699 - The Falling Leaves

Each year, fall inthe North Central region is accompanied by the brilliant colors of the leaveson the trees, followed quickly by the falling leaves accumulating under thetrees. If the same thing happened to binary trees, how large would the piles ofleaves become?

We assume each node in a binary tree "drops" a number of leaves equalto the integer value stored in that node. We also assume that these leaves dropvertically to the ground (thankfully, there's no wind to blow them around).Finally, we assume that the nodes are positioned horizontally in such a mannerthat the left and right children of a node are exactly one unit to the left andone unit to the right, respectively, of their parent. Consider the followingtree:

The nodescontaining 5 and 6 have the same horizontal position (with different verticalpositions, of course). The node containing 7 is one unit to the left of thosecontaining 5 and 6, and the node containing 3 is one unit to their right. Whenthe "leaves" drop from these nodes, three piles are created: theleftmost 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. (Whileit is true that only leaf nodes in a tree would logically have leaves, weignore that in this problem.)

Input 

The input containsmultiple test cases, each describing a single tree. A tree is specified bygiving the value in the root node, followed by the description of the leftsubtree, and then the description of the right subtree. If a subtree is empty,the value -1 is supplied. Thus the tree shown above is specifiedas 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 (whichwould otherwise represent an empty tree).

Output 

For each testcase, 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 an80-character line. Follow the output for each case by a blank line. This formatis 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 

Case1:

7 11 3

 

Case 2:

9 7 21 15

 

代码:

/*

题意：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表示空树.算法：在“建树”的同时计算，无须真正的把树保存下来

*/

 

#include<cstring>

#include<iostream>

using namespacestd;

 

const int maxn =200;

int sum[maxn];

 

void build(intp)// 输入并统计一棵子树，树根水平位置为p

{

    int v;

    cin >> v;

    if(v == -1)

    {

        return; // 空树

    }

    sum[p] += v;

    build(p - 1);

    build(p + 1);

}

 

bool init()// 边读入边统计

{

    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

    sum[pos] = v;

    build(pos - 1); // 左子树

    build(pos + 1); // 右子树

}

 

int main()

{

    int kase = 0;

    while(init())

    {

        int p = 0;

        while(sum[p] == 0)

        {

            p++; // 找最左边的叶子

        }

        cout << "Case "<< ++kase << ":\n" << sum[p++];// 开始输出

        while(sum[p] != 0)

        {

            cout << " "<< sum[p];

            p++;

        }

        cout << "\n\n";

    }

    return 0;

}

注意：

不要将水平位置设为全局变量，也不要传位置的时候使用引用，直接传值即可。