<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:
        queue = deque([(root, root.val, root.val)])
        ans = 0
        while queue:
            node, curr_max, curr_min = queue.popleft()

            ans = max(ans, abs(curr_max - curr_min))
            if node.left:
                queue.append((node.left, max(node.left.val, curr_max), min(node.left.val, curr_min)))
            if node.right:
                queue.append((node.right, max(node.right.val, curr_max), min(node.right.val, curr_min)))
        
        return ans

        # time, space O(N)



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