03-树2 List Leaves（25 point(s)）【Tree】

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该篇博客介绍了如何按层次顺序打印二叉树的叶子节点。给定一棵树的节点数，通过输入每个节点的左右子节点信息，利用广度优先搜索（BFS）策略，遇到没有子节点的节点时将其加入结果列表。示例输入和输出展示了具体操作过程。

03-树2 List Leaves（25 point(s)）

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

思路

广度优先搜索往下搜遇到没有儿子的压入一个 vector 就可以

AC代码

#include <cstdio>
#include <cstring>
#include <ctype.h>
#include <cstdlib>
#include <cmath>
#include <climits>
#include <ctime>
#include <iostream>
#include <algorithm>
#include <deque>
#include <vector>
#include <queue>
#include <string>
#include <map>
#include <stack>
#include <set>
#include <numeric>
#include <sstream>
#include <iomanip>
#include <limits>

#define CLR(a) memset(a, 0, sizeof(a))
#define pb push_back

using namespace std;
typedef long long ll;
typedef long double ld;
typedef unsigned long long ull;
typedef pair <int, int> pii;
typedef pair <ll, ll> pll;
typedef pair<string, int> psi;
typedef pair<string, string> pss;

const double PI = 3.14159265358979323846264338327;
const double E = exp(1);
const double eps = 1e-30;

const int INF = 0x3f3f3f3f;
const int maxn = 1e5 + 5;
const int MOD = 1e9 + 7;

struct Node
{
    int l, r;
}q[10];

queue <int> Q;

vector <int> ans;

void bfs()
{
    int len = Q.size();
    for (int i = 0; i < len; i++)
    {
        int num = Q.front();
        Q.pop();
        if (q[num].l != -1 && q[num].r != -1)
        {
            Q.push(q[num].l);
            Q.push(q[num].r);
        }
        else if (q[num].l != -1)
            Q.push(q[num].l);
        else if (q[num].r != -1)
            Q.push(q[num].r);
        else
            ans.pb(num);
    }
    if (Q.size())
        bfs();
}

int main()
{
    int n;
    scanf("%d", &n);
    char a, b;
    map <int, int> m;
    for (int i = 0; i < n; i++)
    {
        scanf(" %c %c", &a, &b);
        if (a != '-')
        {
            q[i].l = a - '0';
            m[a - '0'] = 1;
        }
        else
            q[i].l = -1;
        if (b != '-')
        {
            q[i].r = b - '0';
            m[b - '0'] = 1;
        }
        else
            q[i].r = -1;
    }
    int root;
    for (int i = 0; i < n; i++)
    {
        if (m[i] == 0)
        {
            root = i;
            break;
        }
    }
    Q.push(root);
    bfs();
    vector <int>::iterator it;
    for (it = ans.begin(); it != ans.end(); it++)
    {
        if (it != ans.begin())
            printf(" ");
        printf("%d", (*it));
    }
    printf("\n");
}