我们在上一篇文章中讨论了AVL插入。在这篇文章中，我们将采用类似的删除方法。
删除的步骤如下。
为了确保给定的树在每次删除后仍保持AVL，我们必须增强标准的BST删除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 left side)
or x (on 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.
令w为要删除的节点
1)对w执行标准的BST删除。
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z的较大高度子级，x为y的较大高度子级。请注意，x和y的定义与此处的插入不同。
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)
与插入类似，以下是在上述4种情况下要执行的操作。请注意，与插入不同，修复节点z不会修复完整的AVL树。修复z之后，我们可能还必须修复z的祖先(请参见此视频讲座以获取证明)
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)右左案例
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
与插入不同，在删除中，在z处进行旋转后，可能必须在z的祖先处进行旋转。因此，我们必须继续追踪路径，直到到达根为止。
例子： _0.jpg)
_1.jpg)
值为32的节点将被删除。删除32之后，我们向上移动并找到第一个非平衡节点44。将其标记为z，将其较高高度的子级标记为y，即62，将y的较高高度的子级标记为x，其x可以为78或50相同的高度。我们已经考虑过78。现在的情况是Right Right，所以我们执行向左旋转。
C实现
以下是AVL树删除的C实现。以下C实现使用递归BST删除作为基础。在递归BST删除中，删除后，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已删除节点的所有祖先。
1)执行正常的BST删除。
2)当前节点必须是已删除节点的祖先之一。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。要检查是左左情况还是左右情况，请获取左子树的平衡因子。如果左子树的平衡因子大于或等于0，则为Left Left情况，否则为Left Right情况。
5)如果平衡因子小于-1，则当前节点不平衡，我们处于“右右”情况或“右左”情况。要检查是右右情况还是右左情况，请获取右子树的平衡因子。如果右子树的平衡因子小于或等于0，则为Right Right情况，否则为Right Left情况。
C++
// C++ program to delete a node from 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 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);
}
Node* insert(Node* node, int key)
{
/* 1. Perform the normal BST rotation */
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 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;
}
/* Given a non-empty binary search tree,
return the node with minimum key value
found in that tree. Note that the entire
tree does not need to be searched. */
Node * minValueNode(Node* node)
{
Node* current = node;
/* loop down to find the leftmost leaf */
while (current->left != NULL)
current = current->left;
return current;
}
// Recursive function to delete a node
// with given key from subtree with
// given root. It returns root of the
// modified subtree.
Node* deleteNode(Node* root, int key)
{
// STEP 1: PERFORM STANDARD BST DELETE
if (root == NULL)
return root;
// If the key to be deleted is smaller
// than the root's key, then it lies
// in left subtree
if ( key < root->key )
root->left = deleteNode(root->left, key);
// If the key to be deleted is greater
// than the root's key, then it lies
// in right subtree
else if( key > root->key )
root->right = deleteNode(root->right, key);
// if key is same as root's key, then
// This is the node to be deleted
else
{
// node with only one child or no child
if( (root->left == NULL) ||
(root->right == NULL) )
{
Node *temp = root->left ?
root->left :
root->right;
// No child case
if (temp == NULL)
{
temp = root;
root = NULL;
}
else // One child case
*root = *temp; // Copy the contents of
// the non-empty child
free(temp);
}
else
{
// node with two children: Get the inorder
// successor (smallest in the right subtree)
Node* temp = minValueNode(root->right);
// Copy the inorder successor's
// data to this node
root->key = temp->key;
// Delete the inorder successor
root->right = deleteNode(root->right,
temp->key);
}
}
// If the tree had only one node
// then return
if (root == NULL)
return root;
// STEP 2: UPDATE HEIGHT OF THE CURRENT NODE
root->height = 1 + max(height(root->left),
height(root->right));
// STEP 3: GET THE BALANCE FACTOR OF
// THIS NODE (to check whether this
// node became unbalanced)
int balance = getBalance(root);
// If this node becomes unbalanced,
// then there are 4 cases
// Left Left Case
if (balance > 1 &&
getBalance(root->left) >= 0)
return rightRotate(root);
// Left Right Case
if (balance > 1 &&
getBalance(root->left) < 0)
{
root->left = leftRotate(root->left);
return rightRotate(root);
}
// Right Right Case
if (balance < -1 &&
getBalance(root->right) <= 0)
return leftRotate(root);
// Right Left Case
if (balance < -1 &&
getBalance(root->right) > 0)
{
root->right = rightRotate(root->right);
return leftRotate(root);
}
return root;
}
// 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, 9);
root = insert(root, 5);
root = insert(root, 10);
root = insert(root, 0);
root = insert(root, 6);
root = insert(root, 11);
root = insert(root, -1);
root = insert(root, 1);
root = insert(root, 2);
/* The constructed AVL Tree would be
9
/ \
1 10
/ \ \
0 5 11
/ / \
-1 2 6
*/
cout << "Preorder traversal of the "
"constructed AVL tree is \n";
preOrder(root);
root = deleteNode(root, 10);
/* The AVL Tree after deletion of 10
1
/ \
0 9
/ / \
-1 5 11
/ \
2 6
*/
cout << "\nPreorder traversal after"
<< " deletion of 10 \n";
preOrder(root);
return 0;
}
// This code is contributed by rathbhupendra C
// C program to delete a node from 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 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);
}
struct Node* insert(struct Node* node, int key)
{
/* 1. Perform the normal BST rotation */
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 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;
}
/* Given a non-empty binary search tree, return the
node with minimum key value found in that tree.
