srcmini本文概述
在二叉搜索树(BST)中, 关键字左子树中的所有关键字必须较小, 而右侧子树中的所有关键字必须较大。所以二叉搜索树根据定义具有不同的键。
如何在每个插入项再插入一个带值的键, 而每次删除都删除一个项的重复项呢？
一种简单的解决方案是允许在右侧使用相同的键(我们也可以选择左侧)。例如, 考虑在空的二叉搜索树中插入键12、10、20、9、11、10、12、12
12
/ \
10 20
/ \ /
9 11 12
/ \
10 12一种
更好的解决方案
是增加每个树节点以与常规字段(如键, 左和右指针)一起存储计数。
在空的二叉搜索树中插入键12、10、20、9、11、10、12、12将创建以下内容。
12(3)
/ \
10(2) 20(1)
/ \
9(1) 11(1)
Count of a key is shown in bracket与上述简单方法相比, 该方法具有以下优点。
1)树木的高度很小, 与重复的数量无关。请注意, 大多数BST操作(搜索, 插入和删除)的时间复杂度为O(h), 其中h是BST的高度。因此, 如果我们能够保持较小的高度, 则可以利用较少的键比较优势。
2)搜索, 插入和删除变得更容易。我们可以使用相同的插入, 搜索和删除算法, 但需进行少量修改(请参见下面的代码)。
3)此方法适用于自平衡BST(AVL树, 红黑树等)。这些树涉及旋转, 旋转可能违反简单解决方案的BST属性, 因为旋转后相同的键可以位于左侧或右侧。
以下是带有每个键计数的普通二叉搜索树的实现。该代码基本上取自
BST中插入和删除的代码
。处理重复项所做的更改将突出显示, 其余代码相同。
C ++
//C++ program to implement basic operations
//(search, insert and delete) on a BST that
//handles duplicates by storing count with
//every node
#include<bits/stdc++.h>
using namespace std;
struct node
{
int key;
int count;
struct node *left, *right;
};
//A utility function to create a new BST node
struct node *newNode( int item)
{
struct node *temp = ( struct node *) malloc ( sizeof ( struct node));
temp->key = item;
temp->left = temp->right = NULL;
temp->count = 1;
return temp;
}
//A utility function to do inorder traversal of BST
void inorder( struct node *root)
{
if (root != NULL)
{
inorder(root->left);
cout <<root->key <<"("
<<root->count <<") " ;
inorder(root->right);
}
}
/* A utility function to insert a new
node with given key in BST */
struct node* insert( struct node* node, int key)
{
/* If the tree is empty, return a new node */
if (node == NULL) return newNode(key);
//If key already exists in BST, //increment count and return
if (key == node->key)
{
(node->count)++;
return node;
}
/* Otherwise, recur down the tree */
if (key <node->key)
node->left = insert(node->left, key);
else
node->right = insert(node->right, key);
/* 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;
}
/* Given a binary search tree and a key, this function deletes a given key and
returns root of modified tree */
struct node* deleteNode( struct node* root, int key)
{
//base case
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
else
{
//If key is present more than once, //simply decrement count and return
if (root->count> 1)
{
(root->count)--;
return root;
}
//ElSE, delete the node
//node with only one child or no child
if (root->left == NULL)
{
struct node *temp = root->right;
free (root);
return temp;
}
else if (root->right == NULL)
{
struct node *temp = root->left;
free (root);
return temp;
}
//node with two children: Get the inorder
//successor (smallest in the right subtree)
struct node* temp = minValueNode(root->right);
//Copy the inorder successor's
//content to this node
root->key = temp->key;
//Delete the inorder successor
root->right = deleteNode(root->right, temp->key);
}
return root;
}
//Driver Code
int main()
{
/* Let us create following BST
12(3)
/ \
10(2) 20(1)
/\
9(1) 11(1) */
struct node *root = NULL;
root = insert(root, 12);
root = insert(root, 10);
root = insert(root, 20);
root = insert(root, 9);
root = insert(root, 11);
root = insert(root, 10);
root = insert(root, 12);
root = insert(root, 12);
cout <<"Inorder traversal of the given tree "
<<endl;
inorder(root);
cout <<"\nDelete 20\n" ;
root = deleteNode(root, 20);
cout <<"Inorder traversal of the modified tree \n" ;
inorder(root);
cout <<"\nDelete 12\n" ;
root = deleteNode(root, 12);
cout <<"Inorder traversal of the modified tree \n" ;
inorder(root);
cout <<"\nDelete 9\n" ;
root = deleteNode(root, 9);
