下落的树叶(The Falling Leaves)

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Description

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.)

Input 

The input contains multiple test cases, each describing a single tree. A tree is specified by giving the value in the root node, followed by the description of the left subtree, and then the description of the right subtree. If a subtree is empty, the value -1 is supplied. Thus the tree shown above is specified as 5 7 -1 6 -1 -1 3 -1 -1. Each actual tree node contains a positive, non-zero value. The last test case is followed by a single -1 (which would otherwise represent an empty tree).

Output 

For each test case, display the case number (they are numbered sequentially, starting with 1) on a line by itself. On the next line display the number of "leaves" in each pile, from left to right, with a single space separating each value. This display must start in column 1, and will not exceed the width of an 80-character line. Follow the output for each case by a blank line. This format is illustrated in the examples below.

Sample Input 

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
  

Sample Output 

Case 1:
7 11 3

Case 2:
9 7 21 15
  

【分析】

       采用递归（先序）方式输入。

用C++语言编写程序，代码如下：

#include<iostream>
#include<cstring>
using namespace std;

const int maxn = 1000;
int sum[maxn];//这里数组不能开太大否则会出现时间超限。
int mostLP = 1001;
//输入并统计一棵子树，树根水平位置为p
void build(int pos) {
	int v;
	cin >> v;
	if (v == -1) return;//空树
	sum[pos] += v;

	if (pos < mostLP)
		mostLP = pos;

	build(pos - 1);
	build(pos + 1);
}
//边读入边统计
bool init() {
	int v;
	cin >> v;
	if (v == -1) return false;
	memset(sum, 0, sizeof(sum));
	int pos = maxn / 2;//树根的水平位置
	mostLP = pos;

	sum[pos] = v;
	build(pos - 1);
	build(pos + 1);
	return true;
}

int main() {
	int kase = 0;
	while (init()) {
		//升级：加入一个变量mostLP，记录数组sum中最左边有被赋值的元素下标
		//这样做的话，就不用再去查找，递归输入的过程中就可以找到最左的位置了
		/*int p = 0;
		while (sum[p] == 0) p++;*///找最左边的叶子
		int p = mostLP;
		cout << "Case " << (++kase) << ":" << endl;
		cout << sum[p];//因为要避免行末多余空格
		for (int i = p + 1; sum[i] != 0; i++)
			cout << " " << sum[i];
		cout << endl << endl;
	}

	return 0;
}

用java语言编写程序，代码如下：

import java.io.BufferedInputStream;
import java.util.Arrays;
import java.util.Scanner;

public class Main {
	public static void main(String[] args) {
		Scanner input = new Scanner(new BufferedInputStream(System.in));
		int kase = 0;
		final int maxn = 1000;
		int[] sum = new int[maxn];
		int mostLP;
		while((mostLP = init(input, sum)) != -1) {
			System.out.println("Case " + (++kase) + ":");
			System.out.print(sum[mostLP]);
			for(int i = mostLP + 1; sum[i] != 0; i++)
				System.out.print(" " + sum[i]);
			System.out.println('\n');
		}
	}
	
	public static int init(Scanner input, int[] sum) {
		int v = input.nextInt();
		if(v == -1) return -1;
		
		Arrays.fill(sum, 0);
		int pos = sum.length / 2;
		int[] mostLP = new int[1];
		mostLP[0] = pos;
		sum[pos] = v;
		build(input, mostLP, sum, pos - 1);
		build(input, mostLP, sum, pos + 1);
		return mostLP[0];
	}
	
	public static void build(Scanner input, int[] mostLP, int[] sum, int p) {
		int v = input.nextInt();
		if(v == -1) return;
		if(mostLP[0] > p)
			mostLP[0] = p;
		sum[p] += v;
		build(input, mostLP, sum, p - 1);
		build(input, mostLP, sum, p + 1);
	}
}