UVA 548（二进制重建和遍历）

J - Tree

Time Limit:3000MS     Memory Limit:0KB     64bit IO Format:%lld
& %llu

Submit Status

Appoint description: 
System Crawler  (2014-05-16)

Description

 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

---

Miguel A. Revilla

1999-01-11

题意：给你二叉树的中序与后序。求从根到叶子全部值之和最小的叶子的值。

依据中序和后序递归建树，直接遍历就可以。

#include<stdio.h>
#include<cstring>
int a[10001],b[10001];
int M,v;
struct tree
{
    int date;
    tree *l,*r;
    tree()
    {
        date=0;
        l=r=NULL;
    }
};
tree* built(int *A,int *B,int n)
{
    if(!n)return NULL;
    tree *now=new tree;
    int i=0;
    for(i=0;i<n;i++)if(A[i]==B[0])break;//找到根在中序遍历中的位置
    if(i>0)now->l=built(A,B+n-i,i);//递归建立左子树
    if(i<n-1)now->r=built(A+i+1,B+1,n-i-1);//递归建立右子树
    now->date=B[0];
    return now;
}
void del(tree *p)
{
    if(!p)return;
    if(p->l)del(p->l);
    if(p->r)del(p->r);
    delete p;
    p=NULL;
}
void dfs(tree *Root,int sum)
{
    if(Root==NULL)return ;
    //printf("%d ",Root->date);
    if(Root->l==NULL&&Root->r==NULL)
    {
        if(sum+Root->date<M)
        {
            M=sum+Root->date;
            v=Root->date;
        }
    }
    dfs(Root->l,sum+Root->date);
    dfs(Root->r,sum+Root->date);
}
int main()
{
    char ch;
    tree *root;
    //freopen("in.txt","r",stdin);
    while(~scanf("%d",&a[0]))
    {
      root=NULL;
      int n=1;
      M=100000001;
      v=0;
      while((ch=getchar())!='\n')
      {
          scanf("%d",a+n);++n;
      }
      for(int i=n-1;i>=0;i--)
      {
           scanf("%d",b+i);
      }
      root=built(a,b,n);
      dfs(root,0);
      printf("%d\n",v);
      del(root);
    }
    return 0;
}

版权声明：本文博主原创文章，博客，未经同意不得转载。