前序+中序得出二叉树，靠着回忆推出的板子
ti

#include<bits/stdc++.h>using namespace std;struct tree{char c;tree *left, *right;tree(char x):c(x), left(NULL), right(NULL){};
};string s1, s2;
int n, m;
int t;
int a[110];tree* build(char x){tree *root;root->c = x;root->left = NULL;root->right = NULL;
}tree* Build(int pre, int in, int len){  //分别表示前序序列开始的位置，中序序列开始的位置，长度 if(len == 1){tree *root = new tree(s1[pre]);return root;}tree *root = new tree(s1[pre]);for(int i = 0; i < len; i ++ ){if(s2[i + in] == s1[pre]){if(i != 0) root->left = Build(pre + 1, in, i);else root->left = NULL;if(pre + i + 1 < pre + len) root->right = Build(pre + i + 1, in + i + 1, len - i - 1);else root->right = NULL;break;}} return root;
}void houxu(tree *root){if(root == NULL) return ;if(root->left != NULL) houxu(root->left);if(root->right != NULL) houxu(root->right);cout<<root->c;return ;
}int main()
{while(cin>>s1>>s2){tree *root = Build(0, 0, s1.length());houxu(root);cout<<endl;}return 0;
}