Let $w$ be the node to be deleted
1) Perform standard BST delete for $w$.
2) Starting from $w$, travel up and find the first unbalanced node. 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.
3) Re-balance the tree by performing appropriate rotations on the subtree rooted with $z$. 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. Note that, unlike insertion, fixing the node $z$ won’t fix the complete AVL tree. After fixing $z$, we may have to fix ancestors of $z$ as well (See this video if you have trouble visualizing the above mentioned cases).
Courtesy of Geeksforgeeks.