pylist

1026. 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:

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: TreeNode) -> int:
        self.V = 0
        self.dfs(root)
        return self.V
        
    def dfs(self, node):
        if node is None:
            return 0, float('inf')
        if node.left is None and node.right is None:
            return node.val, node.val
        
        lmax, lmin = self.dfs(node.left)
        rmax, rmin = self.dfs(node.right)
        
        self.V= max([self.V,
                     node.val - lmin,
                     node.val -rmin,
                     lmax - node.val,
                     rmax -node.val])
        
        treemax = node.val
        treemin = node.val
        
        treemax = max([treemax, lmax, rmax])
        treemin = min([treemin, lmin, rmin])
        return treemax, treemin

Details: