PAT 1102 Invert a Binary Tree [静态二叉树]

最新推荐文章于 2022-07-11 16:03:39 发布

原创最新推荐文章于 2022-07-11 16:03:39 发布 · 170 阅读

0 ·

CC 4.0 BY-SA版权

PAT 专栏收录该内容

100 篇文章

订阅专栏

本文解析了一道关于逆序二叉树的编程题，介绍了如何通过颠倒输入顺序来简化问题，实现层次和中序遍历。代码示例使用C++实现，详细解释了输入处理和遍历算法。

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

------------------------------------这是题目和解题的分割线------------------------------------

我可能是全网唯一一个~~又又又又~~看不懂题目的人，搜了好久看了好多（但都没人解释题目哭了），才终于反应过来，原来是这个意思——根结点隐藏在深处，也就是输入顺序。举个例子，根结点为0的左右结点是1和-，根结点为1的左右结点是-和-，以此类推。也怪我不学无术没写过什么题，以后见着这种输入我就明白了。

回到题目，求左右子树颠倒，那么可以在输入的时候就颠倒输入，之后该层次层次，该中序中序。

#include<cstdio>
#include<queue>

using namespace std;

struct node
{
	int left,right;
}tree[20]; //静态二叉树适合这种输入 

int ifRoot[20] = {0},n;

int charToInt(char c)
{
	if(c=='-') return -1;
	else 
	{
		ifRoot[c-'0'] = 1; //转换数字的时候顺便标记一下出现过的数字 
		return c-'0';
	}	
}

int num = 0;
void levelOrder(int root) //层次遍历 
{
	queue<int> q;
	q.push(root); //根结点入队 
	while(!q.empty())
	{
		int top = q.front(); //取出根结点 
		q.pop();
		printf("%d",top);
		if(num!=n-1) printf(" "); //注意末尾空格 
		num++;
		if(tree[top].left!=-1) q.push(tree[top].left);
		if(tree[top].right!=-1) q.push(tree[top].right);
	}
}

int cnt = 0;
void inOrder(int root) //中序遍历 
{
	if(root!=-1)
	{
		inOrder(tree[root].left);
		printf("%d",root);
		if(cnt!=n-1) printf(" ");  
		cnt++;
		inOrder(tree[root].right);
	}
}

int main()
{
	int i,root;
	char l,r;
	scanf("%d",&n);
	for(i=0;i<n;i++)
	{
		getchar(); //吃掉空行 字符真麻烦 
		scanf("%c %c",&l,&r);
		tree[i].left = charToInt(r); //注意输入的时候left对应r右子树 
		tree[i].right = charToInt(l); //right对应l左子树 
	}
	for(i=0;i<n;i++)
		if(ifRoot[i]==0) break;
	//根结点不会在任何结点的左右子树上，所以0~n-1没出现过的数字就是根结点 
	root = i; 
	levelOrder(root);
	printf("\n");
	inOrder(root);
	return 0;
}