Leetcode 993

同深度，不同父节点就可判断

直接dfs查找，若不存在，返回null

不在第n层就继续往下走

public class isCousins993 {
    TreeNode parX;
    TreeNode parY;
    int deepX;
    int deepY;

    public boolean isCousins(TreeNode root, int x, int y) {
        dfs(root, null, 0, x, y);
        return parX != parY && deepX == deepY;
    }

    private void dfs(TreeNode root, TreeNode par, int deep, int x, int y) {
        if (root == null)
            return;

        if (root.val == x) {
            deepX = deep;
            parX = par;
        } else if (root.val == y) {
            deepY = deep;
            parY = par;
        } else {
            dfs(root.left, root, deep + 1, x, y);
            dfs(root.right, root, deep + 1, x, y);
        }
    }
}