Introduction

In this blog post, we will explore how to solve the LeetCode problem "501" 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

Given the root of a binary search tree (BST) with duplicates, return all the mode(s) (i.e., the most frequently occurred element) in it. If the tree has more than one mode, return them in any order. Assume a BST is defined as follows: The left subtree of a node contains only nodes with keys less than or equal to the node's key. The right subtree of a node contains only nodes with keys greater than or equal to the node's key. Both the left and right subtrees must also be binary search trees. Example 1: Input: root = [1,null,2,2] Output: [2] Example 2: Input: root = [0] Output: [0] Constraints: The number of nodes in the tree is in the range [1, 104]. -105 <= Node.val <= 105 Follow up: Could you do that without using any extra space? (Assume that the implicit stack space incurred due to recursion does not count).

Explanation

Here's the breakdown of the solution:

• Inorder Traversal: Leverage the BST property using an inorder traversal to process nodes in ascending order. This allows us to efficiently track consecutive occurrences of the same value.
• Constant Space Tracking: Utilize a constant number of variables to maintain the current count, maximum count, and the mode(s) found so far. This fulfills the follow-up requirement of minimal space usage.

• Dynamic Mode Update: During traversal, update the mode list whenever a new maximum count is encountered, or append to it if the current count matches the existing maximum count.

• Runtime Complexity: O(N), where N is the number of nodes in the BST. Storage Complexity: O(1) (excluding the output list, and the implicit stack space due to recursion, as specified in the prompt).

Code

    from typing import List, Optional

# 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 findMode(self, root: Optional[TreeNode]) -> List[int]:
        modes = []
        max_count = 0
        current_count = 0
        pre_val = None

        def inorder(node: Optional[TreeNode]):
            nonlocal modes, max_count, current_count, pre_val

            if not node:
                return

            inorder(node.left)

            if pre_val is None or node.val != pre_val:
                current_count = 1
            else:
                current_count += 1

            if current_count > max_count:
                modes = [node.val]
                max_count = current_count
            elif current_count == max_count:
                modes.append(node.val)

            pre_val = node.val

            inorder(node.right)

        inorder(root)
        return modes