/*
struct TreeNode {
int val;
struct TreeNode *left;
struct TreeNode *right;
TreeNode(int x) :
val(x), left(NULL), right(NULL) {
}
};
*/
class Solution {
public:
char* Serialize(TreeNode *root) {
if(root == NULL)
return NULL;
string str;
Serialize(root, str);
char *ret = new char[str.length() + 1];
int i;
for(i = 0; i < str.length(); i++){
ret[i] = str[i];
}
ret[i] = '\0';
return ret;
}
void Serialize(TreeNode *root, string& str){
if(root == NULL){
str += '#';
return ;
}
string r = to_string(root->val);
str += r;
str += ',';
Serialize(root->left, str);
Serialize(root->right, str);
}
TreeNode* Deserialize(char *str) {
if(str == NULL)
return NULL;
TreeNode *ret = Deserialize(&str);
return ret;
}
TreeNode* Deserialize(char **str){//由于递归时,会不断的向后读取字符串
if(**str == '#')
{ //所以一定要用**str,
++(*str); //以保证得到递归后指针str指向未被读取的字符
return nullptr;
}
int num = 0;
while(**str != '\0' && **str != ','){
num =num*10 +((**str) - '0'); //这里需要知道:如果某个节点的val应该是12,那么这个12会在str中占
//两个字节,所以第一个数字1应该是10,再加个位的2,这就是为什么定义num
++(*str);
}
TreeNode *root = new TreeNode(num);
if(**str == '\0')
return root;
else
(*str)++;
root->left = Deserialize(str);
root->right = Deserialize(str);
return root;
}
};
剑指offer面试题37 序列化二叉树
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转载自blog.csdn.net/qq_34793133/article/details/81146533
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