📜  AVL树|设置1(插入)

📅  最后修改于: 2021-05-24 22:33:09             🧑  作者: Mango

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

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

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   T2

b)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   T2

c)对对的情况

z                                y
 /  \                            /   \ 
T1   y     Left Rotate(z)       z      x
    /  \   - - - - - - - ->    / \    / \
   T2   x                     T1  T2 T3  T4
       / \
     T3  T4

d)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:

avlinsert1

avlinsert2-jpg

avlinsert3

avlinsert4

avlinsert5

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 Jaiswal


Python3
# 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 Pathak


C#
// 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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