How do you solve All Possible Full Binary Trees on LeetCode?

All Possible Full Binary Trees (LeetCode #894) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n³) time complexity and O(n) space complexity. Key insight: Full binary trees can only have an odd number of nodes.

What is the brute force solution for All Possible Full Binary Trees?

The brute force approach for All Possible Full Binary Trees is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute-force approach generates all possible trees by recursively trying to create left and right subtrees for each node. This method checks every combination of left and right subtree sizes to form a full binary tree. The optimal Optimal Solution approach improves this to O(n³).

What are the interview tips for All Possible Full Binary Trees?

Explain your thought process clearly when discussing recursive solutions. Consider edge cases, like n being even, early in your solution. Discuss time and space complexity as you develop your solution.

What are common mistakes when solving All Possible Full Binary Trees?

Forgetting to check if n is even and returning an empty list. Not using memoization, leading to excessive recursion.

#894

All Possible Full Binary Trees

Medium

Dynamic ProgrammingTreeRecursionMemoizationBinary TreeDynamic ProgrammingRecursion

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n³)
Space	O(1)	O(n)

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Intuition

Time O(n³)Space O(n)

The optimal solution uses dynamic programming to store previously computed results for each number of nodes. This avoids redundant calculations and speeds up the process significantly.

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Algorithm

3 steps

1Step 1: Create a memoization table to store results for each odd number of nodes.
2Step 2: For each odd number of nodes, generate trees by combining results from smaller subproblems.
3Step 3: Use the memoization table to retrieve results instead of recalculating them.

solution.py21 lines

1class TreeNode:
2    def __init__(self, val=0):
3        self.val = val
4        self.left = None
5        self.right = None
6
7def allPossibleFBT(n):
8    if n % 2 == 0:
9        return []
10    memo = {1: [TreeNode(0)]}
11    for i in range(3, n + 1, 2):
12        memo[i] = []
13        for left in range(1, i, 2):
14            right = i - 1 - left
15            for left_tree in memo[left]:
16                for right_tree in memo[right]:
17                    root = TreeNode(0)
18                    root.left = left_tree
19                    root.right = right_tree
20                    memo[i].append(root)
21    return memo[n]

ℹ

Complexity note: The time complexity is O(n³) due to the nested loops for generating left and right subtrees for each odd number of nodes. The space complexity is O(n) for storing the results in the memoization table.

1Full binary trees can only have an odd number of nodes.
2Dynamic programming helps avoid redundant calculations.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.