draw a tree with elements(50,33,44,22,77,35,60,40)respectively inorder

To draw a binary search tree (BST) using the given elements in the inorder traversal sequence (50, 33, 44, 22, 77, 35, 60, 40), we need to follow a few steps:

Step 1: Start with the first element of the sequence as the root of the BST. In this case, the root will be 50.

Step 2: Compare the next element, 33, with the current root (50). Since 33 is less than 50, it becomes the left child of 50.

Step 3: Now, compare the next element, 44, with the current node (33). Since 44 is greater than 33, it becomes the right child of 33.

Step 4: Proceed with the next element, 22. It is less than both 33 and 50, so it becomes the left child of 33.

Step 5: Move to the next element, 77. Since it is greater than 50, it becomes the right child of 50.

Step 6: Compare the next element, 35, with the current node (44). Since 35 is less than 44, it becomes the left child of 44.

Step 7: Move to the next element, 60, and compare it with the current node (77). Since 60 is less than 77, it becomes the left child of 77.

Step 8: Finally, place the last element, 40, in the left subtree of 60, as it is smaller.

The resulting binary search tree is as follows:

```
50
/ \
33 77
/ \ /
22 44 60
\
35
\
40
```

Keep in mind that this is just one possible representation of the BST based on the given inorder traversal. There can be variations based on the order in which elements are inserted.