7 Serialize and Deserialize Binary Tree 序列化及反序列化二叉树
原题网址:http://www.lintcode.com/zh-cn/problem/serialize-and-deserialize-binary-tree/#
设计一个算法,并编写代码来序列化和反序列化二叉树。将树写入一个文件被称为“序列化”,读取文件后重建同样的二叉树被称为“反序列化”。
如何反序列化或序列化二叉树是没有限制的,你只需要确保可以将二叉树序列化为一个字符串,并且可以将字符串反序列化为原来的树结构。
注意事项
There is no limit of how you deserialize or serialize a binary tree, LintCode will take your output of serialize
as the input of deserialize
, it won't check the result of serialize.
您在真实的面试中是否遇到过这个题?
Yes
样例
给出一个测试数据样例, 二叉树{3,9,20,#,#,15,7}
,表示如下的树结构:
3
/ \
9 20
/ \
15 7
我们的数据是进行BFS遍历得到的。当你测试结果wrong answer时,你可以作为输入调试你的代码。
你可以采用其他的方法进行序列化和反序列化。
标签
思路:序列化二叉树时用广度优先搜素的方法,设置一个队列保存节点,依次遍历。因为空节点无法插入队列,所以用root代替空节点,这样判断当前节点是否为root就可以判断是有效节点还是空节点了。
注意,该程序结束时字符串尾部会插入多余的‘#’和‘,’,要将这些无效字符删掉。
更新:又测试了下发现NULL可以插入指针队列,emmm……
反序列化:依旧是利用队列的先入先出特性,因为要判断新建立的节点是左节点还是右节点,所以再设置一个bool类型的判断标志。
首先建立根节点,然后遍历字符串,如果当前字符为‘,’就continue;
如果为‘#’就判断新建立的是左节点还是右节点,如果是右节点代表其根节点已经全部挂载完成,此时需要队列出队,在下一个节点上继续挂载,同时标志反转;
否则,当前节点为有效数值,首先将这些字符转换成int型数据,然后new出一个二叉树节点,判断挂载在左侧还是右侧,然后标志反转,若是右侧,队列再出队。
AC代码:
注意空字符串要返回""而不是NULL……
以及,int转string可以用to_string(),但我测试代码时用的VS2010,还不能支持这个函数,所以自己定义了一个。
/** * Definition of TreeNode: * class TreeNode { * public: * int val; * TreeNode *left, *right; * TreeNode(int val) { * this->val = val; * this->left = this->right = NULL; * } * } */ class Solution { public: /** * This method will be invoked first, you should design your own algorithm * to serialize a binary tree which denote by a root node to a string which * can be easily deserialized by your own "deserialize" method later. */ string serialize(TreeNode * root) { // write your code here if (root==NULL) { return "";//return NULL出错; } string result; //result=result+to_string(root->val); result=result+int2str(root->val); queue<TreeNode *> level; level.push(root->left); level.push(root->right); while(!level.empty()) { TreeNode *temp=level.front(); level.pop(); if (temp!=NULL) { //result=result+","+to_string(temp->val); result=result+","+int2str(temp->val); level.push(temp->left); level.push(temp->right); } else { result=result+","+"#"; } } int size=result.size(); int id=size-1; while(id>0&&(result[id]=='#'||result[id]==',')) { id--; } result.resize(id+1); return result; } /** * This method will be invoked second, the argument data is what exactly * you serialized at method "serialize", that means the data is not given by * system, it's given by your own serialize method. So the format of data is * designed by yourself, and deserialize it here as you serialize it in * "serialize" method. */ TreeNode * deserialize(string &data) { // write your code here if (data.empty()) { return NULL; } int size=data.size(); int i=0; int ro=0;//根节点数值; while(data[i]!=','&&i<size) { char tm=data[i]; ro=ro*10+tm-'0'; i++; } TreeNode * root=new TreeNode(ro); queue<TreeNode *> level; TreeNode *index=root; bool isLeft=true; for (;i<size;i++) { if (data[i]==',') { continue; } else if (data[i]=='#') { if (isLeft) { //index->left=NULL; isLeft=false; } else { //index->right=NULL; if (!level.empty()) { index=level.front(); level.pop(); } isLeft=true; } } else { int val=0; while(i<size&&data[i]!=',') //注意不能写成data[i]!=','&&i<size,因为下标可能超出范围; { char temp=data[i]; val=val*10+temp-'0'; i++; } TreeNode * tempNode=new TreeNode(val); level.push(tempNode); if (isLeft) { index->left=tempNode; isLeft=false; } else { index->right=tempNode; if (!level.empty()) { index=level.front(); level.pop(); } isLeft=true; } } } return root; } string int2str(int &i) { string str; stringstream stream; stream<<i; str=stream.str();//stream>>str; return str; } };
