331. Verify Preorder Serialization of a Binary Tree

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每日一题 2019 - 01 - 23

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 #.

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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:

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Input: "9,3,4,#,#,1,#,#,2,#,6,#,#"
Output: true

Example 2:

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Input: "1,#"
Output: false

Example 3:

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Input: "9,#,#,1"
Output: false

解法：

这个题让我们判断给定的序列是不是二叉树的先序遍历序列，这里我们需要注意给定的序列的特点：

  • 任意一个非空结点一定有两个子结点，这两个子结点可以为空，也即#
  • 一颗空树的序列为 #，满足题意要求

同时我们还需要注意题中所给的二叉树的一个特点就是：

  • 空结点的个数 = 非空结点的个数 + 1

所以如果我们在遍历序列的过程中发现#的个数大于当前已经遍历过的非空节点的个数 + 1，那么证明这颗树是不存在的，有了这个思路我们就可以完成这个题的解答。

代码：

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class Solution {
public:
    bool isValidSerialization(string preorder) {
        int n = preorder.size();
        if(n == 0) return true;
        vector<string> v;
        int pos = -1;
        for(int i = 0; i < n; i++)
        {
            if(preorder[i] == ',')
            {
                v.push_back(preorder.substr(pos+1, i-pos-1));
                pos = i;
            }
        }
        if(preorder[n-1] != '#') return false;
        v.push_back("#");
        int cnt = 1;
        for(int i = 0; i < v.size(); i++)
        {
            if(cnt <= 0) return false;
            if(v[i] == "#") cnt -= 1;
            else cnt += 1;
        }    
        return cnt == 0;
    }
};
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