1026. Maximum Difference Between Node and Ancestor

Problem

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.
 * function TreeNode(val, left, right) {
 *     this.val = (val===undefined ? 0 : val)
 *     this.left = (left===undefined ? null : left)
 *     this.right = (right===undefined ? null : right)
 * }
 */
/**
 * @param {TreeNode} root
 * @return {number}
 */
var maxAncestorDiff = function(root) {
    return helper(root.val, root.val, root);
};

var helper = function(max, min, node) {
    if (!node) return 0;
    var newMax = node.val > max ? node.val : max;
    var newMin = node.val < min ? node.val : min;
    return Math.max(
        Math.abs(max - node.val),
        Math.abs(min - node.val),
        node.left ? helper(newMax, newMin, node.left) : 0,
        node.right ? helper(newMax, newMin, node.right) : 0,
    );
};

Explain:

use dfs to visit every node in the tree, carry the maximum and minimum number all the way down, the result should be |max - node.val| or |min - node.val|

Complexity:

• Time complexity : O(n).
• Space complexity : O(n).