1102 Invert a Binary Tree (25 分)

最新推荐文章于 2020-03-22 11:20:58 发布

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本文详细介绍了一种用于翻转二叉树的算法，并通过具体的输入输出样例展示了算法的实现过程。文章首先介绍了算法的输入规格，即如何定义二叉树节点及其左右子节点，然后阐述了算法的输出规格，包括层序遍历和中序遍历翻转后的二叉树。通过C++代码实现了二叉树的翻转，包括后序遍历翻转、层序遍历和中序遍历等操作。

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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

#include <iostream>
#include <queue>
using namespace std;

struct Node {
	int lchild;
	int rchild;
}node[110];

bool notRoot[110] = {false};

int n, num;

void print(int id) {
	printf("%d", id);
	
	num++;
	
	if (num < n) printf(" ");
	else printf("\n");
}

void postOrder(int root) {
	if (root == -1) {
		return ;
	}
	
	postOrder(node[root].lchild);
	postOrder(node[root].rchild);
	swap(node[root].rchild, node[root].lchild);
}

void BFS(int root) {
	queue<int> q;
	
	q.push(root);
	
	while (!q.empty()) {
		root = q.front();
		q.pop();
		
		print(root);
		
		if (node[root].lchild != -1) q.push(node[root].lchild);
		if (node[root].rchild != -1) q.push(node[root].rchild);
	}
}

void inOrder(int root) {
	if (root == -1) return ;
	
	inOrder(node[root].lchild);
	print(root);
	inOrder(node[root].rchild);
}

int strToNum(char ch) {
	if (ch >= '0' && ch <= '9') {
		notRoot[ch - '0'] = true;
		return ch - '0';
	}	
	return -1;
}

int findRoot() {
	int i;
	for (i = 0; i < n; i++) {
		if (notRoot[i] == false) return i;
	} 
}

int main() {
	char l, r;
	
	scanf("%d", &n);
	for (int i = 0; i < n; i++) {
		scanf("%*c%c %c", &l, &r);
		node[i].lchild = strToNum(l);
		node[i].rchild = strToNum(r);
	}
	
	int root = findRoot();
	postOrder(root);
	BFS(root);
	num = 0;
	inOrder(root);

	return 0;
}