cout <<"Inorder traversal of the modified tree \n" ;
inorder(root);
return 0;
}
//This code is contributed by Akanksha RaiC
//C program to implement basic operations (search, insert and delete)
//on a BST that handles duplicates by storing count with every node
#include<stdio.h>
#include<stdlib.h>
struct node
{
int key;
int count;
struct node *left, *right;
};
//A utility function to create a new BST node
struct node *newNode( int item)
{
struct node *temp = ( struct node *) malloc ( sizeof ( struct node));
temp->key = item;
temp->left = temp->right = NULL;
temp->count = 1;
return temp;
}
//A utility function to do inorder traversal of BST
void inorder( struct node *root)
{
if (root != NULL)
{
inorder(root->left);
printf ( "%d(%d) " , root->key, root->count);
inorder(root->right);
}
}
/* A utility function to insert a new node with given key in BST */
struct node* insert( struct node* node, int key)
{
/* If the tree is empty, return a new node */
if (node == NULL) return newNode(key);
//If key already exists in BST, icnrement count and return
if (key == node->key)
{
(node->count)++;
return node;
}
/* Otherwise, recur down the tree */
if (key <node->key)
node->left = insert(node->left, key);
else
node->right = insert(node->right, key);
/* 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;
}
/* Given a binary search tree and a key, this function
deletes a given key and returns root of modified tree */
struct node* deleteNode( struct node* root, int key)
{
//base case
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
else
{
//If key is present more than once, simply decrement
//count and return
if (root->count> 1)
{
(root->count)--;
return root;
}
//ElSE, delete the node
//node with only one child or no child
if (root->left == NULL)
{
struct node *temp = root->right;
free (root);
return temp;
}
else if (root->right == NULL)
{
struct node *temp = root->left;
free (root);
return temp;
}
//node with two children: Get the inorder successor (smallest
//in the right subtree)
struct node* temp = minValueNode(root->right);
//Copy the inorder successor's content to this node
root->key = temp->key;
//Delete the inorder successor
root->right = deleteNode(root->right, temp->key);
}
return root;
}
//Driver Program to test above functions
int main()
{
/* Let us create following BST
12(3)
/ \
10(2) 20(1)
/ \
9(1) 11(1) */
struct node *root = NULL;
root = insert(root, 12);
root = insert(root, 10);
root = insert(root, 20);
root = insert(root, 9);
root = insert(root, 11);
root = insert(root, 10);
root = insert(root, 12);
root = insert(root, 12);
printf ( "Inorder traversal of the given tree \n" );
inorder(root);
printf ( "\nDelete 20\n" );
root = deleteNode(root, 20);
printf ( "Inorder traversal of the modified tree \n" );
inorder(root);
printf ( "\nDelete 12\n" );
root = deleteNode(root, 12);
printf ( "Inorder traversal of the modified tree \n" );
inorder(root);
printf ( "\nDelete 9\n" );
root = deleteNode(root, 9);
printf ( "Inorder traversal of the modified tree \n" );
inorder(root);
return 0;
}Java
//Java program to implement basic operations
//(search, insert and delete) on a BST that
//handles duplicates by storing count with
//every node
class GFG
{
static class node
{
int key;
int count;
node left, right;
};
//A utility function to create a new BST node
static node newNode( int item)
{
node temp = new node();
temp.key = item;
temp.left = temp.right = null ;
temp.count = 1 ;
return temp;
}
//A utility function to do inorder traversal of BST
static void inorder(node root)
{
if (root != null )
{
