331. Verify Preorder Serialization of a Binary Tree(难)

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

一、

将元素依次push到vector，如果vector的大小大于等于3，且后两个为#，倒数第三个不为#，则弹出他们，push一个#。

如果最后vector大小为1，且为#，则是合法的前序序列！

class Solution {public:bool isValidSerialization(string preorder) {vector<string> vec;stringstream ss(preorder);string val;while (getline(ss, val, ',')){vec.push_back(val);while (vec.size() >= 3 && vec[vec.size() - 1] == "#"&&vec[vec.size() - 2] == "#" && vec[vec.size() - 3] != "#"){vec.pop_back(); vec.pop_back(); vec.pop_back();vec.push_back("#");}}return vec.size() == 1 && vec[0] == "#";}};

二、

在构建的过程中，记录出度与入度之差，记为diff = outdegree - indegree

当遍历节点时，我们令diff - 1（因为节点提供了一个入度）。如果diff<0,返回false；如果节点非空，再令diff + 2（因为节点提供了2个出度）。

最后判断diff==0;

理解：如果在遍历过程中的某时刻，系统的入度>出度，则说明当前序列中出现过的所有分支节点的“空闲分支”均已用完，后序节点没有办法挂载到之前出现的节点之上，从而判定先序遍历的序列是不合法的。

class Solution {public:bool isValidSerialization(string preorder) {stringstream ss(preorder);string val;int diff = 1;while (getline(ss, val, ',')){diff -= 1;if (diff < 0) return false;if (val != "#")diff += 2;}return diff == 0;}};