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.

Solution

import java.util.StringTokenizer;

public class Solution {
    public boolean isValidSerialization(String preorder) {
        StringTokenizer st = new StringTokenizer(preorder, ",");
        boolean result = isValid(st);
        if (result && !st.hasMoreTokens()) {
            return true;
        } else {
            return false;
        }
    }

    private boolean isValid(StringTokenizer stringTokenizer) {
        if (!stringTokenizer.hasMoreTokens()) {
            return false;
        }

        String current = stringTokenizer.nextToken();
        if (current.equals("#")) {
            return true;
        }

        boolean left = isValid(stringTokenizer);
        if (!left) {
            return false;
        }

        boolean right = isValid(stringTokenizer);
        if (!right) {
            return false;
        }

        return true;
    }
}