PAT 甲级 1102 Invert a Binary Tree

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最新推荐文章于 2020-03-22 11:20:58 发布

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本文介绍了一道经典的算法题目——翻转二叉树。通过输入二叉树的节点信息，实现翻转并输出翻转后的层级遍历和中序遍历序列。文章提供了完整的C++代码实现，展示了如何在输入阶段就颠倒左右节点，简化了后续的遍历过程。

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1102 Invert a Binary Tree （25 point(s)）

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

经验总结：

emmmmm 简单的划水题，可以在输入的时候就把左右结点颠倒，也可以在遍历的时候反着遍历，反正都可以啦~

AC代码

#include <cstdio>
#include <vector>
#include <queue>
#include <algorithm>
using namespace std;
int in[10],level[10],lnum=0;
struct node
{
	int v,left,right;
}tree[10];
void level_order(int s)
{
	lnum=0;
	queue<node> q;
	q.push(tree[s]);
	while(q.size())
	{
		node x=q.front();
		q.pop();
		level[lnum++]=x.v;
		if(x.left!=-1)
			q.push(tree[x.left]);
		if(x.right!=-1)
			q.push(tree[x.right]);
	}
}
void in_order(int root)
{
	if(root==-1)
		return ;
	if(tree[root].left!=-1)
		in_order(tree[root].left);
	in[lnum++]=root;
	if(tree[root].right!=-1)
		in_order(tree[root].right);
}
int main()
{
	int n;
	char a,b;
	bool flag[10]={false};
	scanf("%d",&n);
	getchar();
	for(int i=0;i<n;++i)
	{
		scanf("%c %c",&a,&b);
		getchar();
		if(a!='-')
		{
			tree[i].right=a-'0';
			flag[a-'0']=true;
		}
		else
			tree[i].right=-1;
		if(b!='-')
		{
			tree[i].left=b-'0';
			flag[b-'0']=true;
		}
		else
			tree[i].left=-1;
		tree[i].v=i;
	}
	int root;
	for(root=0;root<n;++root)
	{
		if(flag[root]==false)
			break;
	}
	level_order(root);
	lnum=0;
	in_order(root);
	for(int i=0;i<n;++i)
		printf("%d%c",level[i],i<n-1?' ':'\n');
	for(int i=0;i<n;++i)
		printf("%d%c",in[i],i<n-1?' ':'\n');
	return 0;
}