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[] == '#'){
             if(l == ) return true;
             else return false;
         }
         int i = ;
         stack<int> s;
         s.push();
         while(i<l && preorder[i] != ',') i++;
         i++;
         int te;
         while(i < l)
         {
             if(preorder[i] == '#'){
                 if(s.empty() == ) return false;
                 while(!s.empty())
                 {
                     te = s.top();
                     if(te == ) return false;
                     s.pop();
                     s.push(te + );
                     te = s.top();
                     if(te == ){
                         break;
                     }
                     if(te ==  ){
                         s.pop();
                     }
                 }
                 if(s.empty() ==  && i < l - ) return false;
             }
             else{
                 s.push();
             }
             while(i < l && preorder[i] != ',') i++;
             i++;
         }
         if(s.size() == ) return true;
         return false;
     }
 };