Introduction

In this blog post, we will explore how to solve the LeetCode problem "998" 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

A maximum tree is a tree where every node has a value greater than any other value in its subtree. You are given the root of a maximum binary tree and an integer val. Just as in the previous problem, the given tree was constructed from a list a (root = Construct(a)) recursively with the following Construct(a) routine: If a is empty, return null. Otherwise, let a[i] be the largest element of a. Create a root node with the value a[i]. The left child of root will be Construct([a[0], a[1], ..., a[i - 1]]). The right child of root will be Construct([a[i + 1], a[i + 2], ..., a[a.length - 1]]). Return root. Note that we were not given a directly, only a root node root = Construct(a). Suppose b is a copy of a with the value val appended to it. It is guaranteed that b has unique values. Return Construct(b). Example 1: Input: root = [4,1,3,null,null,2], val = 5 Output: [5,4,null,1,3,null,null,2] Explanation: a = [1,4,2,3], b = [1,4,2,3,5] Example 2: Input: root = [5,2,4,null,1], val = 3 Output: [5,2,4,null,1,null,3] Explanation: a = [2,1,5,4], b = [2,1,5,4,3] Example 3: Input: root = [5,2,3,null,1], val = 4 Output: [5,2,4,null,1,3] Explanation: a = [2,1,5,3], b = [2,1,5,3,4] Constraints: The number of nodes in the tree is in the range [1, 100]. 1 <= Node.val <= 100 All the values of the tree are unique. 1 <= val <= 100

Explanation

Here's the solution to the problem:

• Key Idea: The core idea is to traverse the existing tree and find the correct position to insert the new value val. Since val is appended to the original array, it will be the last element considered when building the new tree. We traverse down the right subtree until we find a node whose value is smaller than val.
• Insertion: When we find such a node, we make the current node the left child of the new node with value val. The original right subtree of the current node becomes the right child of the new node.

• Root Replacement: If the value val is greater than the current root, the current root becomes the left child of the new node, and the new node becomes the new root of the tree.

• Complexity:

  • Runtime: O(H), where H is the height of the tree. In the worst case, H can be N (number of nodes). So O(N)
  • Storage: O(1) - iterative approach consumes constant extra space

Code

    # Definition for a binary tree node.
class TreeNode:
    def __init__(self, val=0, left=None, right=None):
        self.val = val
        self.left = left
        self.right = right

class Solution:
    def insertIntoMaxTree(self, root: TreeNode, val: int) -> TreeNode:
        new_node = TreeNode(val)

        if not root:
            return new_node

        if val > root.val:
            new_node.left = root
            return new_node

        current = root
        while current.right and current.right.val > val:
            current = current.right

        new_node.left = current.right
        current.right = new_node

        return root