PAT 1102. Invert a Binary Tree (25)

1102. Invert a Binary Tree (25)

時間限制

400 ms

內存限制

65536 kB

代碼長度限制

16000 B

判題程序

Standard

作者

CHEN, Yue

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

---

這道題主要考察level order和in order遍歷樹，至於反轉二叉樹倒不是關鍵（只要在建立二叉樹的時候把輸入的左子樹當成右子樹，右子樹當成左子樹即可）。

另外由於序號已知，所以可以直接用int來代替指針指向child或parent。

level order遍歷用隊列來完成，in order用遞歸即可， 代碼如下： 

#include <iostream>
#include <algorithm>
#include <cmath>
#include <queue>
#include <vector>
using namespace std;
typedef struct n{
	int id;
	int parent;
	int leftChild;
	int rightChild;
	n(){
		parent = leftChild = rightChild = -1;
	}
}node;
vector<node> tree;
int root;
bool flag;

void levelorder(int root){
	queue<int> treeQueue;
	treeQueue.push(root);
	while(!treeQueue.empty()){
		int node = treeQueue.front();
		treeQueue.pop();
		if(tree[node].leftChild != -1)
			treeQueue.push(tree[node].leftChild);
		if(tree[node].rightChild != -1)
			treeQueue.push(tree[node].rightChild);
		if(node == root)
			cout<<node;
		else
			cout<<" "<<node;
	}
	cout<<endl;
}
int findRoot(){
	int k = 0;
	while(tree[k].parent != -1){
		k = tree[k].parent;
	}
	return k;
}

void inorder(int x){
	if(x == -1)
		return;
	inorder(tree[x].leftChild);
	if(flag)
		cout<<" "<<x;
	else{
		cout<<x;
		flag = true;
	}
	inorder(tree[x].rightChild);
}
int main(void)
{
	int N;
	cin>>N;
	tree.resize(N);
	for(int i = 0; i < N; i++){
		string str[2];
		cin>>str[0]>>str[1];
		tree[i].id = i;
		if(str[0][0]!='-'){
			tree[i].rightChild = atoi(str[0].c_str());
			tree[atoi(str[0].c_str())].parent = i;
		}
		if(str[1][0]!='-'){
			tree[i].leftChild = atoi(str[1].c_str());
			tree[atoi(str[1].c_str())].parent = i;
		}
	}
	root = findRoot();
	levelorder(root);
	flag = false;
	inorder(root);
	cout<<endl;
	return 0;
}