## 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

 `````` 1 2 3 4 5 6 7 8 9 10 11 12 `````` ``````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

 `````` 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 `````` ``````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); } } ``````