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
二叉树,按照层次从上到下排列。比如根节点在(1,1)的位置,那么其左子树的根节点在(2,0),右子树的根节点在(2,2)。依次类推。最后在同一个垂直线上的叶子掉落在同一堆上,按照从左到右的顺序,依次输出这些堆的叶子数。
这同样是可以用建树来做。不过由于只是要求最后的叶子数,所以可以用递归的方式。而且输入的方式和广义表很相似。直接贴代码吧。
#include<iostream> #include<cstdio> #include<cmath> #include<cstring> using namespace std; int a[1000]; void go(int i, int v) { a[i]+=v; cin>>v; if (v!=-1) go(i-1,v); cin>>v; if (v!=-1) go(i+1,v); } int main () { int n,i,t=1; while(cin>>n) { if (n==-1) break; memset(a,0,sizeof(a)); go(1000/2,n); printf("Case %d:\n",t++); i=0; while(!a[i]) i++; cout<<a[i]; for (i=i+1; i<1000; i++) if (a[i]) cout<<" "<<a[i]; cout<<endl<<endl; } return 0; }