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 (≤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

考点在于，借助队列实现层序遍历（有条件的），还有树的静态实现。
坑点：tmpL,tmpR的输入，复杂格式尽量使用c++风格，不然容易错。

#include <iostream>#include <cstring>#include <queue>#define Null -1typedef struct {    int L;    int R;} tree;tree T[10];int mark[10];void traversal(int r) {    //int r = root;    int tag = 0;    std::queue<int> q;    //if(T[r] == Null) return;    q.push(r);    while(!q.empty()) {        int now = q.front();        q.pop();        if(T[now].L==Null && T[now].R==Null) {            if(tag) printf(" ");            else tag = 1;            printf("%d", now);        }        if(T[now].L != Null) q.push(T[now].L);        if(T[now].R != Null) q.push(T[now].R);    }}int main(){    int N;    scanf("%d", &N);    memset(mark, 0, sizeof(mark));    for(int i=0; i<N; ++i) {        char tmpL, tmpR;        //这里用scanf容易错。        std::cin >> tmpL >> tmpR;        if(tmpL=='-') T[i].L = Null;        else {            T[i].L = tmpL - '0';            mark[T[i].L] = 1;        }        if(tmpR=='-') T[i].R = Null;        else {            T[i].R = tmpR - '0';            mark[T[i].R] = 1;        }    }    int root;    for(root=0; root<N; ++root) {        if(!mark[root]) break;    }    // for(int i=0; i<N; ++i) {    //     printf("%d %d\n", T[i].L, T[i].R);    // }    // printf("\n%d", root);    //层序遍历    if(N) traversal(root);    else printf("0");    return 0;}