程序设计基础13 反转二叉树的方法

1102 Invert a Binary Tree（25 分）

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

一，反转二叉树写法（当然，本题我未用反转二叉树，而是从输入的时候反着赋值）

  

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

二，我的代码

#include
#include
using namespace std;
int N = 0;
int times = 0;
int root = 0;
int flag[15] = { 0 };
struct Node {
	int data;
	int lchild;
	int rchild;
}node[15];
queue que;
void levelOrder(int index) {
	while (!que.empty()) {
		int front = que.front();
		que.pop();
		printf("%d", front);
		times++;
		if (times != N) {
			printf(" ");
		}
		else {
			printf("\n");
		}
		if (node[front].lchild != -1)que.push(node[front].lchild);
		if (node[front].rchild != -1)que.push(node[front].rchild);
	}
}
void inOrder(int index) {
	if (index == -1) {
		return;
	}
	inOrder(node[index].lchild);
	printf("%d", index);
	times++;
	if (times != N) {
		printf(" ");
	}
	inOrder(node[index].rchild);
}
int main() {
	char a;
	char b;
	int num_1 = 0;
	int num_2 = 0;
	scanf("%d", &N);
	getchar();
	for (int i = 0; i < N; i++) {
		node[i].data = i;
		node[i].lchild = -1;
		node[i].rchild = -1;
	}
	for (int i = 0; i < N; i++) {
		a = getchar();
		getchar();
		b = getchar();
		getchar();
		if (a != '-') {
			num_1 = a - '0';
			node[i].rchild = num_1;
			if (flag[num_1] == 0) {
				flag[num_1] = 1;
			}
		}
		if (b != '-') {
			num_2 = b - '0';
			node[i].lchild = num_2;
			if (flag[num_2] == 0) {
				flag[num_2] = 1;
			}
		}
	}
	int k = 0;
	for (k = 0; k < N; k++) {
		if (flag[k] == 0) {
			break;
		}
	}
    root = k;
	que.push(root);
	levelOrder(root);
	times = 0;
	inOrder(root);
	return 0;
}