leetcode 331. Verify Preorder Serialization of a Binary Tree 二叉树的前序序列验证

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

这道题很简单，就是验证一个给定的序列是否是一个二叉树的前序遍历的序列。

嗯嗯，这个还需要思考，看来编程能力还待提升。

建议和leetcode 297. Serialize and Deserialize Binary Tree 二叉树的序列化和反序列化 和 leetcode 654. Maximum Binary Tree 构造最大二叉树 一起学习

代码如下：

import java.util.Stack;public class Solution{    /*     * using a stack, scan left to right     * case 1: we see a number, just push it to the stack     * case 2: we see #, check if the top of stack is also #     * if so, pop #, pop the number in a while loop, until top of stack is not #     * if not, push it to stack     * in the end, check if stack size is 1, and stack top is #     * */    public boolean isValidSerialization1(String preorder)     {        if (preorder == null)             return false;        Stack<String> stack = new Stack<>();        String[] str=preorder.split(",");        for(int i=0;i<str.length;i++)        {            if(str[i].equals("#"))            {                while(stack.isEmpty()==false && stack.peek().equals("#"))                {                    stack.pop();                    if(stack.isEmpty())                        return false;                    stack.pop();                }                stack.push(str[i]);            }else                 stack.push(str[i]);        }        return stack.size()==1 && stack.peek().equals("#");    }    /*     * 其实，只是判断节点的位置和数量是否有误，因此不需要栈，也可以操作。     * 非叶子结点，入度是1，出度是2     * 叶子节点，入度是1，出度是0     * 所以计算diff表示总的（入度-出度），初始化1     * 主要过程中，出现diff<0表示出错，结束后diff==0才可以     * */    public boolean isValidSerialization(String preorder)     {        String[] nodes = preorder.split(",");        int diff = 1;        for (String node: nodes)         {            diff--;            if (diff <= -1)                 return false;            if (node.equals("#")==false)                 diff += 2;        }        return diff == 0;    }}