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):

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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("#");     } }