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

代码如下：（方法一：利用树节点的 初度=入度）

 public class Solution {
     public boolean isValidSerialization(String preorder) {
         if(preorder.equals("#"))
         return true;

       String[] ss=preorder.split(",");
       if(ss[0].equals("#"))
       return false;
       int b=2;
       for(int i=1;i<ss.length;i++)
       {
           if(b<=0)
           return false;
           if(ss[i].equals("#"))
           {b=b-1;}
           else b=b+1;

       }
       if(b!=0)
       return false;
        return true;
     }
 }

代码如下：（方法二：利用一个根节点有两个子节点）

 public class Solution {
     public boolean isValidSerialization(String preorder) {
         if(preorder.length()==0||preorder=="") return false;
       String[] ss=preorder.split(",");
      Stack<String> stack=new Stack<>();

      for(int i=0;i<ss.length;)
      {
          while(stack.size()>=2&&stack.peek().equals("#")&&ss[i].equals("#"))
               {stack.pop();stack.pop();}
          if(stack.size()==1&&stack.peek().equals("#")&&i<ss.length-1)
          return false;
          else stack.push(ss[i++]);
      }

      if(stack.size()==1&&stack.peek().equals("#"))
      return true;

      return false;
     }
 }