Introduction

In this blog post, we will explore how to solve the LeetCode problem "894" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like Chatmagic, we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

Problem Statement

Given an integer n, return a list of all possible full binary trees with n nodes. Each node of each tree in the answer must have Node.val == 0. Each element of the answer is the root node of one possible tree. You may return the final list of trees in any order. A full binary tree is a binary tree where each node has exactly 0 or 2 children. Example 1: Input: n = 7 Output: [[0,0,0,null,null,0,0,null,null,0,0],[0,0,0,null,null,0,0,0,0],[0,0,0,0,0,0,0],[0,0,0,0,0,null,null,null,null,0,0],[0,0,0,0,0,null,null,0,0]] Example 2: Input: n = 3 Output: [[0,0,0]] Constraints: 1 <= n <= 20

Explanation

• Recursion with Memoization: The core idea is to use recursion to build full binary trees. For a given n, we iterate through possible sizes for the left subtree and recursively build the left and right subtrees. Memoization (using a dictionary) avoids redundant computations, significantly improving efficiency.
  • Base Case: When n is 1, we create a single node with value 0 and return it as the only possible full binary tree.
  • Building Trees: For each possible left subtree size, we determine the right subtree size accordingly (n = leftSize + rightSize + 1). We then recursively generate all possible left and right subtrees and combine them to form new full binary trees.

• Runtime Complexity: O(C(n/2)), where C(n/2) is the (n/2)-th Catalan number. Due to memoization, each n value is computed only once. Storage Complexity: O(C(n/2)), to store all possible trees in the memoization dictionary.

Code

    class TreeNode:
    def __init__(self, val=0, left=None, right=None):
        self.val = val
        self.left = left
        self.right = right

def allPossibleFBT(n: int) -> list[TreeNode]:
    """
    Given an integer n, return a list of all possible full binary trees with n nodes.
    Each node of each tree in the answer must have Node.val == 0.
    Each element of the answer is the root node of one possible tree.
    """

    memo = {}  # Memoization dictionary to store results for each n

    def generate_trees(n):
        if n in memo:
            return memo[n]

        if n == 1:
            return [TreeNode(0)]

        if n % 2 == 0:  # Full binary trees must have an odd number of nodes
            return []

        trees = []
        for left_size in range(1, n, 2):
            right_size = n - 1 - left_size
            left_trees = generate_trees(left_size)
            right_trees = generate_trees(right_size)

            for left_tree in left_trees:
                for right_tree in right_trees:
                    root = TreeNode(0)
                    root.left = left_tree
                    root.right = right_tree
                    trees.append(root)

        memo[n] = trees
        return trees

    return generate_trees(n)