LeetCode-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

Credits:
Special thanks to @dietpepsi for adding this problem and creating all test cases.

Analysis:

We use a recursion function to verify whether subtree starting at @start is a valid binary tree or how far it can go to maintain a valid binary tree. This includes a vlid left subtree starting from @start+1 and a valid right subtree staring from @leftEnd+1.

Solution:

public class Solution {
   public boolean isValidSerialization(String preorder) {
        String[] s = preorder.split(",");

        int res = isValidRecur(0, s);
        // This condition after completing recursion is IMPORTANT!
        // Either we found a false subtree, or we ended before going through the
        // whole string.
        // NOTE: it is (res+1) rather (res)
        if (res == -1 || res + 1 < s.length)
            return false;
        return true;
    }

    public int isValidRecur(int start, String[] preorder) {
        if (start >= preorder.length)
            return -1;
        if (preorder[start].equals("#"))
            return start;

        int leftEnd = isValidRecur(start + 1, preorder);
        if (leftEnd == -1)
            return leftEnd;
        int rightEnd = isValidRecur(leftEnd + 1, preorder);
        if (rightEnd == -1)
            return rightEnd;

        return rightEnd;
    }
    
}