Topics

We want to prove the recurrence relation for Catalan numbers, , which is given by:

with the base case .

Approach: Binary Trees

We’ll use the interpretation of as the number of full binary trees with internal nodes. A full binary tree has each node either with exactly two children or being a leaf.

Let’s consider counting full binary trees with internal nodes, which we denote as .

  1. Root Node: Any binary tree with more than one node has a root. In our case, with internal nodes (for ), the root must be an internal node. This root has a left and a right child.

  2. Subtrees: The children of the root are themselves roots of full binary subtrees (or leaves, effectively representing empty subtrees in terms of internal nodes).

  3. Internal Nodes Distribution: If the total tree has internal nodes, and we’ve accounted for the root, there are internal nodes remaining to be distributed between the left and right subtrees.

    Let’s say the left subtree has internal nodes. Then, the right subtree must have internal nodes, where can range from to .

  4. Counting for a Fixed :
    For a fixed count of internal nodes in the left subtree:

    • The number of possible left subtrees is .
    • The number of possible right subtrees is .

    Since the choice of the left and right subtrees are independent, for a specific , the number of combinations is .

  5. Summing Over All :
    To get the total count of trees with internal nodes, we need to consider all possible values of , from to . We sum up the counts for each :

Thus, we have:

This relation holds because it systematically counts all possible binary trees of size by decomposing them at the root and considering all distributions of size between the left and right subtrees. The number of ways to form the left and right subtrees are given by smaller Catalan numbers, leading to the recurrence. This decomposition strategy is a common and powerful technique in combinatorial problems.