本文概述

• C ++
• C
• Java
• Python3
• C#

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 /\ 1020 / \/ 91112 /\ 1012

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 Rai

C

//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 29AjayKumar

Python3

# 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 PranchalK

C#

//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

【如何处理二叉搜索树中的重复项（）】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。
本文作者：奇拉格。如果发现任何不正确的地方, 或者想分享有关上述主题的更多信息, 请发表评论。

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