List Leaves（树的层序遍历）（队列）

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本文介绍了一种遍历树形结构并按特定顺序输出叶子节点的算法。输入包含树的节点数及各节点的孩子节点信息，输出则按照从下到上、从左到右的顺序列出所有叶子节点的索引。通过构建队列并使用标志位来实现遍历过程。

03-树2 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 $N$ ( $\leq 10$ ) 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

题意：按从下到上（优先），从左到右的方向遍历出所有叶子结点

思路：先找出根结点，加入队列，循环，如果当前结点为叶结点，则输出当前节点，再把当前结点的左右孩子（非空）加入队列中

反思:写错了一个符号，找了半天才找到，关键是这个符号的错误跟测试点的解释一点关系都没有。。。。

附上AC代码:

#include <cstdio>   
#include <iostream> 
#include <algorithm> 
#include <cmath> 
#include <cstdlib> 
#include <cstring> 
#include <vector> 
#include <list> 
#include <map> 
#include <stack> 
#include <queue> 
using namespace std; 
#define ll long long
struct Tree {
	int i;
	char l,r;
};
bool f[30];
bool flag;
Tree T[10];
queue<Tree> t; 
int n;
void set()
{
	for(int i = 0;i < n;i++)
		if(f[i] == 0)
		{
			t.push(T[i]);
			break;
		}
	while(!t.empty())
	{
		Tree x = t.front();
		t.pop();
		if(x.l == '-'&&x.r == '-')
		{
			if(flag)
				cout <<' '<< x.i ;
			else
			{
				cout << x.i;
				flag = 1;
			}
		}
		if(x.l != '-')
			t.push(T[x.l-'0']);
		if(x.r != '-')
			t.push(T[x.r-'0']);
	}
}
int main() 
{
	flag = 0;
	memset(f,0,sizeof(f));
	cin >> n;
	for(int i = 0;i < n;i++)
	{
		cin >> T[i].l >> T[i].r;
		if(T[i].l != '-')
			f[T[i].l-'0'] = 1;
		if(T[i].r != '-')
			f[T[i].r-'0'] = 1;
		T[i].i = i;
	}
	set();
    //cout << "AC" <<endl; 
    return 0; 
}