# 【PAT甲级】1004 Counting Leaves (30分)：DFS/BFS树的遍历

## 1004 Counting Leaves (30分)

A family hierarchy is usually presented by a pedigree tree. Your job is to count those family members who have no child.

### Input Specification:

Each input file contains one test case. Each case starts with a line containing 0<N<100, the number of nodes in a tree, and M (<N), the number of non-leaf nodes. Then M lines follow, each in the format:

ID K ID[1] ID[2] ... ID[K]



where ID is a two-digit number representing a given non-leaf node, K is the number of its children, followed by a sequence of two-digit ID's of its children. For the sake of simplicity, let us fix the root ID to be 01.

The input ends with N being 0. That case must NOT be processed.

### Output Specification:

For each test case, you are supposed to count those family members who have no child for every seniority level starting from the root. The numbers must be printed in a line, separated by a space, and there must be no extra space at the end of each line.

The sample case represents a tree with only 2 nodes, where 01 is the root and 02 is its only child. Hence on the root 01 level, there is 0 leaf node; and on the next level, there is 1 leaf node. Then we should output 0 1 in a line.

### Sample Input:

2 1
01 1 02



### Sample Output:

0 1


## 分析：BFS/DFS遍历树

• BFS：

设置layer[]记录每层叶子节点的个数。

节点结构体中设置lay，记录节点所在层数。每次队列push孩子节点时，更新孩子节点所在层数：Node[top].lay+1。若一个结点没有孩子节点，则是叶子，layer[该节点]++。遍历过程中，记录最深节点所在层数。

最后输出层数个数字。

## 代码

• BFS
/*BFS遍历树，数每层叶子节点的个数*/
#include<iostream>
#include<vector>
#include<queue>

using namespace std;

const int maxN=110;
const int maxM=110;
struct node {
int lay;//所在层数（0开始）
vector<int> child;//孩子节点
} Node[maxM];
int layer[maxM]; //每层叶子节点的个数
int N,M;
int maxLayer=0;//最深节点的层数（0开始 )

void BFS() {
queue<int> que;
que.push(1);//根节点

while(!que.empty()) {
int top=que.front();
que.pop();
if(Node[top].lay>maxLayer)
maxLayer=Node[top].lay;

bool leafFlag=true;//是否是叶子节点
for(int i=0; i<Node[top].child.size(); i++) {
leafFlag=false;
int index=Node[top].child[i];
Node[index].lay=Node[top].lay+1;
que.push(index);
}
if(leafFlag)
layer[Node[top].lay]++;
}
}

int main() {
fill(layer,layer+maxM,0);
int id1,chNo,id2;
scanf("%d%d",&N,&M);
for(int i=0; i<M; i++) {
scanf("%d%d",&id1,&chNo);
for(int j=0; j<chNo; j++) {
scanf("%d",&id2);
Node[id1].child.push_back(id2);
}
}

BFS();

for(int i=0; i<=maxLayer; i++) {
if(i!=0) cout<<" ";
cout<<layer[i];
}

return 0;
}


• DFS：（柳神的）
/*DFS遍历*/
#include <iostream>
#include <vector>
#include <algorithm>
using namespace std;
vector<int> v[100];
int book[100], maxdepth = -1;
void dfs(int index, int depth) {
if(v[index].size() == 0) {
book[depth]++;
maxdepth = max(maxdepth, depth);
return ;
}
for(int i = 0; i < v[index].size(); i++)
dfs(v[index][i], depth + 1);
}
int main() {
int n, m, k, node, c;
scanf("%d %d", &n, &m);
for(int i = 0; i < m; i++) {
scanf("%d %d",&node, &k);
for(int j = 0; j < k; j++) {
scanf("%d", &c);
v[node].push_back(c);
}
}
dfs(1, 0);
printf("%d", book[0]);
for(int i = 1; i <= maxdepth; i++)
printf(" %d", book[i]);
return 0;
}


6 3
01 2 02 05
02 2 03 04
05 1 06


posted @ 2020-02-16 15:32  musecho  阅读(...)  评论(...编辑  收藏