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:
"9,3,4,#,#,1,#,#,2,#,6,#,#"
Return true

Example 2:
"1,#"
Return false

Example 3:
"9,#,#,1"
Return false

The question should be rephrased as if the preorder serialization can be reconstructed to a valid binary tree. The approach from the contributor is that in a tree:

All non-null nodes have two out-degrees and one in-degree (2 children and 1 parent) except for the root;
All null nodes have one in-degree.

The difference between  out-degrees and in-degrees should never be negative, and overall should equal 0.

public boolean isValidSerialization(String preorder) {
        if (preorder == null || preorder.length() == 0)
            return false;
        //initialize to 1 because there is no in-degree for root
        int diff = 1;
        String[] nodes = preorder.split(",");
        for (int i = 0; i < nodes.length; i++) {
            if (--diff < 0) return false;
            if (!"#".equals(nodes[i])) diff += 2;
        }
        return diff == 0;
    }