uva548 tree

题意：建立一颗树然后计算每一条到叶子节点路径的中值和最小的，输出叶子节点值

坑爹一个最后del（）多写一个一直RERERE。。

题目：

Tree

You are to determine the value of the leaf node in a given binary tree that is the terminal node of a path of least value from the root of the binary tree to any leaf. The value of a path is the sum of values of nodes along that path.

Input

The input file will contain a description of the binary tree given as the inorder and postorder traversal sequences of that tree. Your program will read two line (until end of file) from the input file. The first line will contain the sequence of values associated with an inorder traversal of the tree and the second line will contain the sequence of values associated with a postorder traversal of the tree. All values will be different, greater than zero and less than 10000. You may assume that no binary tree will have more than 10000 nodes or less than 1 node.

Output

For each tree description you should output the value of the leaf node of a path of least value. In the case of multiple paths of least value you should pick the one with the least value on the terminal node.

Sample Input

3 2 1 4 5 7 6
3 1 2 5 6 7 4
7 8 11 3 5 16 12 18
8 3 11 7 16 18 12 5
255
255

Sample Output

1
3
255


代码：

  1 #include <iostream>
  2 #include <string>
  3 #include <sstream>
  4 #include <limits.h>
  5 #include <malloc.h>
  6 #include <cstdio>
  7 #include <string.h>
  8 using namespace std;
  9 
 10 const int maxn = 50000;
 11 
 12 typedef struct Tnode
 13 {
 14     int data;
 15     struct Tnode* left,*right;
 16 }Node;
 17 
 18 int inorder[maxn];
 19 int postorder[maxn];
 20 int pi,pp;
 21 
 22 int num = 0;
 23 
 24 Node* newnode()
 25 {
 26     Node* T;
 27     T=(Node*)malloc(sizeof(Node));
 28     if(!T)return NULL;
 29     T->left=NULL;T->right=NULL;
 30     T->data=0;
 31     return T;
 32 }
 33 
 34 
 35 Node* build(int left,int right,int* inorder)
 36 {
 37     //cout<<"LEfT : "<<left<<"RIGHT : "<<right<<endl;
 38 
 39     //cout<<"POST :"<<pp<<"  "<<postorder[pp]<<" NExt"<<pp-1<<" "<<postorder[pp-1]<<endl;
 40     //cout<<endl;
 41 
 42 
 43     if(left>right||left<0 || right>=pi)return NULL;
 44     if(left==right)
 45     {
 46         //cout<<"here1"<<endl;
 47         Node* now;
 48         now = newnode();
 49         now->data=inorder[left];
 50         now->left=now->right=NULL;
 51         pp--;
 52         return now;
 53     }
 54 
 55 
 56     for(int i=left;i<=right;i++)
 57     {
 58         if(inorder[i]==postorder[pp-1])
 59         {
 60             //cout<<"here2"<<endl;
 61             Node* now;
 62             now = newnode();now->data= inorder[i];
 63             pp--;
 64             now->left = build(i+1,right,inorder);
 65            // if(!now->left)pp--;
 66             now->right = build (left,i-1,inorder);
 67             return now;
 68         }
 69     }
 70 
 71 }
 72 void print(Node* T)
 73 {
 74     if(T)
 75     {
 76         cout<<T->data<<" ";
 77         print(T->left);
 78         print(T->right);
 79     }
 80 }
 81 int MIN,LAST,ans;
 82 void cal(int sum,Node* T)
 83 {
 84     if(T==NULL)return;
 85     sum+=T->data;
 86     if(T->left==NULL && T->right==NULL)
 87     {
 88         if(sum<MIN)
 89         {
 90             MIN=sum;
 91             LAST= T->data;
 92             ans=LAST;
 93         }
 94         if(sum==MIN && T->data<LAST)
 95         {
 96             LAST=T->data;
 97             ans=LAST;
 98         }
 99     }
100     else
101     {
102         if(T->left)
103             cal(sum,T->left);
104         if(T->right)
105             cal(sum,T->right);
106     }
107 
108 }
109 
110 
111 void del(Node* T)
112 {
113     if(T)
114     {
115         del(T->left);
116         del(T->right);
117         free(T);
118     }
119 }
120 Node* root;
121 int main()
122 {
123     int ti;
124     char tc;
125     while(scanf("%d%c",&ti,&tc)!=EOF)
126     {
127         inorder[pi++]=ti;
128         while(tc!='
')
129         {
130             scanf("%d%c",&ti,&tc);
131             inorder[pi++]=ti;
132         }
133         scanf("%d%c",&ti,&tc);
134         postorder[pp++]=ti;
135         while(tc!='
')
136         {
137             scanf("%d%c",&ti,&tc);
138             postorder[pp++]=ti;
139         }
140 
141         root=build(0,pi-1,inorder);
142        // print(root);
143        // cout<<endl;
144         MIN = INT_MAX;
145         LAST = 0;
146         ans=0;
147         cal(0,root);
148         cout<<ans<<endl;
149         del(root);
150         pi=0;pp=0;
151         memset(inorder,0,sizeof(inorder));
152         memset(postorder,0,sizeof(postorder));
153         //print();
154     }
155     //del(root);
156     return 0;
157 }