331. Verify Preorder Serialization of a Binary Tree

One way to serialize a binary tree is to use pre-order traversal. When we encounter a non-null node, we record the node's value. If it is a null node, we record using a sentinel value such as #.
     _9_
    /   \
   3     2
  / \   / \
 4   1  #  6
/ \ / \   / \
# # # #   # #
For example, the above binary tree can be serialized to the string "9,3,4,#,#,1,#,#,2,#,6,#,#", where # represents a null node.
Given a string of comma separated values, verify whether it is a correct preorder traversal serialization of a binary tree. Find an algorithm without reconstructing the tree.
Each comma separated value in the string must be either an integer or a character '#' representing null pointer.
You may assume that the input format is always valid, for example it could never contain two consecutive commas such as "1,,3".
Example 1:
Input: "9,3,4,#,#,1,#,#,2,#,6,#,#"
Output: true
Example 2:
Input: "1,#"
Output: false

See detailed comments below. Time complexity is O(n), space is also O(n) for the stack.

public class Solution {
    public boolean isValidSerialization(String preorder) {
        // using a stack, scan left to right
        // case 1: we see a number, just push it to the stack
        // case 2: we see #, check if the top of stack is also #
        // if so, pop #, pop the number in a while loop, until top of stack is not #
        // if not, push it to stack
        // in the end, check if stack size is 1, and stack top is #
        if (preorder == null) {
            return false;
        }
        Stack<String> st = new Stack<>();
        String[] strs = preorder.split(",");
        for (int pos = 0; pos < strs.length; pos++) {
            String curr = strs[pos];
            while (curr.equals("#") && !st.isEmpty() && st.peek().equals(curr)) {
                st.pop();
                if (st.isEmpty()) {
                    return false;
                }
                st.pop();
            }
            st.push(curr);
        }
        return st.size() == 1 && st.peek().equals("#");
    }
}