How to Find the Maximum Difference Between Node and Ancestor in Java

The challenge

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 <= 10<sup>5</sup>

Definition for a Binary Tree Node

public class TreeNode {
    int val;
    TreeNode left;
    TreeNode right;
    TreeNode() {}
    TreeNode(int val) { this.val = val; }
    TreeNode(int val, TreeNode left, TreeNode right) {
        this.val = val;
        this.left = left;
        this.right = right;
    }
}

The solution in Java code

class Solution {
    public int maxAncestorDiff(TreeNode root) {
        if (root == null) return 0;
        return helper(root, root.val, root.val);
    }

    public int helper(TreeNode node, int curMax, int curMin) {
        // if encounter leaves, return the max-min along the path
        if (node == null) return curMax - curMin;

        // otherwise: update max and min and..
        // return the max of left and right subtrees

        curMax = Math.max(curMax, node.val);
        curMin = Math.min(curMin, node.val);

        int left = helper(node.left, curMax, curMin);
        int right = helper(node.right, curMax, curMin);

        return Math.max(left, right);
    }
}