AcWing 50. 序列化二叉树 + 层序遍历
原题链接
困难
作者:
skypemifen
,
2021-06-25 01:28:49
,
所有人可见
,
阅读 327
层序遍历
/**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
class Solution {
public:
// Encodes a tree to a single string.
string serialize(TreeNode* root) {
string res;
dfs(root, res);
return res;
}
void dfs(TreeNode* root, string& res){
queue<TreeNode*> q;
q.push(root);
while(q.size()){
auto t = q.front();
q.pop();
if(t) res += to_string(t->val) + ' ';
else {
//弹出元素为空,逃过当前循环
res += "null ";
continue;
}
q.push(t->left);
q.push(t->right);
}
}
// Decodes your encoded data to tree.
TreeNode* deserialize(string data) {
int u = 0;
queue<TreeNode*> q;
auto root = get_next(data, u);
q.push(root);
while(q.size()){
auto t = q.front();
q.pop();
if(t){
auto left = get_next(data, u), right = get_next(data, u);
t->left = left, t->right = right;
q.push(left), q.push(right);
}
}
return root;
}
//得到字符串下一个结点
TreeNode* get_next(string data, int &u){
if(u == data.size()) return NULL;
int k = u;
while(data[k] != ' ') k++;
if(data[u] == 'n'){
u = k + 1;
return NULL;
}
int minus = 1;
if(data[u] == '-') minus = -1, ++u;
int val = 0;
for(int i=u; i<k; i++) val = val * 10 + data[i] - '0';
val *= minus;
u = k + 1;
return new TreeNode(val);
}
};