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
题意:
每个叶子在垂直方向落叶,最后垂直和是多少?
用个辅助数组sum[MAX],来保存竖直方的和值。
MAX/2作为树叶的根节点,然后通过build(pos-1) 和 build(pos+1) 来模拟建立一颗二叉树
最后输出sum不为0的数
#include <iostream> #include <stdio.h> #include <sstream> #include <string> #include <string.h> using namespace std; const int MAX = 2000; int sum[MAX]; void build(int pos) { int v; cin >> v; if(v == -1) //如果是空树则不用考虑 return ; sum[pos] += v; build(pos-1); build(pos+1); } bool init() { int v; cin >> v; if(v == -1){ return false; } memset(sum,0,sizeof(sum)); int pos = MAX/2; sum[pos] += v; build(pos - 1); build(pos + 1); return true; } int main() { int cas = 1; while (init()) { int p = 0; while(sum[p] == 0) p++; cout<<"Case "<<cas++<<":"<<endl; cout<<sum[p++]; while(sum[p] != 0) cout<<" "<<sum[p++]; cout<<endl<<endl; } return 0; }