How do you solve Maximum Difference Between Node and Ancestor on LeetCode?

Maximum Difference Between Node and Ancestor (LeetCode #1026) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(h) space complexity. Key insight: The maximum difference can be efficiently calculated using DFS by tracking min and max values.

What is the brute force solution for Maximum Difference Between Node and Ancestor?

The brute force approach for Maximum Difference Between Node and Ancestor is Brute Force, with O(n²) time complexity and O(1) space complexity. We can explore all possible ancestor-node pairs and calculate their differences. This approach is straightforward but inefficient, as it checks every node against every ancestor. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Maximum Difference Between Node and Ancestor?

Maximum Difference Between Node and Ancestor uses the following concepts: Tree, Depth-First Search, Binary Tree. The recommended patterns to study are: Depth-First Search, Tree Traversal.

What are the interview tips for Maximum Difference Between Node and Ancestor?

Clarify the definition of ancestor in the context of the problem. Discuss the brute force approach before jumping to optimization to show your thought process. Use examples to illustrate your understanding of the problem and potential edge cases.

What are common mistakes when solving Maximum Difference Between Node and Ancestor?

Not considering all ancestor nodes when calculating differences. Failing to update min and max values correctly during traversal.

#1026

Maximum Difference Between Node and Ancestor

Medium

TreeDepth-First SearchBinary TreeDepth-First SearchTree Traversal

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

💡

Intuition

Time O(n)Space O(h)

Instead of checking all ancestors for every node, we can use a depth-first search (DFS) to track the minimum and maximum values encountered along the path from the root to each node. This allows us to compute the maximum difference in a single pass.

⚙️

Algorithm

3 steps

1Step 1: Perform a DFS traversal of the tree, starting from the root.
2Step 2: At each node, update the minimum and maximum values encountered on the path from the root.
3Step 3: Calculate the difference between the current node's value and the minimum/maximum values, and update the maximum difference accordingly.

solution.py16 lines

1# Full working Python code
2class TreeNode:
3    def __init__(self, val=0, left=None, right=None):
4        self.val = val
5        self.left = left
6        self.right = right
7
8def maxAncestorDiff(root):
9    def dfs(node, min_val, max_val):
10        if not node:
11            return max_val - min_val
12        min_val = min(min_val, node.val)
13        max_val = max(max_val, node.val)
14        return max(dfs(node.left, min_val, max_val), dfs(node.right, min_val, max_val))
15
16    return dfs(root, root.val, root.val)

ℹ

Complexity note: The time complexity is linear because we visit each node exactly once. The space complexity is O(h) due to the recursion stack, where h is the height of the tree.

1The maximum difference can be efficiently calculated using DFS by tracking min and max values.
2Understanding the ancestor relationship in trees is crucial for solving this problem.

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