UVA 548 Tree 建树

题意：

输入中序和后序的权值，输出哪个叶子使它到根的路径上权和最小。

思路：

输入后建树，然后dfs求最小的叶子。

 #include<iostream>
 #include<cstdio>
 #include<queue>
 #include<cstring>
 #include<sstream>
 using namespace std;

 int mid[], post[];
 int judge[];
 struct node {
     int index;
     node *left, *right;
     node()
     {
         left = right = NULL;
     }
 };
 int n;
 bool input(int* a)//输入函数
 {
     string line;
     memset(a, , sizeof(a));
     if (!getline(cin, line))
         return false;
     stringstream ss(line);
     int x;
     n = ;
     while (ss >> x)
         a[n++] = x;
     //printf("n=%d\n", n);//
     return n>;
 }

 int cnt, res;
 node* build(node* root, int l, int r)
 {
     int i;
     int t = post[--cnt];
     for (i = l; i<r; i++)
     {
         if (t == mid[i])
             break;
     }
     root = new node();
     root->index = t;
     /* 注意下面是先建右边然后建左边
     因为后序往前走(--cnt)

     */
     if (i<r - )
         root->right = build(root->right, i + , r);
     if (i>l)
         root->left = build(root->left, l, i);

     return root;
 }

 void preorder(node *root)
 {
     printf("%d ", root->index);
     if (root->left != NULL)
         preorder(root->left);
     if (root->right != NULL)
         preorder(root->right);
 }
 int best, best_num;
 void dfs(node *root, int sum)
 {
     sum += root->index;
     if (!root->left && !root->right)
     {
         if (best_num>sum || (sum == best_num&&root->index<best))
         {
             best = root->index;
             //printf("best=%d\n", best);
             best_num = sum;
         }
     }
     if (root->left != NULL)
         dfs(root->left, sum);
     if (root->right != NULL)
         dfs(root->right, sum);
 }

 int main()
 {
     while (input(mid))
     {
         input(post);
         cnt = n;
         //printf("cnt=%d\n", cnt);
         memset(judge, , sizeof(judge));
         node *root = NULL;
         root = build(root, , n);
         best_num = ;
         //preorder(root);
         //printf("\n");
         dfs(root,);
         printf("%d\n",best);
     }
     return ;
 }