leetcode 331. Verify Preorder Serialization of a Binary Tree

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331. Verify Preorder Serialization of a Binary Tree

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Total Accepted: 10790 Total Submissions: 34071 Difficulty: Medium

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

 

题意：

判断给的字符串 是不是一个合法的 前序遍历

 

思路：

用栈，如果 一个节点的 左子树 和 右子树 都 true 的话，该节点 向上返回true

 

false的情况：

1. 某个节点已经 左子树 和 右子树 都遍历过了，后续又 涉及到该节点的儿子（如 1,#,#,#）

2. 根节点是 已经 访问结束,后续还有其它节点 （#,1  或 1,#,#,1）

 

150 / 150 test cases passed.	Status:  Accepted
Runtime: 8 ms

 

 

class Solution {
public:
    bool isValidSerialization(string preorder) {
        int l = preorder.length();
        if(preorder[0] == '#'){
            if(l == 1) return true;
            else return false;
        }
        int i = 0;
        stack<int> s;
        s.push(0);
        while(i<l && preorder[i] != ',') i++;
        i++;
        int te;
        while(i < l)
        {
            if(preorder[i] == '#'){
                if(s.empty() == 1) return false;
                while(!s.empty())
                {
                    te = s.top();
                    if(te == 2) return false;
                    s.pop();
                    s.push(te + 1);
                    te = s.top();
                    if(te == 1){
                        break;
                    }
                    if(te == 2 ){
                        s.pop();
                    }
                }
                if(s.empty() == 1 && i < l - 1) return false;
            }
            else{
                s.push(0);
            }
            while(i < l && preorder[i] != ',') i++;
            i++;
        }
        if(s.size() == 0) return true;
        return false;
    }
};