Introduction

In this blog post, we will explore how to solve the LeetCode problem "1026" 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 tree, find the maximum value v for which there exist different nodes a and b where v = |a.val - b.val| and a is an ancestor of b. A node a is an ancestor of b if either: any child of a is equal to b or any child of a is an ancestor of b. Example 1: Input: root = [8,3,10,1,6,null,14,null,null,4,7,13] Output: 7 Explanation: We have various ancestor-node differences, some of which are given below : |8 - 3| = 5 |3 - 7| = 4 |8 - 1| = 7 |10 - 13| = 3 Among all possible differences, the maximum value of 7 is obtained by |8 - 1| = 7. Example 2: Input: root = [1,null,2,null,0,3] Output: 3 Constraints: The number of nodes in the tree is in the range [2, 5000]. 0 <= Node.val <= 105

Explanation

Here's the breakdown of the solution:

• Key Idea: Perform a Depth-First Search (DFS) traversal of the tree. For each node, maintain the minimum and maximum values encountered along the path from the root to that node. The maximum difference between the current node's value and the path's minimum and maximum will be a candidate for the overall maximum difference.
• DFS Traversal: Recursively explore the tree, updating the minimum and maximum values as we descend. At each node, calculate the absolute difference between the node's value and the current minimum and maximum, and update the global maximum difference if needed.

• Optimization: By passing the minimum and maximum values down the tree during the DFS, we avoid redundant calculations and efficiently track the relevant ancestor values.

• Complexity: O(N) runtime, O(H) storage where N is the number of nodes and H is the height of the tree. In the worst case, H can be equal to N.

Code

    class TreeNode:
    def __init__(self, val=0, left=None, right=None):
        self.val = val
        self.left = left
        self.right = right

def maxAncestorDiff(root: TreeNode) -> int:
    max_diff = 0

    def dfs(node, current_min, current_max):
        nonlocal max_diff

        if not node:
            return

        max_diff = max(max_diff, abs(node.val - current_min), abs(node.val - current_max))

        current_min = min(current_min, node.val)
        current_max = max(current_max, node.val)

        dfs(node.left, current_min, current_max)
        dfs(node.right, current_min, current_max)

    dfs(root, root.val, root.val)
    return max_diff