Given $n$, there is only one shape for a complete binary tree (CBT) with $n$ nodes.
For any deterministic traversal, the correspondence between a node's position in a CBT and its position during the traversal of that CBT is completely fixed. So if a deterministic traversal is given, we can reconstruct the CBT uniquely. This applies not only to any one of pre-order, post-order, or in-order traversals, it also applies to breath-first-traversal (with the left child visited before the right child), or any other deterministic traversal as mentioned by Hendrik Jan in his comment.
Here is an example. The shape above is the only shape for a CBT with 12 nodes, which are, 1 root (at depth 0), 2 nodes at depth 1, 4 nodes at depth 2 and 5 nodes at depth 3. A pre-order traversal of that CBT visits the nodes in the following order.
- the root node.
- the first node of depth 1.
- the first node of depth 2.
- the first node of depth 3.
- the second node of depth 3.
- the second node of depth 2.
- the third node of depth 3.
- the fourth node of depth 3.
- the second node of depth 1.
- the third node of depth 2.
- the fifth node of depth 3.
- the fourth node of depth 2.
Give the list above, we know the root node is the 1st-node visited, the first node of depth 1 is the 2nd-node visited, ..., the fourth node of depth 3 is the 8th-node visited and the fifth node of depth 3 is the 11th-node visited.