Pre-order to post-order traversal

Question

If the pre-order traversal of a binary search tree is 6, 2, 1, 4, 3, 7, 10, 9, 11, how to get the post-order traversal?

Answer 1

I know this is old but there is a better solution.

We don't have to reconstruct a BST to get the post-order from the pre-order.

Here is a simple python code that does it recursively:

import itertools

def postorder(preorder):
    if not preorder:
        return []
    else:
        root = preorder[0]
        left = list(itertools.takewhile(lambda x: x < root, preorder[1:]))
        right = preorder[len(left) + 1:]
        return postorder(left) + postorder(right) + [root]

if __name__ == '__main__':
    preorder = [20, 10, 6, 15, 30, 35]
    print(postorder(preorder))

Output:

 [6, 15, 10, 35, 30, 20]

Explanation:

We know that we are in pre-order. This means that the root is at the index 0 of the list of the values in the BST. And we know that the elements following the root are:

• first: the elements less than the root, which belong to the left subtree of the root
• second: the elements greater than the root, which belong to the right subtree of the root

We then just call recursively the function on both subtrees (which still are in pre-order) and then chain left + right + root (which is the post-order).