Maximum Difference Between Node and Ancestor
<p>Given the <code>root</code> of a binary tree, find the maximum value <code>v</code> for which there exist <strong>different</strong> nodes <code>a</code> and <code>b</code> where <code>v = |a.val - b.val|</code> and <code>a</code> is an ancestor of <code>b</code>.</p>

<p>A node <code>a</code> is an ancestor of <code>b</code> if either: any child of <code>a</code> is equal to <code>b</code>&nbsp;or any child of <code>a</code> is an ancestor of <code>b</code>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/1000-1099/1026.Maximum%20Difference%20Between%20Node%20and%20Ancestor/images/tmp-tree.jpg" style="width: 400px; height: 390px;" />
<pre>
<strong>Input:</strong> root = [8,3,10,1,6,null,14,null,null,4,7,13]
<strong>Output:</strong> 7
<strong>Explanation: </strong>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.</pre>

<p><strong class="example">Example 2:</strong></p>
<img alt="" src="https://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/1000-1099/1026.Maximum%20Difference%20Between%20Node%20and%20Ancestor/images/tmp-tree-1.jpg" style="width: 250px; height: 349px;" />
<pre>
<strong>Input:</strong> root = [1,null,2,null,0,3]
<strong>Output:</strong> 3
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li>The number of nodes in the tree is in the range <code>[2, 5000]</code>.</li>
	<li><code>0 &lt;= Node.val &lt;= 10<sup>5</sup></code></li>
</ul>
Tree; Depth-First Search; Binary Tree
/**
 * Definition for a binary tree node.
 * type TreeNode struct {
 *     Val int
 *     Left *TreeNode
 *     Right *TreeNode
 * }
 */
func maxAncestorDiff(root *TreeNode) (ans int) {
	var dfs func(*TreeNode, int, int)
	dfs = func(root *TreeNode, mi, mx int) {
		if root == nil {
			return
		}
		ans = max(ans, max(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)
	}
	dfs(root, root.Val, root.Val)
	return
}

func abs(x int) int {
	if x < 0 {
		return -x
	}
	return x
}