How do you solve Maximum Binary Tree on LeetCode?

Maximum Binary Tree (LeetCode #654) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(n) space complexity. Key insight: Finding the maximum value efficiently is crucial for performance.

What is the brute force solution for Maximum Binary Tree?

The brute force approach for Maximum Binary Tree is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute-force approach involves repeatedly finding the maximum value in the array and recursively creating the left and right subtrees. This is straightforward but inefficient due to repeated scans of the same subarrays. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Maximum Binary Tree?

Maximum Binary Tree uses the following concepts: Array, Divide and Conquer, Stack, Tree, Monotonic Stack, Binary Tree. The recommended patterns to study are: Divide and Conquer, Recursion.

What are the interview tips for Maximum Binary Tree?

Explain your thought process clearly while coding. Consider edge cases and test them during the interview. Discuss the time and space complexity of your solution.

What are common mistakes when solving Maximum Binary Tree?

Not handling the base case of empty arrays properly. Forgetting to return the constructed tree from recursive calls.

#654

Maximum Binary Tree

Medium

ArrayDivide and ConquerStackTreeMonotonic StackBinary TreeDivide and ConquerRecursion

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

💡

Intuition

Time O(n)Space O(n)

The optimal solution uses a recursive approach but avoids repeated scans by leveraging the maximum value's index directly. This reduces the time complexity significantly.

⚙️

Algorithm

3 steps

1Step 1: Create a helper function that takes the start and end indices of the current subarray.
2Step 2: Find the maximum value and its index in the current subarray.
3Step 3: Create a root node and recursively build the left and right subtrees using the indices.

solution.py16 lines

1class TreeNode:
2    def __init__(self, val):
3        self.val = val
4        self.left = None
5        self.right = None
6
7def constructMaximumBinaryTree(nums):
8    def buildTree(start, end):
9        if start > end:
10            return None
11        max_index = start + nums[start:end + 1].index(max(nums[start:end + 1]))
12        root = TreeNode(nums[max_index])
13        root.left = buildTree(start, max_index - 1)
14        root.right = buildTree(max_index + 1, end)
15        return root
16    return buildTree(0, len(nums) - 1)

ℹ

Complexity note: This complexity is linear because we only traverse the array once to find the maximum value for each recursive call, leading to a total of O(n) operations.

1Finding the maximum value efficiently is crucial for performance.
2Recursive tree construction mirrors the problem's divide-and-conquer nature.

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