03-树2. List Leaves (25)

/*
并查集 和 树的递归遍历找叶子节点
*/
#include <cstdio>
#include <sstream>
#include <cstring>
#include <iostream>
#include <string>
#include <vector>
#include <queue>
#include <algorithm>
#include <sstream>
#include <cmath>
#include <stack>

using namespace std;

#define  N 11

int n;

struct mydata{
int no;
int left;
int right;
};

typedef struct node{
int data;
struct node* left;
struct node* right;
node(int _data){
data = _data;
left = NULL;
right = NULL;
}
}Bnode;

//int str2int(string s)
//{
//  if (s == "-")
//    return -1;
//  stringstream  ss;
//  ss << s;
//  int tmp;
//  ss >> tmp;
//  return tmp;
//}

mydata a
;

int father
;
int find(int x)
{
if (x == father[x])
return x;
return father[x] = find(father[x]);
}

void merge(int x, int y)
{
father[find(y)] = find(x);
}

Bnode* root;

Bnode* createtree(Bnode* root ,int rootNum)
{
if (rootNum == -1)
return NULL;
root = new node(rootNum);
if (a[rootNum].left == -1)
{
root->left = NULL;
}
else{
root->left = createtree(root->left, a[rootNum].left);
}
if (a[rootNum].right == -1)
{
root->right = NULL;
}
else{
root->right = createtree(root->right, a[rootNum].right);
}
return root;
}

vector<int> v;

void inorder(Bnode* root)
{
queue<Bnode*> que;
que.push(root);
while (!que.empty())
{
Bnode* bt = que.front();
que.pop();
if (bt->left == NULL && bt->right == NULL)
v.push_back(bt->data);
else {
if (bt->left != NULL)
{
que.push(bt->left);
}
if(bt->right != NULL)
{
que.push(bt->right);
}
}
}
}

int main()
{
//freopen("in", "r", stdin);
scanf("%d\n", &n);
int i;
for (i = 0; i < n; i++)
{
father[i] = i;
}
mydata dt;
char c1, c2;
for (i = 0; i < n; i++)
{
scanf("%c %c\n", &c1, &c2);
dt.no = i;
if (c1 == '-')
{
dt.left = -1;
}
else{
dt.left = (int)(c1 - '0');
if (find(i) != find(dt.left))
{
merge(i, dt.left);
}
}
if (c2 == '-')
{
dt.right = -1;
}
else{
dt.right = (int)(c2 - '0');
if (find(i) != find(dt.right))
{
merge(i, dt.right);
}
}
a[i] = dt;
}

int rootNum = find(0);
root = NULL;
root = createtree(root, rootNum);

inorder(root);
int len = v.size();
printf("%d", v[0]);
for (i = 1; i < len; i++)
{
printf(" %d", v[i]);
}
printf("\n");
return 0;
}