The Falling Leaves （二叉搜索树+stl）

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我是靠谱客的博主迷你小霸王，最近开发中收集的这篇文章主要介绍The Falling Leaves （二叉搜索树+stl），觉得挺不错的，现在分享给大家，希望可以做个参考。

概述

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

题意：一组数据根据前序遍历原则输入，当某个节点的左儿子或右儿子为空时，它的左儿子或右儿子的值为 -1。以根节点为准线，它的左儿子就以准线为基准减一个位置，右儿子加一个位置。求从左到右各个位置上的节点之和为多少。

思路：先将数据依次输入，还原成一颗树。以根节点为y轴，左儿子减一，右儿子加一。并用map记录每个位置的节点和，遍历map，依次输出。

代码如下：

#include<cstdio>
#include<cstring>
#include<map>
#include<iostream>
using namespace std;
map<int,int>Q;
void sove(int t)
{
    int x;
    scanf("%d",&x);//输入节点值
    if(x!=-1) //不为 -1 时
    {
        Q[t]+=x;//求该位置和
        sove(t-1); //它的左儿子
        sove(t+1); //它的右儿子
    }
}
int main()
{
    int k=1;
    int x;
    map<int,int>::iterator it;
    while(~scanf("%d",&x)&&(x!=-1))
    {
        Q.clear();//清空
        Q[1]+=x;//根的位置为 1
        sove(0);//根的左儿子
        sove(2);//根的右儿子
        printf("Case %d:n",k++);
        int flag=0;
        for(it=Q.begin();it!=Q.end();it++) //格式化输出
        {
            if(flag==0)
                cout<<it->second;
            else
                cout<<" "<<it->second;
            flag=1;
        }
        cout<<endl<<endl;
    }
}