Note that the entire tree does not need to be
searched. */
struct Node * minValueNode(struct Node* node)
{
struct Node* current = node;
/* loop down to find the leftmost leaf */
while (current->left != NULL)
current = current->left;
return current;
}
// Recursive function to delete a node with given key
// from subtree with given root. It returns root of
// the modified subtree.
struct Node* deleteNode(struct Node* root, int key)
{
// STEP 1: PERFORM STANDARD BST DELETE
if (root == NULL)
return root;
// If the key to be deleted is smaller than the
// root's key, then it lies in left subtree
if ( key < root->key )
root->left = deleteNode(root->left, key);
// If the key to be deleted is greater than the
// root's key, then it lies in right subtree
else if( key > root->key )
root->right = deleteNode(root->right, key);
// if key is same as root's key, then This is
// the node to be deleted
else
{
// node with only one child or no child
if( (root->left == NULL) || (root->right == NULL) )
{
struct Node *temp = root->left ? root->left :
root->right;
// No child case
if (temp == NULL)
{
temp = root;
root = NULL;
}
else // One child case
*root = *temp; // Copy the contents of
// the non-empty child
free(temp);
}
else
{
// node with two children: Get the inorder
// successor (smallest in the right subtree)
struct Node* temp = minValueNode(root->right);
// Copy the inorder successor's data to this node
root->key = temp->key;
// Delete the inorder successor
root->right = deleteNode(root->right, temp->key);
}
}
// If the tree had only one node then return
if (root == NULL)
return root;
// STEP 2: UPDATE HEIGHT OF THE CURRENT NODE
root->height = 1 + max(height(root->left),
height(root->right));
// STEP 3: GET THE BALANCE FACTOR OF THIS NODE (to
// check whether this node became unbalanced)
int balance = getBalance(root);
// If this node becomes unbalanced, then there are 4 cases
// Left Left Case
if (balance > 1 && getBalance(root->left) >= 0)
return rightRotate(root);
// Left Right Case
if (balance > 1 && getBalance(root->left) < 0)
{
root->left = leftRotate(root->left);
return rightRotate(root);
}
// Right Right Case
if (balance < -1 && getBalance(root->right) <= 0)
return leftRotate(root);
// Right Left Case
if (balance < -1 && getBalance(root->right) > 0)
{
root->right = rightRotate(root->right);
return leftRotate(root);
}
return root;
}
// 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, 9);
root = insert(root, 5);
root = insert(root, 10);
root = insert(root, 0);
root = insert(root, 6);
root = insert(root, 11);
root = insert(root, -1);
root = insert(root, 1);
root = insert(root, 2);
/* The constructed AVL Tree would be
9
/ \
1 10
/ \ \
0 5 11
/ / \
-1 2 6
*/
printf("Preorder traversal of the constructed AVL "
"tree is \n");
preOrder(root);
root = deleteNode(root, 10);
/* The AVL Tree after deletion of 10
1
/ \
0 9
/ / \
-1 5 11
/ \
2 6
*/
printf("\nPreorder traversal after deletion of 10 \n");
preOrder(root);
return 0;
} Java
// Java program for deletion 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 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 rotation */
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 // Equal 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
Wunbalanced */
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;
}
/* Given a non-empty binary search tree, return the
node with minimum key value found in that tree.
