AcWing 50. 序列化二叉树--各种遍历
原题链接
困难
作者:
小明同学hh
,
2020-09-15 19:13:54
,
所有人可见
,
阅读 497
/**
* 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_s(root,res);
return res;
}
/* 前序
void dfs_s(TreeNode* root,string &res) {
if(root==NULL) {
res+="null ";
return;
}
res += (to_string(root->val)+' ');
dfs_s(root->left,res);
dfs_s(root->right,res);
}*/
/* 后序
void dfs_s(TreeNode* root,string &res) {
if(!root) {
res+=" null";
return;
}
dfs_s(root->left,res);
dfs_s(root->right,res);
res+=(' '+to_string(root->val));
} */
/* 层序*/
void dfs_s(TreeNode* root,string &res) {
queue<TreeNode*> q;
q.push(root);
while(q.size()) {
auto t=q.front();
q.pop();
if(!t) {
res+="null ";
}
else {
res+= (to_string(t->val)+' ');
q.push(t->left);
q.push(t->right);
}
}
return;
}
// Decodes your encoded data to tree.
TreeNode* deserialize(string data) {
// 后序 int u=data.size()-1;
// 前序 int u=0;
// 层序 int u=0;
return dfs_d(data,u);
}
/*层序*/
TreeNode* dfs_d(string data,int &u) {
if(data[u]=='n') return NULL;
int k=u;
while(data[k]!=' ') k++;
int val=0;
for(int i=0;i<k;i++) {
if(data[i]=='-') continue;
val = val*10+data[i]-'0';
}
u=k+1;
if(data[0]=='-') val=-val;
TreeNode* root=new TreeNode(val);
queue<TreeNode*> q;
q.push(root);
while(q.size()) {
auto t=q.front();
q.pop();
int k=u;
while(data[k]!=' ') k++;
if(data[u]=='n') t->left=NULL;
else {
int val=0;
for(int i=u;i<k;i++) {
if(data[i]=='-') continue;
val = val*10+data[i]-'0';
}
if(data[u]=='-') val=-val;
t->left = new TreeNode(val);
q.push(t->left);
}
u=k+1;
k=u;
while(data[k]!=' ') k++;
if(data[u]=='n') t->right=NULL;
else {
int val=0;
for(int i=u;i<k;i++) {
if(data[i]=='-') continue;
val=val*10+data[i]-'0';
}
if(data[u]=='-') val=-val;
t->right=new TreeNode(val);
q.push(t->right);
}
u=k+1;
}
return root;
}
/* 后序
TreeNode* dfs_d(string data,int &u) {
if(u==3) return NULL;
int k=u;
while(data[k]!=' ') k--;
if(data[k+1]=='n') {
u=k-1;
return NULL;
}
int val=0;
bool minus=false;
for(int i=k+1;i<=u;i++) {
if(data[i]=='-') {
minus=true;
continue;
}
val=val*10+data[i]-'0';
}
if(minus) val=-val;
u=k-1;
TreeNode* root=new TreeNode(val);
root->right=dfs_d(data,u);
root->left=dfs_d(data,u);
return root;
}*/
/* 前序
TreeNode* dfs_d(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 val=0;
bool minus=false;
for(int i=u;i<k;i++) {
if(data[i]=='-') {
minus=true;
continue;
}
val = val*10+data[i]-'0';
}
u=k+1;
if(minus) val=-val;
auto root=new TreeNode(val);
root->left = dfs_d(data,u);
root->right = dfs_d(data,u);
return root;
}*/
};