Verify Preorder Serialization of a Binary Tree

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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 #.

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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:
“9,3,4,#,#,1,#,#,2,#,6,#,#”
Return true

Example 2:
“1,#”
Return false

Example 3:
“9,#,#,1”
Return false

解法是利用栈，逐个消去节点的子树：

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"9,3,4,#,#,1,#,#,2,#,6,#,#"

9    3    4    #     #        9    3    #
true true true false false -> true true false

9    3    #     1    #     #        9    3    #     #        9    #     
true true false true false false -> true true false false -> true false

9    #     2    #     6    #     #        9    #     2    #     #        9    #     #        #
true false true false true false false -> true false true false false -> true false false -> false

AC代码，使用了vector作为栈：

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class Solution {
public:
    bool isValidSerialization(string preorder) {
        string s;
        vector<bool> sk;
        if (preorder[0] == '#' && preorder.length() == 1)
        {
            return true;
        }
        if(preorder.length() < 3 || preorder[0] == '#')
        {
            return false;
        }
        preorder.push_back(',');
        for (int i = 0 ; i < preorder.length(); ++i)
        {
            if (preorder[i] == ',')
            {
                if (s.length() && s[0] != '#')
                {
                    sk.push_back(true);
                }
                s.clear();
                if (sk.size() > 2)
                {
                    size_t preLen = 0;
                    while(preLen != sk.size() && sk.size() >= 2)
                    {
                        preLen = sk.size();
                        if (sk[sk.size()-1] == false && sk[sk.size()-2] == false)
                        {
                            sk.pop_back();
                            if (sk.size() == 0)
                            {
                                return false;
                            }
                            sk.pop_back();
                            if (sk.size() == 0)
                            {
                                return false;
                            }
                            sk.pop_back();
                            sk.push_back(false);
                            if (sk[0] == false && i < preorder.length() - 1)
                            {
                                return false;
                            }
                        }
                    }
                }
            }
            else if (preorder[i] == '#')
            {
                sk.push_back(false);
            }
            else
            {
                s.push_back(preorder[i]);
            }
        }
        if (sk.size() == 1)
        {
            return true;
        }
        return false;
    }
};