leetcode - 1026. Maximum Difference Between Node and Ancestor

Description

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 <= 10^5

Solution

Solved after hints.

For each node, find the smallest and largest value of its descendants, and the result will be either node.val - smallest or largest - node.val.

Go with postorder traversal, and use a map to store the values of the nodes.

Time complexity: $o(n)$
Space complexity: $o(n)$

Code

# 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_info = {None: [100001, -1]}
        stack = [(root, 0)]
        res = 0
        while stack:
            node, status = stack.pop()
            if status == 0:
                stack.append((node, 1))
                if node.right:
                    stack.append((node.right, 0))
                if node.left:
                    stack.append((node.left, 0))
            else:
                left_min, left_max = nodes_info[node.left]
                right_min, right_max = nodes_info[node.right]
                nodes_info[node] = [min(left_min, right_min, node.val), max(left_max, right_max, node.val)]
                res = max(res, max([nodes_info[node][1] - node.val, node.val - nodes_info[node][0]]))
        return res