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 程式師世界 >> 編程語言 >> C語言 >> C++ >> C++入門知識 >> UVA 548(二叉樹重建與遍歷)

UVA 548(二叉樹重建與遍歷)

編輯:C++入門知識

J - Tree Time Limit:3000MS Memory Limit:0KB 64bit IO Format:%lld & %llu Submit Status Appoint description:

Description

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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

題意:給你二叉樹的中序與後序,求從根到葉子所有值之和最小的葉子的值。

根據中序和後序遞歸建樹,直接遍歷即可。

#include
#include
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;i0)now->l=built(A,B+n-i,i);//遞歸建立左子樹
    if(ir=built(A+i+1,B+1,n-i-1);//遞歸建立右子樹
    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->datedate;
            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;
}



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