<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://assets.leetcode.com/uploads/2020/11/09/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://assets.leetcode.com/uploads/2020/11/09/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>

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution:
    def maxAncestorDiff(self, root: Optional[TreeNode]) -> int:
        global MX 

        def ss(root,mx,mn):
            global MX 
            if root:
                MX = max(MX, abs(root.val-mx), abs(root.val-mn))
                ss(root.left, max(mx,root.val),min(mn,root.val))
                ss(root.right, max(mx,root.val),min(mn,root.val)) 
        mx = root.val 
        MX = 0 
        ss(root,mx,mx)  
        return MX 
17800