A1102 Invert a Binary Tree

Posted on | In |

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.

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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

题目要求输出反转后的二叉树层序和终须遍历序列

由于给的是节点编号 所以用二叉树的静态写法更方便

在后序遍历访问到根节点时 交换它的左右孩子 即可反转二叉树 

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#include<bits/stdc++.h> 
#define MAX 12
using namespace std;
typedef long long ll;

int N,sum=0,num=0;
int notR[MAX];
struct node
{
	int left;
	int right;
	
};
node nodes[MAX];
void bfs(int root){
	queue<int> que;
	que.push(root);
	while(!que.empty()){
		int t=que.front();
		cout<<t;
		que.pop();
		num++;
		if(num<N) cout<<" ";
		if(nodes[t].left!=-1) que.push(nodes[t].left);
		if(nodes[t].right!=-1) que.push(nodes[t].right);		
	}
	
}
void porder(int root){
	if(root==-1) return;
	porder(nodes[root].left);
	porder(nodes[root].right);
	swap(nodes[root].left,nodes[root].right);
} 
void inorder(int root){ 
	if(root==-1) return;
	inorder(nodes[root].left);
	cout<<root;
	sum++;
	if(sum<N) cout<<" "; 
	inorder(nodes[root].right);
}
int strto(char a){
	if(a=='-') return -1;
	else {
		int b=a-'0';
		notR[b]=true;//!
		return b;
	}
}
int findR(){
	int i=-1;
	for(i=0;i<N;i++){
		if(notR[i]==false){
			break;
		}
	}
	return i;
}
int main(){
	cin>>N;
	for(int i=0;i<N;i++){
		char a,b;
		getchar();
		scanf("%c %c",&a,&b); 
		int l=strto(a);
		int r=strto(b);
		nodes[i].left=l;nodes[i].right=r; 
	}
	int root=findR();
	porder(root);
	bfs(root);
	cout<<endl;
	inorder(root);
	
	
	
}