Note that the entire tree does not need to be
searched. */
Node minValueNode(Node node)
{
Node current = node;
/* loop down to find the leftmost leaf */
while (current.left != null)
current = current.left;
return current;
}
Node deleteNode(Node root, int key)
{
// STEP 1: PERFORM STANDARD BST DELETE
if (root == null)
return root;
// If the key to be deleted is smaller than
// the root's key, then it lies in left subtree
if (key < root.key)
root.left = deleteNode(root.left, key);
// If the key to be deleted is greater than the
// root's key, then it lies in right subtree
else if (key > root.key)
root.right = deleteNode(root.right, key);
// if key is same as root's key, then this is the node
// to be deleted
else
{
// node with only one child or no child
if ((root.left == null) || (root.right == null))
{
Node temp = null;
if (temp == root.left)
temp = root.right;
else
temp = root.left;
// No child case
if (temp == null)
{
temp = root;
root = null;
}
else // One child case
root = temp; // Copy the contents of
// the non-empty child
}
else
{
// node with two children: Get the inorder
// successor (smallest in the right subtree)
Node temp = minValueNode(root.right);
// Copy the inorder successor's data to this node
root.key = temp.key;
// Delete the inorder successor
root.right = deleteNode(root.right, temp.key);
}
}
// If the tree had only one node then return
if (root == null)
return root;
// STEP 2: UPDATE HEIGHT OF THE CURRENT NODE
root.height = max(height(root.left), height(root.right)) + 1;
// STEP 3: GET THE BALANCE FACTOR OF THIS NODE (to check whether
// this node became unbalanced)
int balance = getBalance(root);
// If this node becomes unbalanced, then there are 4 cases
// Left Left Case
if (balance > 1 && getBalance(root.left) >= 0)
return rightRotate(root);
// Left Right Case
if (balance > 1 && getBalance(root.left) < 0)
{
root.left = leftRotate(root.left);
return rightRotate(root);
}
// Right Right Case
if (balance < -1 && getBalance(root.right) <= 0)
return leftRotate(root);
// Right Left Case
if (balance < -1 && getBalance(root.right) > 0)
{
root.right = rightRotate(root.right);
return leftRotate(root);
}
return root;
}
// 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, 9);
tree.root = tree.insert(tree.root, 5);
tree.root = tree.insert(tree.root, 10);
tree.root = tree.insert(tree.root, 0);
tree.root = tree.insert(tree.root, 6);
tree.root = tree.insert(tree.root, 11);
tree.root = tree.insert(tree.root, -1);
tree.root = tree.insert(tree.root, 1);
tree.root = tree.insert(tree.root, 2);
/* The constructed AVL Tree would be
9
/ \
1 10
/ \ \
0 5 11
/ / \
-1 2 6
*/
System.out.println("Preorder traversal of "+
"constructed tree is : ");
tree.preOrder(tree.root);
tree.root = tree.deleteNode(tree.root, 10);
/* The AVL Tree after deletion of 10
1
/ \
0 9
/ / \
-1 5 11
/ \
2 6
*/
System.out.println("");
System.out.println("Preorder traversal after "+
"deletion of 10 :");
tree.preOrder(tree.root);
}
}
// This code has been contributed by Mayank JaiswalPython3
# Python code to delete 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 insertion,
# deletion operations
class AVL_Tree(object):
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
# Recursive function to delete a node with
# given key from subtree with given root.
# It returns root of the modified subtree.
def delete(self, root, key):
# Step 1 - Perform standard BST delete
if not root:
return root
elif key < root.val:
root.left = self.delete(root.left, key)
elif key > root.val:
root.right = self.delete(root.right, key)
else:
if root.left is None:
temp = root.right
root = None
return temp
elif root.right is None:
temp = root.left
root = None
return temp
temp = self.getMinValueNode(root.right)
root.val = temp.val
root.right = self.delete(root.right,
temp.val)
# If the tree has only one node,
# simply return it
if root is None:
return root
# 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 self.getBalance(root.left) >= 0:
return self.rightRotate(root)
# Case 2 - Right Right
if balance < -1 and self.getBalance(root.right) <= 0:
return self.leftRotate(root)
# Case 3 - Left Right
if balance > 1 and self.getBalance(root.left) < 0:
root.left = self.leftRotate(root.left)
return self.rightRotate(root)
# Case 4 - Right Left
if balance < -1 and self.getBalance(root.right) > 0:
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 getMinValueNode(self, root):
if root is None or root.left is None:
return root
return self.getMinValueNode(root.left)
def preOrder(self, root):
if not root:
return
print("{0} ".format(root.val), end="")
self.preOrder(root.left)
self.preOrder(root.right)
myTree = AVL_Tree()
root = None
nums = [9, 5, 10, 0, 6, 11, -1, 1, 2]
for num in nums:
root = myTree.insert(root, num)
# Preorder Traversal
print("Preorder Traversal after insertion -")
myTree.preOrder(root)
print()
# Delete
key = 10
root = myTree.delete(root, key)
# Preorder Traversal
print("Preorder Traversal after deletion -")
myTree.preOrder(root)
print()
# This code is contributed by Ajitesh PathakC#
// C# program for deletion in AVL Tree
using System;
public 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 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 rotation */
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 // Equal 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
Wunbalanced */
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;
}
/* Given a non-empty binary search tree, return the
node with minimum key value found in that tree.
