
// See the diagram given above.
Node rightRotate(Node y)
{
Node x = y.left;
Node T2 = x.right;
// Perform rotation
x.right = y;
y.left = T2;
// Update heights
y.height = max(height(y.left),
height(y.right)) + 1;
x.height = max(height(x.left),
height(x.right)) + 1;
// Return new root
return x;
}
// A utility function to left
// rotate subtree rooted with x
// See the diagram given above.
Node leftRotate(Node x)
{
Node y = x.right;
Node T2 = y.left;
// Perform rotation
y.left = x;
x.right = T2;
// Update heights
x.height = max(height(x.left),
height(x.right)) + 1;
y.height = max(height(y.left),
height(y.right)) + 1;
// Return new root
return y;
}
// Get Balance factor of node N
int getBalance(Node N)
{
if (N == null)
return 0;
return height(N.left) - height(N.right);
}
Node insert(Node node, int key)
{
/* 1. Perform the normal BST insertion */
if (node == null)
return (new Node(key));
if (key < node.key)
node.left = insert(node.left, key);
else if (key > node.key)
node.right = insert(node.right, key);
else // Duplicate keys not allowed
return node;
/* 2. Update height of this ancestor node */
node.height = 1 + max(height(node.left),
height(node.right));
/* 3. Get the balance factor of this ancestor
node to check whether this node became
unbalanced */
int balance = getBalance(node);
// If this node becomes unbalanced, then there
// are 4 cases Left Left Case
if (balance > 1 && key < node.left.key)
return rightRotate(node);
// Right Right Case
if (balance < -1 && key > node.right.key)
return leftRotate(node);
// Left Right Case
if (balance > 1 && key > node.left.key)
{
node.left = leftRotate(node.left);
return rightRotate(node);
}
// Right Left Case
if (balance < -1 && key < node.right.key)
{
node.right = rightRotate(node.right);
return leftRotate(node);
}
/* return the (unchanged) node pointer */
return node;
}
// A utility function to print preorder traversal
// of the tree.
// The function also prints height of every node
void preOrder(Node node)
{
if (node != null)
{
Console.Write(node.key + " ");
preOrder(node.left);
preOrder(node.right);
}
}
// Driver code
public static void Main(String[] args)
{
AVLTree tree = new AVLTree();
/* Constructing tree given in the above figure */
tree.root = tree.insert(tree.root, 10);
tree.root = tree.insert(tree.root, 20);
tree.root = tree.insert(tree.root, 30);
tree.root = tree.insert(tree.root, 40);
tree.root = tree.insert(tree.root, 50);
tree.root = tree.insert(tree.root, 25);
/* The constructed AVL Tree would be
30
/ \
20 40
/ \ \
10 25 50
*/
Console.Write("Preorder traversal" +
" of constructed tree is : ");
tree.preOrder(tree.root);
}
}
// This code has been contributed
// by PrinciRaj1992输出：
Preorder traversal of the constructed AVL tree is
30 20 10 25 40 50时间复杂度：旋转操作(左右旋转)需要恒定的时间，因为在那里仅更改了几个指针。更新高度并获得平衡因子也需要花费固定的时间。因此，AVL插入的时间复杂度与BST插入的时间复杂度相同，为O(h)，其中h是树的高度。由于AVL树是平衡的，因此高度为O(Logn)。因此，AVL插入的时间复杂度为O(Logn)。
与红黑树的比较
AVL树和其他类似Red Black的自平衡搜索树对于在O(log n)时间内完成所有基本操作很有用。与红黑树相比，AVL树更加平衡，但是它们可能会在插入和删除过程中引起更多旋转。因此，如果您的应用程序涉及许多频繁的插入和删除操作，则应优先选择Red Black树。而且，如果插入和删除操作的频率较低，而搜索操作的频率较高，则AVL树应优先于Red Black Tree。
以下是要删除的帖子。
AVL树|套装2(删除)
以下是一些使用自平衡搜索树的帖子。
中位数为整数流(运行整数)
大小为k的所有子数组的最大值
计算右侧较小的元素
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