7-4 List Leaves

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本文介绍了一种通过广度优先搜索遍历树结构的方法，并实现按从上到下、从左到右的顺序打印所有叶子节点的功能。输入包括树的节点数及各节点的左右子节点信息，输出为按照指定顺序排列的所有叶子节点的索引。

7-4 List Leaves（25 分）

Given a tree, you are supposed to list all the leaves in the order of top down, and left to right.

Input Specification:

Each input file contains one test case. For each case, the first line gives a positive integer N (≤10) which is the total number of nodes in the tree -- and hence the nodes are numbered from 0 to N−1. Then N lines follow, each corresponds to a node, and gives the indices of the left and right children of the node. If the child does not exist, a "-" will be put at the position. Any pair of children are separated by a space.

Output Specification:

For each test case, print in one line all the leaves' indices in the order of top down, and left to right. There must be exactly one space between any adjacent numbers, and no extra space at the end of the line.

Sample Input:

8
1 -
- -
0 -
2 7
- -
- -
5 -
4 6

Sample Output:

4 1 5

AC代码如下：

#include <iostream>
#include <string>
#include <queue>
using namespace std;

struct Node
{
    int data;
    char left;
    char right;
}T[20];
int n = 0;
int root = -1;
void build()
{
    char cl, ch;
    cin >> n;
    int i, j, k;
    int book [n];

    for (i = 0; i < n; i++)
    {
        book[i] = 0;
    }

    for (j = 0; j < n; j++)
    {
        cin >> cl >> ch;
        T[j].data = j;
        if (cl == '-')
        {
            T[j].left = -1;
        }
        else
        {
            T[j].left = cl - '0';
            book[cl-'0'] = 1;
        }
        if (ch == '-')
        {
            T[j].right = -1;
        }
        else
        {
            T[j].right = ch - '0';
            book[ch-'0'] = 1;
        }
    }
    for (k = 0; k < n; k++)
    {
        if (book[k] != 1)
        {
            root = k;
            break;
        }
    }
}

void traversal()
{
    queue <int> q;
    q.push(T[root].data);
    int arr[20];
    int i = 0;
    while (!q.empty())
    {
        int item = q.front();
        arr[i] = item;
        q.pop();
        if (T[item].left != -1)
        {
            q.push(T[item].left);
        }
        if (T[item].right != -1)
        {
            q.push(T[item].right);
        }
        i++;
    }
    bool flag = false;
    for (int j = 0; j < i; j++)
    {
        if (T[arr[j]].left == -1 && T[arr[j]].right == -1)
        {
            if(flag)
                cout << " ";
            cout << T[arr[j]].data;
            flag = true;
        }
    }
}
int main()
{
    build();
    traversal();
    return 0;
}