All Possible Full Binary Trees Leetcode 894 [Python]

LeetCode problem, All Possible Full Binary Trees with the problem number 894 falls under the category of Trees and is of medium difficulty.

We will provide you with a Python solution to solve this problem efficiently.

Problem Overview

The problem statement is as follows: Given an integer n, you need to return a list of all possible full binary trees with n nodes.

Each node in the answer must have a Node.val equal to 0. A full binary tree is a binary tree where each node has either 0 or 2 children.

Let’s start by understanding the problem, its constraints, and how to approach it.

Example:

Before we dive into the problem-solving process, let’s look at a couple of examples:

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

To solve this problem, we’ll go through different approaches, including a brute-force approach and a more efficient solution, and explain the reasoning behind our approach.

Understanding Constraints

The constraints provided in the problem statement are essential for determining the approach we will take to solve the problem.

In this case, the most critical constraint is that n can vary from 1 to 20. This means that our solution needs to be efficient, as brute-forcing all possibilities would be impractical for larger values of n.

All Possible Full Binary Trees LeetCode Problem Solution

Now, let’s delve into the solutions for this problem.

We’ll explore both the brute-force approach and an optimized version that uses dynamic programming and memoization to improve efficiency.

1. Bruteforce Approach

The brute-force approach involves trying out all possible combinations to generate full binary trees with n nodes.

We’ll use recursion to explore different combinations of left and right subtrees.

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

class Solution:
    def allPossibleFBT(self, n: int):
        # Base cases: n equals 0, 1, or 2
        if n == 0:
            return []
        if n == 1:
            return [TreeNode(0)]
        if n == 2:
            return []

        result = []

        for l in range(n - 1):
            r = n - 1 - l
            left_trees = self.allPossibleFBT(l)
            right_trees = self.allPossibleFBT(r)

            for left_tree in left_trees:
                for right_tree in right_trees:
                    result.append(TreeNode(0, left_tree, right_tree))

        return result

In this solution, we create a Solution class with a method called allPossibleFBT.

The input n represents the number of nodes we need to form full binary trees for.

We start with the base cases.

If n is 0, there are no valid trees, so we return an empty list.

If n is 1, we create a tree with a single node, and if n is 2, we return an empty list as it’s not possible to create a full binary tree with just two nodes.

For n greater than 2, we iterate through the possible number of nodes in the left subtree, represented by the variable l.

The right subtree will then have n - 1 - l nodes.

We recursively call the allPossibleFBT function for both the left and right subtrees to get lists of all possible trees for those subtrees.

Then, we combine these subtrees by iterating through them and forming different combinations.

We create a new tree node with a value of 0, connecting the left and right subtrees.

The result is a list of all possible full binary trees with n nodes.

2. Efficient Approach with Memoization

While the brute-force approach works, it is not efficient for larger values of n.

To optimize the solution, we can use dynamic programming and memoization to avoid redundant calculations.

We’ll store the results for different values of n to avoid recomputing them.

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

class Solution:
    def allPossibleFBT(self, n: int):
        dp = {0: [], 1: [TreeNode()]}

        def backtrack(n):
            if n in dp:
                return dp[n]

            result = []

            for l in range(n):
                r = n - 1 - l
                left_trees = backtrack(l)
                right_trees = backtrack(r)

                for left_tree in left_trees:
                    for right_tree in right_trees:
                        result.append(TreeNode(0, left_tree, right_tree))
            dp[n] = result
            return result

        return backtrack(n)

In this optimized solution, we introduce a dynamic programming dictionary dp to store the results of different values of n.

This way, if we encounter the same n value again in our recursion, we can directly retrieve the precomputed results instead of recomputing them.

The backtrack function is the core of the solution.

It takes an integer n as input and returns a list of all possible full binary trees with n nodes.

If the result for n is already cached in dp, we return it immediately.

Otherwise, we proceed with the computation.

Within the backtrack function, we use two nested loops to iterate through different combinations of the number of nodes in the left and right subtrees.

For each combination, we retrieve the lists of all possible trees for the left and right subtrees by making recursive calls to backtrack.

We then combine these left and right subtrees to create new full binary trees by creating a root node with a value of 0 and attaching the left and right subtrees.

These newly formed trees are added to the result list.

Finally, we cache the result list for the current n value in the dp dictionary before returning it.

This memoization approach significantly improves the efficiency of the solution.

Time and Space Complexity

Time Complexity

The time complexity of the optimized solution is much better than the brute-force approach, thanks to memoization.

When computing the list of full binary trees for n, we break it down into subproblems.

For each possible value of l (

number of nodes in the left subtree), we compute the list of trees for the left and right subtrees.

This recursive process ensures that we don’t recompute the same subproblems.

Therefore, the time complexity can be considered as O(2^n), where n is the input value.

Space Complexity

The space complexity is determined by the space used for memoization.

We use the dp dictionary to store results for different values of n.

In the worst case, the dictionary might have entries for all n values from 0 to the input n.

Therefore, the space complexity is O(n).

Reasoning Behind Our Approach

The reasoning behind our approach is to break down the problem into subproblems and use memoization to avoid redundant calculations.

By considering the constraints of the problem, we aimed to provide an efficient solution that can handle larger values of n.

This approach allows us to find all possible full binary trees with n nodes in a structured and optimized manner.

Conclusion

In this blog post, we tackled the LeetCode problem All Possible Full Binary Trees with a Python solution.

We explored both a brute-force approach and an optimized solution using dynamic programming and memoization.

While the brute-force approach is straightforward, it is not efficient for larger values of n.

The optimized solution provides a more efficient way to find all possible full binary trees, thanks to memoization.

We hope this guide has been helpful in understanding the problem and its solutions.

If you have any questions, suggestions, or comments, please feel free to share them in the comments section below.

You can also check out the LeetCode problem description for further details and practice.

Happy coding!