UVA699 UVALive5471 The Falling Leaves【树权和】

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 on the right:
    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

Regionals 1998 >> North America - North Central NA

问题链接：UVA699 UVALive5471 The Falling Leaves
问题简述：
    给一棵二叉树，每个节点都有一个水平位置：左儿子在它左边1个单位，右儿子在右边1个单位。从左向右输出每个水平位置的所有结点的权值之和。按照递归方式输入，-1表示空 。
问题分析：
    把树建在一维数组中是一个好办法。
程序说明：（略）
参考链接：（略）
题记：（略）

AC的C++程序如下：

/* UVA699 UVALive5471 The Falling Leaves */

#include <bits/stdc++.h>

using namespace std;

const int N = 200;
int sum[N];

// 输入并统计一棵子树，树根位置为k
void build(int k)
{
  int a;
  cin >> a;
  if(a == -1) return; // 空树

  sum[k] += a;
  build(k - 1);
  build(k + 1);
}

// 边读入数据边建树边统计
bool init() {
  int a;
  scanf("%d", &a);
  if(a == -1) return false;

  memset(sum, 0, sizeof(sum));
  int pos = N / 2;      // 树根位置
  sum[pos] = a;
  build(pos - 1); // 左子树
  build(pos + 1); // 右子树
  return true;
}

int main()
{
    int caseno = 0;
    while(init()) {
        int k = 0;
        while(sum[k] == 0) k++; // 找最左边的叶子

        // 输出结果
        printf("Case %d:\n%d", ++caseno, sum[k++]);
        while(sum[k])
            printf(" %d", sum[k++]);
        printf("\n\n");
    }

    return 0;
}

标签：node,int,tree,sum,UVA699,UVALive5471,each,Falling,leaves
来源： https://www.cnblogs.com/tigerisland45/p/10449960.html