331. Verify Preorder Serialization of a Binary Tree

Task:

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

Some Questions:

 Is the value in each node single digit?

What's the length of the string?

Solution:

At the first glance, I have no idea about the problem. So I decided to simulate some cases to find whether there are some regulations. After trying several cases I found that the value before two consecutive '#'s should be a leaf and there are two consecutive '#'s after a leaf. So I think out a solution using stack.

Here is the steps of the algorithm: 

1)push a '*' when we encounter a number;

2)push a '#' when we encounter a '#', 

then judge whether the top two elements in the stack are '#',  -------------- question 1

if yes for question 1, 

judge whether the third element is '*'  ---------------- question 2

if yes for question 2, pop the top three element, then push a '#'.

if no for question 2, then return false.

if no for question 1, then continue.

In the end, if there is only one element in the stack and it is '#', then return true, else return false.

This algorithm runs in O(n) time complexity and O(n) space complexity.

Code:

class Solution {
public:
    bool isValidSerialization(string preorder) {
        stack<char>stk;
        int i,len=preorder.length();
        for(i=0;i<len;i++)
        {
            if(isdigit(preorder[i]))
            {
                while(i<len&&isdigit(preorder[i]))
                {
                    i++;
                }
                i--;
                stk.push('*');
            }
            else if(preorder[i]=='#')
            {
                stk.push('#');
                while(stk.size()>=2&&stk.top()=='#')
                {
                    stk.pop();
                    if(stk.top()=='#')//get the last two '#', change '*' to '#'
                    {
                        stk.pop();
                        if(stk.empty())return false;
                        if(stk.top()=='#')return false;
                        stk.pop();
                        stk.push('#');
                    }
                    else
                    {
                        stk.push('#');
                        break;
                    }
                }
            }
        }
        
        return stk.size()==1&&stk.top()=='#';
    }
};