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GeeksforGeeks » Trees specific questions

Given a node of BST, make it the root

(5 posts)

1. 

   g1
   guest
   Posted 6 months ago #
   
   

   Given a node of a BST, modify it in such a way that the given node becomes the root. The tree should still be BST. One way I could get is store the Inoder traversal of the tree. Find that node in the traversal and recursively make the BST.

2. 

   rampalli
   guest
   Posted 5 months ago #
   
   

   Let given a node of BST (B) is n. These are the steps to be followed
   1. Start constructing new BST tree(B_new) with n as root.
   2. Do level order traversal in given BST (B) and add each node to the new BST(B_new) tree.
   3. As BST do not allow the duplicate nodes, skip the node n in given BST(B) i.e. don't add this node to new BST while constructing.
   At the end we will get new BST as given node n is the root.
   For example:
   35
   20 40
   10 25 30 50
   is the given BST. Now make 30 as the root.
   Sol: Step1 take 30 as root and construct new BST.
   30
   Step2: Now do level order traversal and add each visited node to new BST. Level order traversal of given BST is 35 20 40 10 25 30 50

   30 -> 30 -> 30 --> 30
   35 20 35 20 35 20 35
   40 10 40
   do the same procedure, finally we get
   30
   20 35
   10 25 40
   50

   Thanks.

3. 

   camster
   Member
   Posted 5 months ago #
   
   

   void RearrangePointersLeft(nodemary*& root, nodemary*& target){
        nodemary* tmp;
   
   	 tmp = root;
   	 tmp->left = NULL;
   	 root = target;
   	 root->right = tmp;
   }
   
   void RearrangePointersRight(nodemary*& root, nodemary*& target){
                        nodemary* tmp;
   
   	 tmp = root;
   	 tmp->right = NULL;
   	 root = target;
   	 root->left = tmp;
   }
   
   void ModifyBST(nodemary*& root, nodemary*& target){
   	if (root == NULL){
   		return;
   	}
   
   	if (root->value < target->value){
   		ModifyBST(root->right, target);
                                             RearrangePointersRight(root, target);
   	}
   	else if (root->value > target->value){
   		ModifyBST(root->left, target);
   		RearrangePointersLeft(root, target);
   	}
   }

4. 

   vpshastry
   guest
   Posted 5 months ago #
   
   

   The above procedure is similar to balancing a BST( right rotate/left rotate ) wrt the node and the root.
   root
   / \
   n1 n2
   / \ / \
   newroot n3 n4 n5
   / \
   n6 n7

   after the procedure
   newroot
   / \
   n6 root
   / \
   n1 n2
   / \ / \
   n7 n3 n4 n5

5. 

   vpshastry
   guest
   Posted 5 months ago #
   
   

   The above procedure is similar to balancing a BST( right rotate/left rotate ) wrt the node and the root.
   .................................................. root
   .............................................. / \
   .......................................... n1 n2
   .................................... / \ / \
   ............................. newroot n3 n4 n5
   ........................... / \
   ........................... n6 n7

   after the procedure
   ................................................ newroot
   ................................................ / \
   ............................................ n6 root
   ....................................................... / \
   ....................................................... n1 n2
   ................................................... / \ / \
   ................................................ n7 n3 n4 n5

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