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.
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
1 3 255
题意:给你二叉树的中序与后序,求从根到叶子所有值之和最小的叶子的值。
根据中序和后序递归建树,直接遍历即可。
#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;
}
UVA 548(二叉树重建与遍历),布布扣,bubuko.com
原文地址:http://blog.csdn.net/u013268685/article/details/27554837