# 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

```.wp-block-code{border:0;padding:0}.wp-block-code>div{overflow:auto}.shcb-language{border:0;clip:rect(1px,1px,1px,1px);-webkit-clip-path:inset(50%);clip-path:inset(50%);height:1px;margin:-1px;overflow:hidden;padding:0;position:absolute;width:1px;word-wrap:normal;word-break:normal}.hljs{box-sizing:border-box}.hljs.shcb-code-table{display:table;width:100%}.hljs.shcb-code-table>.shcb-loc{color:inherit;display:table-row;width:100%}.hljs.shcb-code-table .shcb-loc>span{display:table-cell}.wp-block-code code.hljs:not(.shcb-wrap-lines){white-space:pre}.wp-block-code code.hljs.shcb-wrap-lines{white-space:pre-wrap}.hljs.shcb-line-numbers{border-spacing:0;counter-reset:line}.hljs.shcb-line-numbers>.shcb-loc{counter-increment:line}.hljs.shcb-line-numbers .shcb-loc>span{padding-left:.75em}.hljs.shcb-line-numbers .shcb-loc::before{border-right:1px solid #ddd;content:counter(line);display:table-cell;padding:0 .75em;text-align:right;-webkit-user-select:none;-moz-user-select:none;-ms-user-select:none;user-select:none;white-space:nowrap;width:1%}```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;
}
}```Code language: Java (java)```

## 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);
}
}
```Code language: Java (java)```
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