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  1. Prepare
  2. Data Structures
  3. Balanced Trees
  4. Self Balancing Tree

Self Balancing Tree

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An AVL tree (Georgy Adelson-Velsky and Landis' tree, named after the inventors) is a self-balancing binary search tree. In an AVL tree, the heights of the two child subtrees of any node differ by at most one; if at any time they differ by more than one, rebalancing is done to restore this property.

We define balance factor for each node as :

balanceFactor = height(left subtree) - height(right subtree)

The balance factor of any node of an AVL tree is in the integer range [-1,+1]. If after any modification in the tree, the balance factor becomes less than −1 or greater than +1, the subtree rooted at this node is unbalanced, and a rotation is needed.

(https://en.wikipedia.org/wiki/AVL_tree)

You are given a pointer to the root of an AVL tree. You need to insert a value into this tree and perform the necessary rotations to ensure that it remains balanced.

Input Format

You are given a function,

node *insert(node * root,int new_val)
{


}

'node' is defined as :

struct node
{
int val;            //value
struct node* left;  //left child
struct node* right; //right child
int ht;             //height of the node
} node;

You only need to complete the function.

Note: All the values in the tree will be distinct. Height of a Null node is -1 and the height of the leaf node is 0.


Output Format

Insert the new value into the tree and return a pointer to the root of the tree. Ensure that the tree remains balanced.

Sample Input

    3
  /  \
 2    4
       \
        5

The value to be inserted is 6.

Sample Output

    3
  /  \
 2    5
     / \
    4   6

Explanation

After inserting 6 in the tree. the tree becomes:

    3 (Balance Factor = -2)
  /  \
 2    4 (Balance Factor = -2)
       \
        5 (Balance Factor = -1)
         \
          6 (Balance Factor = 0)

Balance Factor of nodes 3 and 4 is no longer in the range [-1,1]. We need to perform a rotation to balance the tree. This is the right right case. We perform a single rotation to balance the tree.

After performing the rotation, the tree becomes :

                              3 (Balance Factor = -1)
                            /   \
      (Balance Factor = 0) 2     5 (Balance Factor = 0)
                                / \
           (Balance Factor = 0)4   6 (Balance Factor = 0)

Author

vatsalchanana

Difficulty

Medium

Max Score

50

Submitted By

14095

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