The code is based on the
"Data Structure" written in this article 03-Tree 2 List Leaves
Problem Solving Idea: It is enough to traverse the tree in layer order.
1. The use of the flag variable cannot be written by myself.
if(t.Left==Null&&t.Right==Null){
if(flag){
cout<<" ";
}
else{
flag = true;
}
cout<<t.Data;
}
2. At the beginning, the Data in the structure TreeNode was not written, resulting in not knowing what data to output.
3.using namespace std;//If you write it will report an
error'cout ' was not declared in this scope 4.pop has no return value, I mistakenly thought that the element returned was the popup.
#include<iostream>
#include<queue>
#define Null -1
#define MAXSIZE 10 //不需要加分号
using namespace std;//如果写会报错‘cout’ was not declared in this scope
typedef int Tree;
struct TreeNode{
int Data; //使用这个Data来存放结构数组T内每个节点存储的位置，便于输出结果
Tree Left;
Tree Right;
}T[MAXSIZE];
Tree BuildTree(struct TreeNode T[]);
void Traversal(int root);
Tree BuildTree(struct TreeNode T[]){
int Root=Null,N; //刚开始将节点置为空，若为空树的时候可返回Null
char cl,cr;
scanf("%d",&N);
if(N){
//加入对树为空的判断
int check[MAXSIZE]={
0};//置为零
for(int i=0;i<N;++i){
//把N个节点读入树
//C语言输入：scanf("\n%c %c",&cl,&cr);//&T[i].Data,&cl,&cr是不是不需要& 蒙
//C++输入见下一行，似乎更简单了
cin>>cl>>cr;
T[i].Data = i;
if(cl!='-'){
//将单引号错写为双引号会报错comparison between pointer and integer
T[i].Left = cl - '0';
check[T[i].Left] = 1;
}
else{
T[i].Left = Null;
}
if(cr!='-'){
T[i].Right = cr - '0';
check[T[i].Right] = 1;
}
else{
T[i].Right = Null;
}
}
int i;
for(i=0;i<N;++i){
if(!check[i]) break;
}
Root = i;
}
return Root;
}
void Traversal(int root){
if(root==Null) {
cout<<"-";
return; //结束函数下运行
}
bool flag = false;
struct TreeNode t;
queue<struct TreeNode> q;
q.push(T[root]);
while(!q.empty()){
t = q.front();
q.pop(); //队列中的pop没有返回值么
if(t.Left==Null&&t.Right==Null){
if(flag){
cout<<" ";
}
else{
flag = true;
}
cout<<t.Data;
}
//以下两个if 我没十分搞清楚应该放在循环里还是外面，如果放在里面第一次循环pop出根节点之后队列为空，不再循环了
if(t.Left!=Null){
q.push(T[t.Left]);
}
if(t.Right!=Null){
q.push(T[t.Right]);
}
}
}
int main(){
int r;
r = BuildTree(T);
Traversal(r);
return 0;
}