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If it is a binary search tree, you could do in-order with going to right first and then to left.
This will make in-order traverse in descending order and stop traversal at count = 3.

If this is not a binary search tree, then convert the tree to a max heap - should be O(n) I believe. then pop three times to get the third highest.

Keep an array of 3 elements. Initialized to Integer.minimum.

Traverse the BT (inorder) and push the elements in to a position it fits in the 3 element array/linked list(sorted) and remove the last element. After the whole traversal is complete then take the last index in the array..... simple.

This way you can solve for any order - of course it taxes memory a bit.

in a BST or a BT? in BST, go to highest node and go back two. : O(lg(n)) if tree is balanced
in BT just scan all element using some traversal (in order, post order, ...) and keep an array with the three highest numbers O(n)