331. 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.

首先是根据树的性质解题，一个数字，肯定对应两个节点，所以遇到数字，count+=2，遇到#不操作。同时在每次遍历的之后，减去自身。如果最后count=0，说明符合规则，这个解题方法无视Preorder，Inorder，和Postorder。代码如下：

public boolean isValidSerialization(String preorder) {    String[] nodes = preorder.split(",");    int diff = 1;    for (String node: nodes) {        if (--diff < 0) return false;        if (!node.equals("#")) diff += 2;    }    return diff == 0;}

另一种方法，遇到两个连续的#，pop两次，第一次相当于把左节点pop出去，第二次相当于把夫节点pop出去，这是如果peek不是#，把#push进来充当上上个夫节点的右节点，如果peek还是#，接着push。代码如下：

public class Solution {    public boolean isValidSerialization(String preorder) {        if (preorder == null && preorder.length() == 0) {            return false;        }        String[] strs = preorder.split(",");        Stack<String> stack = new Stack<String>();        for (String str: strs) {            while (str.equals("#") && !stack.isEmpty() && stack.peek().equals(str)) {                stack.pop();                if (stack.isEmpty()) {                    return false;                }                stack.pop();            }            stack.push(str);        }        return stack.size() == 1 && stack.peek().equals("#");    }}