[leetcode] 331. Verify Preorder Serialization of a Binary Tree-优快云博客

本文介绍了一种不重建树即可验证字符串是否为正确二叉树前序遍历序列的方法。通过计数节点和空节点的数量，确保序列的有效性。

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

解法一：

感觉规律还是难发现。

1. 最后一个肯定是"#"

2. 数字的个数比"#"的个数少一个

3. 在计数过程中，"#"个数不能超过数字个数。

我们使用一个cnt来标记，遇到数字加一，遇到#减一，可见cnt不能小于0，最后的结果还要等于0。

class Solution {
public:
    bool isValidSerialization(string preorder) {
        if (preorder.empty()) return false;
        istringstream in(preorder);
        vector<string> nodes;
        string str; 
        while(getline(in,str,',')) nodes.push_back(str);
        
        int cnt = 0;
        for(int i=0; i<nodes.size()-1;i++){
            if(nodes[i]=="#") cnt--;
            else cnt++;
            if(cnt<0) return false;
        }
        return cnt==0&&nodes.back()=="#";
        
    }
};