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

Example 2:
"1,#"
Return false

Example 3:
"9,#,#,1"
Return false

Understand the problem:
The key of the problem is if a preorder traversal of a binary tree is valid, a leaf node must have the sequence like "number, #, #". Therefore, we can start from leaf nodes of tree, remove the leaf nodes by replacing the number, #, # by a single # until the tree becomes empty. 

Code (Java):

public class Solution {
    public boolean isValidSerialization(String preorder) {
        if (preorder == null || preorder.length() == 0) {
            return true;
        }
        
        String[] nodes = preorder.split(",");
        Stack<String> stack = new Stack<>();
        
        for (String node : nodes) {
            if (node.equals("#")) {
                while (!stack.isEmpty() && stack.peek().equals("#")) {
                    stack.pop();
                    if (stack.isEmpty()) {
                        return false;
                    }
                    
                    stack.pop();
                }
            }
            
            stack.push(node);
        }
        
        return stack.size() == 1 && stack.peek().equals("#");
    }
}