#1026

Medium

auto_awesomeBinary-tree traversal and state tracking

LeetCode Problem Workspace

Maximum Difference Between Node and Ancestor

Find the maximum absolute difference between a node and its ancestor in a binary tree.

Problem Statement

Given the root of a binary tree, you need to find the maximum value v where v = |a.val - b.val| and a is an ancestor of b. A node a is considered an ancestor of b if either: any child of a is b, or any child of a is an ancestor of b.

For example, consider the tree rooted at [8,3,10,1,6,null,14,null,null,4,7,13]. The maximum difference between any ancestor and descendant node is 7, where |8 - 1| = 7.

Examples

Example 1

Input: root = [8,3,10,1,6,null,14,null,null,4,7,13]

Output: 7

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

Example details omitted.

Constraints

The number of nodes in the tree is in the range [2, 5000].
0 <= Node.val <= 105

Solution Approach

DFS Traversal with Min and Max Tracking

Perform a Depth-First Search (DFS) traversal of the tree. For each node, track the minimum and maximum values found in its subtree. The result is the maximum absolute difference between any ancestor and its descendants.

Track Min and Max at Each Node

At each node during traversal, keep track of the minimum and maximum values encountered on the path from the root to that node. The difference between the current node's value and these tracked values will give potential answers.

Optimal Performance with State Management

To avoid redundant computations, manage the state of the minimum and maximum values as you recurse down the tree. This allows the algorithm to compute the maximum difference efficiently.

Complexity Analysis

Metric	Value
Time	Depends on the final approach
Space	Depends on the final approach

The time complexity of this solution is O(n) because we traverse each node once. The space complexity is O(h) where h is the height of the tree, due to the recursive call stack. In the worst case (skewed tree), h can be O(n), making the space complexity O(n).

What Interviewers Usually Probe

Look for a clear understanding of tree traversal and recursive solutions.
Ensure the candidate can explain how state tracking is essential for finding the maximum difference.
Evaluate if the candidate can optimize the solution and explain the trade-offs of space and time complexities.

Common Pitfalls or Variants

Common pitfalls

Failing to correctly track both the minimum and maximum values as you traverse down the tree.
Misunderstanding the ancestor relationship, which could lead to incorrect calculations of the node differences.
Not handling edge cases such as very small or unbalanced trees effectively.

Follow-up variants

Finding the maximum difference between nodes for a non-binary tree structure.
Implementing the solution iteratively without recursion.
Handling a tree where the node values are all the same, potentially simplifying the problem.

FAQ

How do I approach the Maximum Difference Between Node and Ancestor problem?

Use a DFS traversal while tracking the minimum and maximum values at each node to find the maximum absolute difference between ancestor and descendant nodes.

What is the time complexity of the Maximum Difference Between Node and Ancestor problem?

The time complexity is O(n), where n is the number of nodes, as each node is visited exactly once during DFS traversal.

Can this problem be solved iteratively?

Yes, it can be solved iteratively using a stack to simulate the recursive DFS approach.

How does the ancestor relationship affect the solution?

The ancestor relationship is crucial because it ensures that the value difference must always be between a node and its ancestor, not just any two nodes.

What is the space complexity of this solution?

The space complexity is O(h), where h is the height of the tree, due to the recursion stack. In the worst case, h can be O(n) for a skewed tree.

terminal

Solution

Solution 1: DFS

For each node, to find the maximum difference with its ancestor nodes, we only need to find the difference between the maximum and minimum values of the ancestor nodes. The maximum difference among all nodes and their ancestor nodes is the answer.

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# 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 maxAncestorDiff(self, root: Optional[TreeNode]) -> int:
        def dfs(root: Optional[TreeNode], mi: int, mx: int):
            if root is None:
                return
            nonlocal ans
            ans = max(ans, abs(mi - root.val), abs(mx - root.val))
            mi = min(mi, root.val)
            mx = max(mx, root.val)
            dfs(root.left, mi, mx)
            dfs(root.right, mi, mx)

        ans = 0
        dfs(root, root.val, root.val)
        return ans

Continue Practicing

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