PTA刷题]树List Leaves
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 NN (\le 10≤10) which is the total number of nodes in the tree -- and hence the nodes are numbered from 0 to N-1N−1. Then NN 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#include <iostream>
#include <queue>
using namespace std;#define Null -1
struct IntNode{
int Left;
int Right;
}T1[11];
int flag=0;
int BuildInt(IntNode T[]);
void TraverseLeaves(int root);
int main(){
int R1;
R1=BuildInt(T1);
TraverseLeaves(R1);
return 0;
}
int BuildInt(IntNode T[]){
int root=Null;
int N,check[11];
char lc,rc;
cin>>N;
if(N>0){
for(int i=0;i<N;i++)
check[i]=0;
for(int i=0; i<N; i++){
cin>>lc>>rc;
if(lc!='-'){
T[i].Left=lc-'0';
check[T[i].Left]=1;
}else
T[i].Left=Null;
if(rc!='-'){
T[i].Right=rc-'0';
check[T[i].Right]=1;
}else
T[i].Right=Null;
}
for(int i=0;i<N;i++){
if(check[i]==0)
{ root=i;
break;
}
}
}
return root;
}
void TraverseLeaves(int root)
{
queue<int> theQ;
int node;
theQ.push(root);
while(!theQ.empty()){
node=theQ.front();
theQ.pop();
if(T1[node].Left==Null && T1[node].Right==Null )
{
if(flag==0){
cout<<node;
flag=1;
}else
cout<<' '<<node;
}
else{
if(T1[node].Left!=Null)
theQ.push(T1[node].Left);
if(T1[node].Right!=Null)
theQ.push(T1[node].Right);
}
}
}
考查的是对树的存储和遍历。
不一定非得建树，根据其节点之间的联系建立起关系，根据其关系找叶子结点，就相当于一个双向链表。
这里用到了和“树的同构”一题里相同的存储树的方法。
遍历树的时候要用的是层序遍历，以满足题目的“从上到下从左到右的叶子节点的输出”的要求。