Maximum Difference Between Node and Ancestor

theabbie

Abhishek Chaudhary

Posted on June 10, 2022

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
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theabbie
Abhishek Chaudhary

Posted on June 10, 2022

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