inorder(root.left);
System.out.print(root.key + "(" +
root.count + ") " );
inorder(root.right);
}
}
/* A utility function to insert a new
node with given key in BST */
static node insert(node node, int key)
{
/* If the tree is empty, return a new node */
if (node == null ) return newNode(key);
//If key already exists in BST, //increment count and return
if (key == node.key)
{
(node.count)++;
return node;
}
/* Otherwise, recur down the tree */
if (key <node.key)
node.left = insert(node.left, key);
else
node.right = insert(node.right, key);
/* 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. */
static node minValueNode(node node)
{
node current = node;
/* loop down to find the leftmost leaf */
while (current.left != null )
current = current.left;
return current;
}
/* Given a binary search tree and a key, this function deletes a given key and
returns root of modified tree */
static node deleteNode(node root, int key)
{
//base case
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
else
{
//If key is present more than once, //simply decrement count and return
if (root.count> 1 )
{
(root.count)--;
return root;
}
//ElSE, delete the node
//node with only one child or no child
if (root.left == null )
{
node temp = root.right;
root= null ;
return temp;
}
else if (root.right == null )
{
node temp = root.left;
root = null ;
return temp;
}
//node with two children: Get the inorder
//successor (smallest in the right subtree)
node temp = minValueNode(root.right);
//Copy the inorder successor's
//content to this node
root.key = temp.key;
//Delete the inorder successor
root.right = deleteNode(root.right, temp.key);
}
return root;
}
//Driver Code
public static void main(String[] args)
{
/* Let us create following BST
12(3)
/ \
10(2) 20(1)
/\
9(1) 11(1) */
node root = null ;
root = insert(root, 12 );
root = insert(root, 10 );
root = insert(root, 20 );
root = insert(root, 9 );
root = insert(root, 11 );
root = insert(root, 10 );
root = insert(root, 12 );
root = insert(root, 12 );
System.out.print( "Inorder traversal of " +
"the given tree " + "\n" );
inorder(root);
System.out.print( "\nDelete 20\n" );
root = deleteNode(root, 20 );
System.out.print( "Inorder traversal of " +
"the modified tree \n" );
inorder(root);
System.out.print( "\nDelete 12\n" );
root = deleteNode(root, 12 );
System.out.print( "Inorder traversal of " +
"the modified tree \n" );
inorder(root);
System.out.print( "\nDelete 9\n" );
root = deleteNode(root, 9 );
System.out.print( "Inorder traversal of " +
"the modified tree \n" );
inorder(root);
}
}
//This code is contributed by 29AjayKumarPython3
# Python3 program to implement basic operations
# (search, insert and delete) on a BST that handles
# duplicates by storing count with every node
# A utility function to create a new BST node
class newNode:
# Constructor to create a new node
def __init__( self , data):
self .key = data
self .count = 1
self .left = None
self .right = None
# A utility function to do inorder
# traversal of BST
def inorder(root):
if root ! = None :
inorder(root.left)
print (root.key, "(" , root.count, ")" , end = " " )
inorder(root.right)
# A utility function to insert a new node
# with given key in BST
def insert(node, key):
# If the tree is empty, return a new node
if node = = None :
k = newNode(key)
return k
# If key already exists in BST, increment
# count and return
if key = = node.key:
(node.count) + = 1
return node
# Otherwise, recur down the tree
if key <node.key:
node.left = insert(node.left, key)
else :
node.right = insert(node.right, key)
