Find second largest element in the binary search tree

Find second largest element - review peer's code

/*
10:32 pm

Find second largest element in binary search tree

stack iterative traversal of first 2 elements
time o(n) space o(n)

if tree has less than 2 elements return null

Ex 1 res=5
    10*  no right child, so no larger element, 10 is largest possible element
  5*     5,10
  
    10*  no right child, so no larger element, 10 is largest possible element
  5*      5,10  since 5 has right child, right child must be larger 
1   8*      8,10 ret 8
  
  
    10*  no right child, so no larger element, 10 is largest possible element
  5*      5,10  since 5 has right child, right child must be larger 
1   8*      8,10 child is left which is smaller, does not affect result
   6  
  
    10*  no right child, so no larger element, 10 is largest possible element
  5*      5,10  since 5 has right child, right child must be larger 
1   8*      8 has right child, has to be larger
   6 9     9,10

result=9
     10//10
    9//9,10 
   8//left is smaller, queue is full, so doesn't matter
  7
  
  ascending pq
         
         3
       2
      1
      
       4
    2 
 1     3
  [4, 2, 3]   
     

         10         
   5            15*
      8    13*
             14*
             
             3*
           2*
         1*
*/
10X,13X,14,15
public int secondLargestBSTNode(TreeNode root){
    //null check, return -1
    int result = -1;
    PriorityQueue<Integer>pq=new PriorityQueue<>();   

    while(root != null){//2
        pq.offer(root.val);//1,2,3
        if(pq.size()>2){
            pq.poll();//2,3
        }
        if(root.right != null){
            root = root.right;//
        }
        else{
            root = root.left;//null
        }
    }
    //need check pq.size()==2
    return pq.poll();//14
}

time
tree ologn
k=2
o*k (log n)