## DEV Community

Abhishek Chaudhary

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# Maximum Difference Between Node and Ancestor

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`

SOLUTION:

``````# 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:
nodes = [(root, root.val, root.val)]
mdiff = 0
while len(nodes) > 0:
curr, currmin, currmax = nodes.pop()
mdiff = max(mdiff, currmax - currmin)
if curr.left:
nodes.append((curr.left, min(currmin, curr.left.val), max(currmax, curr.left.val)))
if curr.right:
nodes.append((curr.right, min(currmin, curr.right.val), max(currmax, curr.right.val)))
return mdiff
``````