[Leetcode] 331. Verify Preorder Serialization of a Binary Tree 解题报告

题目：

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

思路：

我们定义一个栈，对于字符串中每个代表结点的字符，做如下操作：

1）如果是代表空节点的#，则看栈顶元素是否也是#,如果是，则说明我们遇到了一个叶子结点，此时模拟不断消除该叶子结点（弹出栈顶的两个元素，并将#压入栈），直到不存在当前需要消除的叶子结点。这是因为在前序遍历中，叶子结点在字符串中一定是以连续三个字符“value,#,#”的形式存在。

2）如果不满足上述消除叶子结点的条件，则将该字符入栈。

如果字符串是合法的二叉树前序遍历，则最终栈中会只有一个元素“#”，返回true。否则返回false。

代码：

class Solution {public:    bool isValidSerialization(string preorder) {        int start = 0, end = 0;        stack<string> st;        while(start < preorder.size()) {            end = start;            while(end < preorder.size() && preorder[end] != ',') {                ++end;            }            string word = preorder.substr(start, end - start);            if(word != "#") {                st.push(word);            }            else {                while(!st.empty() && st.top() == "#") {                    st.pop();                    if(st.empty()) {                        return false;                    }                    st.pop();                }                st.push("#");            }            start = end + 1;        }        return st.size() == 1 && st.top() == "#";    }};