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 #.

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     _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"

思路

如果我们利用栈的方法来解决，我们可以从左到右对字符串进行扫描。如果我们看到一个数字，将其入栈；如果看到#号，检查栈顶是否也为#号，如果是，一直做pop操作直到栈顶不是#号，如果不是，则将#号入栈。扫描完后检查栈的大小为1且栈顶为#即可。

我们还可以用图的出度入度来解决。

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public class Solution {
    public boolean isValidSerialization(String preorder) {
        String[] array = preorder.split(",");
        //The in-degree of root is 0, out-degree is 2. Compansation.
        int count = -1;
        for(int i = 0; i < array.length; i++){
            count++;
            if (count > 0) return false;
            //Positive for in-degree, negative for out-degree.
            //Normal node: in-degree:1, out-degree:2
            if (!array[i].equals("#")) count=count-2;
        }
        return count==0;
    }
}

参考资料：https://discuss.leetcode.com/topic/35977/simple-python-solution-using-stack-with-explanation