A binary search tree is built by inserting these chars in this order: G R O U N D H O G Assume that nodes greater than or equal to the parent will go right, less than the parent will go left. What is the post-order traversal of the tree? Enter JUST the characters, all caps, no spaces:

To solve this problem, we first need to build the binary search tree (BST) using the given order of characters: G R O U N D H O G.

1. Start with the first character 'G'. This becomes the root of the tree.
2. The next character is 'R'. Since 'R' is greater than 'G', it goes to the right of 'G'.
3. The next character is 'O'. 'O' is less than 'R' but greater than 'G', so it goes to the left of 'R'.
4. The next character is 'U'. 'U' is greater than 'R', so it goes to the right of 'R'.
5. The next character is 'N'. 'N' is less than 'U' but greater than 'O', so it goes to the right of 'O' and left of 'U'.
6. The next character is 'D'. 'D' is less than 'N' and 'O', so it goes to the left of 'N'.
7. The next character is 'H'. 'H' is greater than 'G' but less than 'O', so it goes to the right of 'G' and left of 'O'.
8. The next character is 'O'. 'O' is equal to the existing 'O' in the tree, so it goes to the right of the existing 'O'.
9. The last character is 'G'. 'G' is equal to the root 'G', so it goes to the right of the root 'G'.

Now, to find the post-order traversal of the tree, we follow the rule: Left, Right, Root.

The post-order traversal of the tree is: D N O G O H U R G