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Merge two BSTs

(5 posts)

1. 

   BADSUN
   guest
   Posted 1 year ago #
   
   

   Problem: Merge two BSTs .

   #include<iostream>
   
   class node{
    public:
     int d;
     node* l;
     node* r;
    node(int x):d(x),l(0),r(0){}
   };
   void add(int x,node *&a){//lower or equal value to the left BST
     if(!a)a=new node(x);
     else if(x>a->d)add(x,a->r);
     else add(x,a->l);
   }
   void print(node *a){
     if(a){
       print(a->l);
       std::cout<<a->d<<" ";
       print(a->r);
     }
   }
   void merge(node *&a,node *&b){//merge b into a
     if(b==0)return;
     else if(a==0)a=b;
     else if(a->d<b->d){
       node* t = b->l;
       b->l=0;
       merge(a,t);
       merge(a->r,b);
     }
     else if(a->d>=b->d){
       node *t = b->r;
       b->r=0;
       merge(a->l,b);
       merge(a,t);
     }
   }
   int main(){
   #define NL std::cout<<"\n"
     node *a=0, *b=0;
     add(8,a);add(3,a);add(1,a);add(6,a);add(4,a);
      add(7,a);add(10,a);add(14,a);add(13,a);
     add(8,b);add(3,b);add(1,b);add(6,b);add(4,b);
     add(7,b);
     print(a);NL;
     print(b);NL;
     merge(a,b);
     b=0;
     print(a);
     return 0;
   }
   

2. 

   nutcracker
   guest
   Posted 1 year ago #
   
   

   #include "stdafx.h"
   #include "stdlib.h"
   #include "stdio.h"
   
   struct node{
     int d;
     struct node* l;
     struct node* r;
   };
   
   struct node *createNode(int x)
   {
     struct node * pTemp =(struct node *) malloc(sizeof(struct node));
     pTemp->d =x;pTemp->l=NULL; pTemp->r = NULL;
     return pTemp;
   }
   void insert(struct node *pIns,struct node **pRoot){
   	if (!(*pRoot)) { *pRoot = pIns; return;}
   	struct node *pTemp = *pRoot;
   	struct node *pParent = NULL;
   	while (pTemp)
   	{
   		if (pIns->d < pTemp->d)
   		{
   			pParent = pTemp;
   			pTemp = pTemp->l;
   			continue;
   		}
   		else
   		{
   			pParent = pTemp;
   			pTemp = pTemp->r;
   			continue;
   		};
   	}
   	if (pIns->d <pParent->d) pParent->l = pIns;
   	else pParent->r = pIns;
   }
   
   void merge(struct node **aa,struct node *b){//merge b into a
   	struct node *a=*aa;
   	if(b==0)return;
   	else if(a==0)a=b;
   	else if ((b->l == NULL)&&(b->r == NULL))
   	  insert(b, &a);
   	else if(a->d < b->d){
       struct node* t = b->l;
       b->l=NULL;
       merge(&a,t);
       merge(&a->r,b);
     }
     else if(a->d >= b->d){
       struct node *t = b->r;
       b->r=NULL;
       merge(&a->l,b);
       merge(&a,t);
     }
     *aa=a;
   }
   int main(){
   
     struct node *pRoot= NULL;
     struct node *pRoot1=NULL;
       insert(createNode(3),&pRoot1);
   	insert(createNode(8),&pRoot1);
    // insert(createNode(4),&pRoot1);
   	insert(createNode(1),&pRoot1);
   
     //insert(createNode(6),&pRoot1);
   
   	insert(createNode(5),&pRoot);
   	insert(createNode(9),&pRoot);
    // insert(createNode(7),&pRoot);
   	insert(createNode(2),&pRoot);
   //  insert(createNode(10),&pRoot);
       merge(&pRoot, pRoot1);
   
   	return 0;
   }
   

3. 

   madhuri
   guest
   Posted 1 year ago #
   
   

   this has a simple solution.
   Find the preorder traversal of both lists , this gives the list in ascending order.
   Copy it in an array and then call BST with the array
   Running time
   O(n) + O(m) + c(m+n) + O(m+n)
   where n- no of elements in tree 1
   m= no of elements in tree2
   c- constant time to copy each element to an array
   O(m+n) - time for BST algo to construct a BST

4. 

   monikaluils
   guest
   Posted 7 months ago #
   
   

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5. 

   Amol
   guest
   Posted 7 months ago #
   
   

   Check this solution
   http://dzmitryhuba.blogspot.com/2011/03/merge-binary-search-trees-in-place.html

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