How do you solve Maximum Binary Tree II on LeetCode?

Maximum Binary Tree II (LeetCode #998) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(1) space complexity. Key insight: Understanding the properties of a maximum binary tree is crucial.

What is the brute force solution for Maximum Binary Tree II?

The brute force approach for Maximum Binary Tree II is Brute Force, with O(n²) time complexity and O(n) space complexity. We can reconstruct the original array from the tree by performing a traversal. Once we have the array, we can append the new value and reconstruct the tree again. This is straightforward but inefficient. The optimal Optimal Solution approach improves this to O(n).

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

Maximum Binary Tree II uses the following concepts: Tree, Binary Tree. The recommended patterns to study are: Tree, Recursion.

What are the interview tips for Maximum Binary Tree II?

Always clarify the problem requirements before jumping to a solution. Think about edge cases, such as inserting into an empty tree. Practice explaining your thought process as you code.

What are common mistakes when solving Maximum Binary Tree II?

Failing to recognize when to create a new root node. Not handling the case where the new value is less than the current root.

#998

Maximum Binary Tree II

Medium

TreeBinary TreeTreeRecursion

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

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Intuition

Time O(n)Space O(1)

Instead of reconstructing the entire array, we can directly insert the new value into the tree. We traverse the tree to find the appropriate position for the new value while maintaining the properties of the maximum binary tree.

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Algorithm

2 steps

1Step 1: If the new value is greater than the root's value, create a new root with the new value, and make the current root its left child.
2Step 2: If the new value is less than or equal to the root's value, recursively insert the new value into the right subtree.

solution.py13 lines

1class TreeNode:
2    def __init__(self, val=0, left=None, right=None):
3        self.val = val
4        self.left = left
5        self.right = right
6
7    def insertIntoMaxTree(self, root, val):
8        if val > root.val:
9            new_root = TreeNode(val)
10            new_root.left = root
11            return new_root
12        root.right = self.insertIntoMaxTree(root.right, val)
13        return root

ℹ

Complexity note: The time complexity is O(n) because in the worst case, we may traverse the entire height of the tree. The space complexity is O(1) as we are not using any additional data structures.

1Understanding the properties of a maximum binary tree is crucial.
2Directly manipulating the tree structure can be more efficient than reconstructing arrays.

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