Cousins in Binary Tree

In a binary tree, the root node is at depth 0, and children of each depth k node are at depth k+1.

Two nodes of a binary tree are cousins if they have the same depth, but have different parents.

We are given the root of a binary tree with unique values, and the values x and y of two different nodes in the tree.

Return true if and only if the nodes corresponding to the values x and y are cousins.

Example 1:

Input: root = [1,2,3,4], x = 4, y = 3
Output: false

Example 2:

Input: root = [1,2,3,null,4,null,5], x = 5, y = 4
Output: true

Example 3:

Input: root = [1,2,3,null,4], x = 2, y = 3
Output: false

Constraints:

The number of nodes in the tree will be between 2 and 100.
Each node has a unique integer value from 1 to 100.

Solution:

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
class Solution {
    int[] parent = new int[101];
    int[] depth = new int[101];
    
    public boolean isCousins(TreeNode root, int x, int y) {
        pre(root, 0, null);
        return parent[x] != parent[y] && depth[x] == depth[y];
    }
    
    private void pre(TreeNode root, int d, TreeNode p) {
        if (root == null) return;
        if (p != null) {
            parent[root.val] = p.val;
        }
        depth[root.val] = d;
        pre(root.left, d + 1, root);
        pre(root.right, d + 1, root);
    }
}