프로그램 설계 기초 13 반전 트리 방법

3202 단어 pat 세월
1102 Invert a Binary Tree(25점)
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

하나, 반전 두 갈래 나무 쓰기 (물론 이 문제는 반전 두 갈래 나무를 사용하지 않고 입력할 때 반으로 값을 부여한다)
  
void posOrder(int root) {
	if (root == -1) {
		return;
	}
	posOrder(Node[root].lchild);
	posOrder(Node[root].rchild);
	swap(Node[root].lchild, Node[root].rchild);
}

내 코드
#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("
"); } 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; }

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