Note that the entire tree does not need to be
searched. */
Node minValueNode(Node node)
{
Node current = node;
/* loop down to find the leftmost leaf */
while (current.left != null)
current = current.left;
return current;
}
Node deleteNode(Node root, int key)
{
// STEP 1: PERFORM STANDARD BST DELETE
if (root == null)
return root;
// If the key to be deleted is smaller than
// the root's key, then it lies in left subtree
if (key < root.key)
root.left = deleteNode(root.left, key);
// If the key to be deleted is greater than the
// root's key, then it lies in right subtree
else if (key > root.key)
root.right = deleteNode(root.right, key);
// if key is same as root's key, then this is the node
// to be deleted
else
{
// node with only one child or no child
if ((root.left == null) || (root.right == null))
{
Node temp = null;
if (temp == root.left)
temp = root.right;
else
temp = root.left;
// No child case
if (temp == null)
{
temp = root;
root = null;
}
else // One child case
root = temp; // Copy the contents of
// the non-empty child
}
else
{
// node with two children: Get the inorder
// successor (smallest in the right subtree)
Node temp = minValueNode(root.right);
// Copy the inorder successor's data to this node
root.key = temp.key;
// Delete the inorder successor
root.right = deleteNode(root.right, temp.key);
}
}
// If the tree had only one node then return
if (root == null)
return root;
// STEP 2: UPDATE HEIGHT OF THE CURRENT NODE
root.height = max(height(root.left),
height(root.right)) + 1;
// STEP 3: GET THE BALANCE FACTOR
// OF THIS NODE (to check whether
// this node became unbalanced)
int balance = getBalance(root);
// If this node becomes unbalanced,
// then there are 4 cases
// Left Left Case
if (balance > 1 && getBalance(root.left) >= 0)
return rightRotate(root);
// Left Right Case
if (balance > 1 && getBalance(root.left) < 0)
{
root.left = leftRotate(root.left);
return rightRotate(root);
}
// Right Right Case
if (balance < -1 && getBalance(root.right) <= 0)
return leftRotate(root);
// Right Left Case
if (balance < -1 && getBalance(root.right) > 0)
{
root.right = rightRotate(root.right);
return leftRotate(root);
}
return root;
}
// 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()
{
AVLTree tree = new AVLTree();
/* Constructing tree given in the above figure */
tree.root = tree.insert(tree.root, 9);
tree.root = tree.insert(tree.root, 5);
tree.root = tree.insert(tree.root, 10);
tree.root = tree.insert(tree.root, 0);
tree.root = tree.insert(tree.root, 6);
tree.root = tree.insert(tree.root, 11);
tree.root = tree.insert(tree.root, -1);
tree.root = tree.insert(tree.root, 1);
tree.root = tree.insert(tree.root, 2);
/* The constructed AVL Tree would be
9
/ \
1 10
/ \ \
0 5 11
/ / \
-1 2 6
*/
Console.WriteLine("Preorder traversal of "+
"constructed tree is : ");
tree.preOrder(tree.root);
tree.root = tree.deleteNode(tree.root, 10);
/* The AVL Tree after deletion of 10
1
/ \
0 9
/ / \
-1 5 11
/ \
2 6
*/
Console.WriteLine("");
Console.WriteLine("Preorder traversal after "+
"deletion of 10 :");
tree.preOrder(tree.root);
}
}
/* This code contributed by PrinciRaj1992 */输出：
Preorder traversal of the constructed AVL tree is
9 1 0 -1 5 2 6 10 11
Preorder traversal after deletion of 10
1 0 -1 9 5 2 6 11
时间复杂度：旋转操作(向左和向右旋转)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系数也需要花费固定的时间。因此，AVL删除的时间复杂度与BST删除相同，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删除的时间复杂度为O(Log n)。
参考：
https://www.cs.purdue.edu/homes/ayg/CS251/slides/chap7b.pdf
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