Avl tree deletion example pdf

Here, the keys are the same, but since they are inserted in a different order, they take different positions, and the height of the tree remains balanced. Hence search will not take more than O log n for any element of the tree. In AVL trees, we keep a check on the height of the tree during insertion operation. Modifications are made to maintain the balanced height without violating the fundamental properties of Binary Search Tree.

To make the AVL Tree balance itself, when inserting or deleting a node from the tree, rotations are performed. A single right rotation is performed. A single left rotation is performed. This type of rotation is identified when a node has a balanced factor as -2, and its right-child has a balance factor as Insert operation is almost the same as in simple binary search trees.

After every insertion, we balance the height of the tree. Insert operation takes O log n worst time complexity. In the above example, insert Step 2 : Once the node is added, the balance factor of each node is updated.

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Abhay Chaudhari. Steps to follow for deletion. To make sure that the given tree remains AVL after every deletion, we must augment the standard BST delete operation to perform some re-balancing.

Let w be the node to be deleted 1 Perform standard BST delete for w. Let z be the first unbalanced node, y be the larger height child of z, and x be the larger height child of y.

Note that the definitions of x and y are different from insertion here. There can be 4 possible cases that needs to be handled as x, y and z can be arranged in 4 ways. Following are the possible 4 arrangements: a y is left child of z and x is left child of y Left Left Case b y is left child of z and x is right child of y Left Right Case c y is right child of z and x is right child of y Right Right Case d y is right child of z and x is left child of y Right Left Case Like insertion, following are the operations to be performed in above mentioned 4 cases.

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