# 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.
def minValueNode(node):
current = node
# loop down to find the leftmost leaf
while current.left ! = None :
current = current.left
return current
# Given a binary search tree and a key, # this function deletes a given key and
# returns root of modified tree
def deleteNode(root, key):
# base case
if root = = None :
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
elif key> root.key:
root.right = deleteNode(root.right, key)
# if key is same as root's key
else :
# If key is present more than once, # simply decrement count and return
if root.count> 1 :
root.count - = 1
return root
# ElSE, delete the node node with
# only one child or no child
if root.left = = None :
temp = root.right
return temp
elif root.right = = None :
temp = root.left
return temp
# node with two children: Get the inorder
# successor (smallest in the right subtree)
temp = minValueNode(root.right)
# Copy the inorder successor's content
# to this node
root.key = temp.key
# Delete the inorder successor
root.right = deleteNode(root.right, temp.key)
return root
# Driver Code
if __name__ = = '__main__' :
# Let us create following BST
# 12(3)
# /\
# 10(2) 20(1)
# /\
# 9(1) 11(1)
root = None
root = insert(root, 12 )
root = insert(root, 10 )
root = insert(root, 20 )
root = insert(root, 9 )
root = insert(root, 11 )
root = insert(root, 10 )
root = insert(root, 12 )
root = insert(root, 12 )
print ( "Inorder traversal of the given tree" )
inorder(root)
print ()
print ( "Delete 20" )
root = deleteNode(root, 20 )
print ( "Inorder traversal of the modified tree" )
inorder(root)
print ()
print ( "Delete 12" )
root = deleteNode(root, 12 )
print ( "Inorder traversal of the modified tree" )
inorder(root)
print ()
print ( "Delete 9" )
root = deleteNode(root, 9 )
print ( "Inorder traversal of the modified tree" )
inorder(root)
# This code is contributed by PranchalKC#
//C# program to implement basic operations
//(search, insert and delete) on a BST that
//handles duplicates by storing count with
//every node
using System;
class GFG
{
public class node
{
public int key;
public int count;
public node left, right;
};
//A utility function to create
//a new BST node
static node newNode( int item)
{
node temp = new node();
temp.key = item;
temp.left = temp.right = null ;
temp.count = 1;
return temp;
}
//A utility function to do inorder
//traversal of BST
static void inorder(node root)
{
if (root != null )
{
inorder(root.left);
Console.Write(root.key + "(" +
root.count + ") " );
inorder(root.right);
}
}
/* A utility function to insert a new
node with given key in BST */
static node insert(node node, int key)
{
/* If the tree is empty, return a new node */
if (node == null ) return newNode(key);
//If key already exists in BST, //increment count and return
if (key == node.key)
{
(node.count)++;
return node;
}
/* Otherwise, recur down the tree */
if (key <node.key)
node.left = insert(node.left, key);
else
node.right = insert(node.right, key);
/* 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. */
static node minValueNode(node node)
{
node current = node;
/* loop down to find the leftmost leaf */
while (current.left != null )
current = current.left;
return current;
}
/* Given a binary search tree and a key, this function deletes a given key and
returns root of modified tree */
static node deleteNode(node root, int key)
{
//base case
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
else
{
//If key is present more than once, //simply decrement count and return
if (root.count> 1)
{
(root.count)--;
return root;
}
//ElSE, delete the node
node temp = null ;
//node with only one child or no child
if (root.left == null )
{
temp = root.right;
root = null ;
return temp;
}
else if (root.right == null )
{
temp = root.left;
root = null ;
return temp;
}
//node with two children: Get the inorder
//successor (smallest in the right subtree)
temp = minValueNode(root.right);
//Copy the inorder successor's
//content to this node
root.key = temp.key;
//Delete the inorder successor
root.right = deleteNode(root.right, temp.key);
}
return root;
}
//Driver Code
public static void Main(String[] args)
{
/* Let us create following BST
12(3)
/ \
10(2) 20(1)
/\
9(1) 11(1) */
node root = null ;
root = insert(root, 12);
root = insert(root, 10);
root = insert(root, 20);
root = insert(root, 9);
root = insert(root, 11);
root = insert(root, 10);
root = insert(root, 12);
root = insert(root, 12);
Console.Write( "Inorder traversal of " +
"the given tree " + "\n" );
inorder(root);
Console.Write( "\nDelete 20\n" );
root = deleteNode(root, 20);
Console.Write( "Inorder traversal of " +
"the modified tree \n" );
inorder(root);
Console.Write( "\nDelete 12\n" );
root = deleteNode(root, 12);
Console.Write( "Inorder traversal of " +
"the modified tree \n" );
inorder(root);
Console.Write( "\nDelete 9\n" );
root = deleteNode(root, 9);
Console.Write( "Inorder traversal of " +
"the modified tree \n" );
inorder(root);
}
}
//This code is contributed by Rajput-Ji输出如下：
Inorder traversal of the given tree
9(1) 10(2) 11(1) 12(3) 20(1)
Delete 20
Inorder traversal of the modified tree
9(1) 10(2) 11(1) 12(3)
Delete 12
Inorder traversal of the modified tree
9(1) 10(2) 11(1) 12(2)
Delete 9
Inorder traversal of the modified tree
10(2) 11(1) 12(2)我们将很快讨论允许重复的AVL和红黑树。
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