1102. Invert a Binary Tree (25)

1102 . Invert a Binary Tree (25)

The following is from Max Howell @twitter:

Google: 90% of our engineers use the software you wrote (Homebrew), but you can’t invert a binary tree on a whiteboard so fuck off.

Now it’s your turn to prove that YOU CAN invert a binary tree!

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 from 0 to N-1, 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 the first line the level-order, and then in the second line the in-order traversal sequences of the inverted tree. 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:

3 7 2 6 4 0 5 1

6 5 7 4 3 2 0 1

二叉树的相关操作，用到并查集，递归，队列

#include <cstdio>
#include <cstdlib>
#include <iostream>
#include <vector>
#include <queue>

using namespace std;

#define N 20

int n ;

struct data
{
int left ;
int right ;
};

data dt
;

int fa
;

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

void merg(int x , int y)
{
fa[find(y)] = find(x);
}

typedef struct node{
int data;
struct node* left;
struct node* right ;
node(int _data = - 1)
{
data = _data ;
}
}Bnode ;

Bnode* createTree(Bnode* root , int num)
{
root = new node(num);
if(dt[num].left == -1)
root->left = NULL ;
else
root->left = createTree( root->left , dt[num].left);

if(dt[num].right == -1)
root->right = NULL ;
else
root->right = createTree( root->right , dt[num].right);

return root ;
}

void levelorder(Bnode* root)
{
queue<Bnode*> que;
que.push(root) ;
bool flag = false;

while( !que.empty())
{
Bnode* bt = que.front();
que.pop();
if(flag == false)
{
printf("%d",bt->data);
flag = true;
}else{
printf(" %d",bt->data);
}
if(bt->right != NULL)
{
que.push(bt->right);
}
if(bt->left != NULL)
{
que.push(bt->left);
}
}
printf("\n");
}

Bnode* reverse(Bnode* root)
{
if(root == NULL)
return NULL;
reverse(root->left);
reverse(root->right);
Bnode* tmp = root->left ;
root->left = root->right;
root->right = tmp ;
return root ;
}

vector<int> in;

void inorder(Bnode* root2)
{
if(root2 != NULL)
{
inorder(root2->left);
in.push_back(root2->data);
inorder(root2->right);
}
}

int main()
{
//freopen("in.txt","r",stdin);
scanf("%d\n",&n);
int i ;
char a , b ;
for(i = 0 ; i< n ;i++)
{
fa[i] = i ;
}
for(i = 0 ; i < n ;i++)
{
scanf("%c %c\n",&a,&b);
if(a == '-')
{
dt[i].left = -1 ;
}else{
dt[i].left = a  - '0' ;
if(find(i) != find(dt[i].left))
merg(i,dt[i].left);
}
if(b == '-')
{
dt[i].right = -1 ;
}else{
dt[i].right = b - '0' ;
if(find(i) != find(dt[i].right))
merg(i,dt[i].right);
}
}

int rootNum = find(0) ;

Bnode* root = NULL;

root = createTree( root , rootNum);

levelorder(root);

Bnode* root2 = reverse(root);
in.clear();
inorder(root2);

printf("%d",in[0]);
for(i = 1;i<n;i++)
{
printf(" %d",in[i]);
}
printf("\n");
return 0 ;
}