Introduction

In this blog post, we will explore how to solve the LeetCode problem "938" 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 node of a binary search tree and two integers low and high, return the sum of values of all nodes with a value in the inclusive range [low, high]. Example 1: Input: root = [10,5,15,3,7,null,18], low = 7, high = 15 Output: 32 Explanation: Nodes 7, 10, and 15 are in the range [7, 15]. 7 + 10 + 15 = 32. Example 2: Input: root = [10,5,15,3,7,13,18,1,null,6], low = 6, high = 10 Output: 23 Explanation: Nodes 6, 7, and 10 are in the range [6, 10]. 6 + 7 + 10 = 23. Constraints: The number of nodes in the tree is in the range [1, 2 * 104]. 1 <= Node.val <= 105 1 <= low <= high <= 105 All Node.val are unique.

Explanation

Here's the breakdown:

• Leverage BST Properties: Exploit the inherent ordering of a Binary Search Tree (BST). This allows us to selectively explore branches, significantly reducing the search space. If the current node's value is less than low, we only need to explore the right subtree. If the current node's value is greater than high, we only need to explore the left subtree.
• Recursive Traversal: Implement a recursive function to traverse the tree. The function checks if the current node's value falls within the range [low, high]. If it does, the value is added to the running sum.

• Pruning the Search Space: Conditionally traverse the left and right subtrees based on the current node's value relative to low and high. This pruning step is crucial for efficiency.

• Time & Space Complexity: O(N) in the worst case (skewed tree), but can be closer to O(log N) for balanced trees. Space complexity is O(H) due to the recursion stack, where H is the height of the tree.

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 rangeSumBST(self, root: TreeNode, low: int, high: int) -> int:
        """
        Calculates the sum of values of nodes within the inclusive range [low, high] in a BST.
        """
        total_sum = 0

        def dfs(node):
            nonlocal total_sum  # Allows modification of the outer scope variable

            if not node:
                return

            if low <= node.val <= high:
                total_sum += node.val

            if node.val > low:
                dfs(node.left)  # Explore left subtree if node value is greater than low

            if node.val < high:
                dfs(node.right) # Explore right subtree if node value is less than high

        dfs(root)
        return total_sum