# 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 <= 105`

## 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);
}
}
```
Tags:
5 1 vote
Article Rating
Subscribe
Notify of 