1026. Maximum Difference Between Node and Ancestor - Medium

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

 

dfs, time = O(n), space = O(n)

class Solution {
    int maxDiff = 0;
    
    public int maxAncestorDiff(TreeNode root) {
        dfs(root, root.val, root.val);
        return maxDiff;
    }
    
    public void dfs(TreeNode root, int min, int max) {
        if(root == null) {
            return;
        }
        
        max = Math.max(max, root.val);
        min = Math.min(min, root.val);
        maxDiff = Math.max(maxDiff, max - min);
        
        dfs(root.left, min, max);
        dfs(root.right, min, max);
    }
}