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;
}
}