原来的代码……留个纪念。
/** * Definition of TreeNode: * class TreeNode { * public: * int val; * TreeNode *left, *right; * TreeNode(int val) { * this->val = val; * this->left = this->right = NULL; * } * } */ class Solution { public: /** * This method will be invoked first, you should design your own algorithm * to serialize a binary tree which denote by a root node to a string which * can be easily deserialized by your own "deserialize" method later. */ string serialize(TreeNode * root) { // write your code here if (root==NULL) { return ""; } string result; //result=result+to_string(root->val); result=result+int2str(root->val); queue<TreeNode *> level; if (root->left==NULL&&root->right!=NULL) { level.push(root); level.push(root->right); } else if (root->left!=NULL&&root->right==NULL) { level.push(root->left); level.push(root); } else if (root->left!=NULL&&root->right!=NULL) { level.push(root->left); level.push(root->right); } while(!level.empty()) { TreeNode *temp=level.front(); level.pop(); if (temp!=root) { //result=result+","+to_string(temp->val); result=result+","+int2str(temp->val); if (temp->left!=NULL&&temp->right!=NULL) { level.push(temp->left); level.push(temp->right); } else if (temp->left==NULL&&temp->right!=NULL) { level.push(root); level.push(temp->right); } else if (temp->left!=NULL&&temp->right==NULL) { level.push(temp->left); level.push(root); } else { level.push(root); level.push(root); } } else { result=result+","+"#"; } } int size=result.size(); int id=size-1; while(id>0&&(result[id]=='#'||result[id]==',')) { id--; } result.resize(id+1); return result; } /** * This method will be invoked second, the argument data is what exactly * you serialized at method "serialize", that means the data is not given by * system, it's given by your own serialize method. So the format of data is * designed by yourself, and deserialize it here as you serialize it in * "serialize" method. */ TreeNode * deserialize(string &data) { // write your code here if (data.empty()) { return NULL; } int size=data.size(); int i=0; int ro=0;//根节点数值; while(data[i]!=','&&i<size) { char tm=data[i]; ro=ro*10+tm-'0'; i++; } TreeNode * root=new TreeNode(ro); queue<TreeNode *> level; TreeNode *index=root; bool isLeft=true; for (;i<size;i++) { if (data[i]==',') { continue; } else if (data[i]=='#') { if (isLeft) { //index->left=NULL; isLeft=false; } else { //index->right=NULL; if (!level.empty()) { index=level.front(); level.pop(); } isLeft=true; } } else { int val=0; while(i<size&&data[i]!=',') //注意不能写成data[i]!=','&&i<size,因为下标可能超出范围; { char temp=data[i]; val=val*10+temp-'0'; i++; } TreeNode * tempNode=new TreeNode(val); level.push(tempNode); if (isLeft) { index->left=tempNode; isLeft=false; } else { index->right=tempNode; if (!level.empty()) { index=level.front(); level.pop(); } isLeft=true; } } } return root; } string int2str(int &i) { string str; stringstream stream; stream<<i; str=stream.str();//stream>>str; return str; } };
其他参考: