UVA 699 The Falling Leaves

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本文介绍了解决UVA699TheFallingLeaves问题的方法，该问题是关于计算二叉树中各位置落叶堆大小的问题。通过递归前序遍历的方式，实现了对二叉树节点的处理，并记录每条垂直线上叶子的数量。

UVA 699 The Falling Leaves

Time Limited:3seconds

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

#include <stdio.h>
int b[1000];
int pre( int i, int a )
{
    b[i] += a;
    int c;
    scanf("%d", &c);
    if( c != -1 )
        pre( i-1, c );
    scanf("%d",&c);
    if( c != -1 )
        pre( i+1, c );
    return 0;
}
int main()
{
    int a = 0;
    int n = 0;
    int i = 500;
    while( scanf("%d", &a) )
    {
           if( a == -1)
              break;
           n++;
           for( int t=0; t<1000; t++ )
                b[t] = 0;
           pre( i, a );
           printf("Case %d:\n",n);
           for( int t=0; t<1000; t++ )
           {
                if( b[t] != 0 )
                {
                    if( b[t+1] != 0 )
                        printf("%d ",b[t]);
                    else
                        printf("%d\n",b[t]);
                }
           }     
    }
    return 0;
}
剛看到這道題似乎需要建樹來處理，但其實仔細分析就會發現其實只是preorder的traversal罷了，一個recursion function就可以解決了