<笔记>快速排序的切分
二叉树遍历
http://www.nowcoder.com/practice/6e732a9632bc4d12b442469aed7fe9ce
#include<iostream>
#include<unordered_map>
using namespace std;
const int N = 30;
unordered_map<char, int> pos, l, r // pos记录中序遍历所在元素的位置;
string pre, in;
string post;
void dfs(char root){
if(l.count(root)) dfs(l[root]);
if(r.count(root)) dfs(r[root]);
post += root;
}
char dfs(int il, int ir, int pl, int pr){
char root = pre[pl];
int k = pos[root];
if(k > il) l[root] = dfs(il, k - 1, pl + 1, pl + 1 + k - 1 - il);
if(k < ir) r[root] = dfs(k + 1, ir, pl + 1 + k - il, pr);
return root;
}
int main(){
while(cin>> pre >> in){
pos.clear(), l.clear(), r.clear();
int n = pre.size();
for(int i = 0; i < in.size(); i ++ ) pos[in[i]] = i;
char root = dfs(0, n - 1, 0, n - 1);
post.clear();
dfs(root);
cout << post << endl;
}
}
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