leetcode--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.

 

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利用栈数字入栈，＃出栈

public class Solution {
    public boolean isValidSerialization(String preorder) {
        Stack<String> stack=new Stack<String>();
        String []num=preorder.split(",");
        int i=0;
        if(num[0].equals("#")&&num.length>1)return false;
        for(;i<num.length;i++){
            if(!num[i].equals("#")){
                stack.push(num[i]);
            }else if(stack.size()>0&&num[i].equals("#")){
                stack.pop();
            }
            else if(stack.size()==0&&num[num.length-1].equals("#")&&i==num.length-1){
                return true;
            }else{
            return false;
            }
        }
        return false;
    }
}