Introduction

All possible binary-operator groupings for a sequence of operands correspond to all full binary tree shapes over those operands. The standard way to generate them in Python is recursive: pick every possible split point, generate all left trees, generate all right trees, and combine every left-right pair into a new parent node.

Represent the Expression Tree Recursively

A simple tree representation can just be nested tuples:

def make_node(left, right):
    return (left, right)

If the operands are ['a', 'b', 'c'], then the two possible groupings become:

• '(('a', 'b'), 'c')'
• '('a', ('b', 'c'))'

Those are two different full binary trees.

Generate All Trees by Splitting the Operand List

The recursive generator looks like this:

1def generate_trees(items):
2    if len(items) == 1:
3        return [items[0]]
4
5    trees = []
6    for split in range(1, len(items)):
7        left_items = items[:split]
8        right_items = items[split:]
9
10        left_trees = generate_trees(left_items)
11        right_trees = generate_trees(right_items)
12
13        for left in left_trees:
14            for right in right_trees:
15                trees.append((left, right))
16
17    return trees
18
19print(generate_trees(['a', 'b', 'c']))

This is the core algorithm. Every split point creates a set of left and right subproblems, and every combination of their results produces one valid grouping tree.

Add the Operator Symbol for Printing

If you want human-readable expressions rather than tuple trees, add a formatter:

1def format_tree(tree, op='*'):
2    if isinstance(tree, str):
3        return tree
4    left, right = tree
5    return f"({format_tree(left, op)} {op} {format_tree(right, op)})"
6
7for tree in generate_trees(['a', 'b', 'c', 'd']):
8    print(format_tree(tree, '+'))

Now the algorithm generates actual parenthesized groupings instead of only structural tuples.

Understand the Growth Rate

The number of such trees is given by the Catalan numbers, which grow quickly. That means this generator is fine for small operand counts and becomes expensive for larger ones.

This is not a bug in the code. It is a property of the combinatorics. If you ask for every grouping, the result set itself becomes large very fast.

Use Memoization for Repeated Subproblems

If performance matters, memoize subproblems by the operand slice:

1from functools import lru_cache
2
3@lru_cache(maxsize=None)
4def generate_cached(items):
5    items = tuple(items)
6    if len(items) == 1:
7        return (items[0],)
8
9    trees = []
10    for split in range(1, len(items)):
11        for left in generate_cached(items[:split]):
12            for right in generate_cached(items[split:]):
13                trees.append((left, right))
14    return tuple(trees)

Memoization does not change the combinatorial growth of the final answer, but it avoids recomputing the same subtrees repeatedly.

The Structure and the Printed Expression Are Different Layers

It is often better to generate tree structure first and pretty-print it later. That separation keeps the combinatorics code simpler and makes it easier to reuse the same tree generator for evaluation, visualization, or symbolic processing.

Common Pitfalls

• Assuming the number of grouping trees grows linearly with the number of operands.
• Mixing tree generation with string formatting too early and making the recursion harder to understand.
• Forgetting the base case for a single operand.
• Recomputing the same subproblems without memoization when performance matters.
• Expecting a unique result when different parenthesizations are genuinely different trees.

Small Inputs Are the Best Way to Validate the Generator

Before generating trees for larger operand lists, test the recursion on three or four operands where you already know the expected groupings. That makes logic errors much easier to spot early.

Summary

• All binary-operator groupings correspond to full binary tree shapes.
• Generate them recursively by splitting the operand list at every possible position.
• Represent trees structurally first, then format them as expressions if needed.
• Memoization can reduce repeated work on subproblems.
• The result count follows Catalan growth, so large operand lists become expensive quickly.