BOJ 1991 트리 순회

트리 순회 [실버 1]

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문제 링크

https://www.acmicpc.net/problem/1991

풀이

#include<iostream>
using namespace std;

int N;

typedef struct node {
	char left;
	char right;
}node;

node tree[26];
int cnt[26];

void preorder(int idx) {
	// print current
	cout << char('A' + idx);
	// left
	if (tree[idx].left != '.') {
		preorder(tree[idx].left - 'A');
	}
	// right
	if (tree[idx].right != '.') {
		preorder(tree[idx].right - 'A');
	}
}

void inorder(int idx) {
	// left
	if (tree[idx].left != '.') {
		inorder(tree[idx].left - 'A');
	}
	// print current
	cout << char('A' + idx);
	// right
	if (tree[idx].right != '.') {
		inorder(tree[idx].right - 'A');
	}
}

void postorder(int idx) {
	// left
	if (tree[idx].left != '.') {
		postorder(tree[idx].left - 'A');
	}
	// right
	if (tree[idx].right != '.') {
		postorder(tree[idx].right - 'A');
	}
	// print current
	cout << char('A' + idx);

}
int main(void)
{

	cin >> N;
	char parent, left, right;
	for (int i = 0; i < N; i++)
	{
		cin >> parent >> left >> right;
		tree[parent - 'A'].left = left;
		tree[parent - 'A'].right = right;

		cnt[parent - 'A']++;
	}

	int rootIdx = -1;
	for (int i = 0; i < N; i++)
	{
		if (cnt[i] == 1) {
			rootIdx = i;
			break;
		}
	}

	preorder(rootIdx);
	cout << "\n";
	inorder(rootIdx);
	cout << "\n";
	postorder(rootIdx);

	return 0;
}