Description
| The Falling Leaves |
Each year, fall in the North Central region is accompanied by the brilliant colors of the leaves on the trees, followed quickly by the falling leaves accumulating under the trees. If the same thing happened to binary trees, how large would the piles of leaves become?
We assume each node in a binary tree "drops" a number of leaves equal to the integer value stored in that node. We also assume that these leaves drop vertically to the ground (thankfully, there‘s no wind to blow them around). Finally, we assume that the nodes
are positioned horizontally in such a manner that the left and right children of a node are exactly one unit to the left and one unit to the right, respectively, of their parent. Consider the following tree:
The nodes containing 5 and 6 have the same horizontal position (with different vertical positions, of course). The node containing 7 is one unit to the left of those containing 5 and 6, and the node containing 3 is one unit to their right. When the "leaves" drop from these nodes, three piles are created: the leftmost one contains 7 leaves (from the leftmost node), the next contains 11 (from the nodes containing 5 and 6), and the rightmost pile contains 3. (While it is true that only leaf nodes in a tree would logically have leaves, we ignore that in this problem.)
5 7 -1 6 -1 -1 3 -1 -1 8 2 9 -1 -1 6 5 -1 -1 12 -1 -1 3 7 -1 -1 -1 -1
Case 1: 7 11 3 Case 2: 9 7 21 15
给出一颗树，让所有的节点垂直下落，计算每堆节点的值之和
#include<stdio.h>
#include<string.h>
int ans[10005],A;
struct tree
{
int lca,date;
tree *l,*r;
tree()
{
lca=0;
l=r=NULL;
}
};
tree *build(int num,int ok)
{
if(ok)scanf("%d",&A);
if(A==-1)return NULL;
tree *p=new tree;
p->date=A;
p->lca=num;
p->l=build(num-1,1);
p->r=build(num+1,1);
return p;
}
void dfs(tree *p)
{
if(!p)return ;
ans[p->lca]+=p->date;
p->date=0;
dfs(p->l);
dfs(p->r);
}
int cas=1;
int main()
{
//freopen("in.txt","r",stdin);
tree *root=NULL;
while(~scanf("%d",&A))
{
if(A==-1)return 0;
root=NULL;
root=build(5000,0);
memset(ans,0,sizeof(ans));
dfs(root);
printf("Case %d:\n",cas++);
int i;
for(i=0;i<=10000;i++)
if(ans[i]){printf("%d",ans[i]);break;}
for(i+=1;i<=10000;i++)
if(ans[i])printf(" %d",ans[i]);
printf("\n\n");
}
return 0;
}
UVA 699（二叉树建树与遍历）,布布扣,bubuko.com
原文地址：http://blog.csdn.net/u013268685/article/details/27678805
