設計一個算法,並編寫代碼來序列化和反序列化二叉樹。將樹寫入一個文件被稱為“序列化”,讀取文件后重建同樣的二叉樹被稱為“反序列化”。
如何反序列化或序列化二叉樹是沒有限制的,你只需要確保可以將二叉樹序列化為一個字符串,並且可以將字符串反序列化為原來的樹結構。
注意事項
There is no limit of how you deserialize or serialize a binary tree, LintCode will take your output ofserialize as the input of deserialize, it won't check the result of serialize.
樣例
給出一個測試數據樣例, 二叉樹{3,9,20,#,#,15,7},表示如下的樹結構:
3
/ \
9 20
/ \
15 7
我們的數據是進行BFS遍歷得到的。當你測試結果wrong answer時,你可以作為輸入調試你的代碼。
你可以采用其他的方法進行序列化和反序列化。
思路:在這里使用先根遍歷來實現;
本題目難點在於,里面穿插關於字符串和整數間的互相轉換。
在序列化時,空節點的表示,不同節點值之間的分割。
在反序列化時,字符串每個字符遍歷時的控制條件和操作,以及將字符串轉換為整數;
參考:http://blog.csdn.net/waltonhuang/article/details/51979479
/**
* 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.
*/
/*
思路:感覺上用BFS更容易解決,在這里使用先根遍歷來實現;
參考:http://blog.csdn.net/waltonhuang/article/details/51979479
本題目難點在於,里面穿插關於字符串和整數間的互相轉換。
在序列化時,空節點的表示,不同節點值之間的分割。
在反序列化時,字符串每個字符遍歷時的控制條件和操作,以及將字符串轉換為整數;
*/
string serialize(TreeNode *root) {
// write your code here
string s = "";
writeTree(s, root);
return s;
}
void writeTree(string &s, TreeNode* root){
if (root == NULL){
s += "# ";
return;
}
s += (to_string(root->val) + ' ');
writeTree(s, root->left);
writeTree(s, root->right);
}
/**
* 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
int pos = 0;
return readTree(data, pos);
}
TreeNode* readTree(string data, int& pos){
if (data[pos] == '#'){
pos += 2;
return NULL;
}
int nownum = 0;
while (data[pos] != ' '){
//這里' '是為了分離不同的數字;
nownum = nownum * 10 + (data[pos] - '0');
pos++;
}
pos++;
TreeNode* nowNode = new TreeNode(nownum);
nowNode->left = readTree(data, pos);
nowNode->right = readTree(data, pos);
return nowNode;
}
};
