Verify Preorder Serialization of a Binary Tree

Problem

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

Solution 

要注意到叶子节点有一个特殊属性， 就是后面必定有两个连续的 ‘＃’。这个属性反之也成立，就是说如果有两个连续的‘＃’， 那它前面一定是个数字使之成为叶子节点。

这样就可以找到叶子节点后把它删除，插入一个 ‘＃’ ， 再一直继续就好了～～ 

有点像  topological sort , 只是现在不是一层层的减，而是一个个的减，减到最后只剩下一个 ‘＃’ 就说明是  valid preorder traversal

这个解法参看了 http://algobox.org/verify-preorder-serialization-of-a-binary-tree/

class Solution {public:    bool isValidSerialization(string preorder) {        if(preorder.empty()) return false;        if(preorder.size() == 1 && preorder == "#") return true;                stack<char> stk;        preorder.push_back(',');        for( int i = 0; i < preorder.size(); i+=2 ) {            if(preorder[i] == '#') {                if(stk.empty() ) return false;                                while( !stk.empty() && stk.top() == '#'){                    stk.pop();                    if(stk.empty() || stk.top() != '*') return false;                    stk.pop();                }                stk.push('#');                            }            else {                stk.push('*');                int j = preorder.find_first_of(',', i);                i = j - 1;            }        }        return stk.size()==1 && stk.top() == '#';    }};

另一个解法是利用出度和入度

https://leetcode.com/discuss/84257/simplest-python-solution-with-explanation-stack-recursion

We just need to remember how many empty slots we have during the process.

Initially we have one ( for the root ).

for each node we check if we still have empty slots to put it in.

• a null node occupies one slot.
• a non-null node occupies one slot before he creates two more. the net gain is one.

class Solution(object):    def isValidSerialization(self, preorder):        """        :type preorder: str        :rtype: bool        """        # remember how many empty slots we have        # non-null nodes occupy one slot but create two new slots        # null nodes occupy one slot        p = preorder.split(',')        #initially we have one empty slot to put the root in it        slot = 1        for node in p:            # no empty slot to put the current node            if slot == 0:                return False            # a null node?            if node == '#':                # ocuppy slot                slot -= 1            else:                # create new slot                slot += 1        #we don't allow empty slots at the end        return slot==0