标签:ref etc i++ HERE yourself continue node 面试 需要
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。
对二进制树进行反序列化或序列化的方式没有限制,LintCode 将您的 serialize 输出作为 deserialize 的输入,它不会检查序列化的结果。
在线测评地址:点击此处前往
样例 1:
输入:{3,9,20,#,#,15,7}
输出:{3,9,20,#,#,15,7}
解释:
二叉树 {3,9,20,#,#,15,7},表示如下的树结构:
3
/ 9 20
/ 15 7
它将被序列化为 {3,9,20,#,#,15,7}
输入:{1,2,3}
输出:{1,2,3}
解释:
二叉树 {1,2,3},表示如下的树结构:
1
/ 2 3
它将被序列化为 {1,2,3}
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。
考点:
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.
*/
vector<string> split(const string &str, string delim) {
vector<string> results;
int lastIndex = 0, index;
while ((index = str.find(delim, lastIndex)) != string::npos) {
results.push_back(str.substr(lastIndex, index - lastIndex));
lastIndex = index + delim.length();
}
if (lastIndex != str.length()) {
results.push_back(str.substr(lastIndex, str.length() - lastIndex));
}
return results;
}
string serialize(TreeNode *root) {
if (root == NULL) {
return "{}";
}
vector<TreeNode *> q;
q.push_back(root);
for(int i = 0; i < q.size(); i++) {
TreeNode * node = q[i];
if (node == NULL) {
continue;
}
q.push_back(node->left);
q.push_back(node->right);
}
while (q[q.size() - 1] == NULL) {
q.pop_back();
}
string sb="";
sb += "{";
sb += to_string(q[0]->val);
for (int i = 1; i < q.size(); i++) {
if (q[i] == NULL) {
sb += (",#");
}
else {
sb += ",";
sb += to_string(q[i]->val);
}
}
sb += "}";
return sb;
}
/**
* 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 == "{}") return NULL;
vector<string> vals = split(data.substr(1, data.size() - 2), ",");
TreeNode *root = new TreeNode(atoi(vals[0].c_str()));
queue<TreeNode *> Q;
Q.push(root);
bool isLeftChild= true;
for (int i = 1; i < vals.size(); i++) {
if (vals[i] != "#") {
TreeNode *node = new TreeNode(atoi(vals[i].c_str()));
if (isLeftChild) Q.front()->left = node;
else Q.front()->right = node;
Q.push(node);
}
if (!isLeftChild) {
Q.pop();
}
isLeftChild = !isLeftChild;
}
return root;
}
};
【LeetCode/LintCode】 题解丨谷歌面试题:二叉树的序列化和反序列化
标签:ref etc i++ HERE yourself continue node 面试 需要
原文地址:https://www.cnblogs.com/lintcode/p/13555337.html