20秋季 7-3 Left-View of Binary Tree (25 分)

题目大意：

给定中序和先序，输出left-view，其实就是输出层序遍历，每层输出一个即可。类似题目：A1020

思路：

①构建树；

②层序遍历树。

#include <iostream>
#include <queue>
#include <algorithm>
using namespace std;
const int maxn = 25;

int in[maxn], pre[maxn];
struct node {
	int data;
	int layer;
	node* lchild;
	node* rchild;
};

node* create(int preL, int preR, int inL, int inR) {
	if (preL > preR)
		return NULL;
	node* root = new node;
	root->data = pre[preL];
	int k;
	for (k = inL; k <= inR; k++) {
		if (in[k] == root->data)
			break;
	}
	int numLeft = k - inL;
	root->lchild = create(preL+1, preL + numLeft, inL, k - 1);
	root->rchild = create(preL + numLeft + 1, preR, k + 1, inR);
	return root;
}
bool hashTable[maxn] = { false };// 判断该层结点有没被输出过，这样可以保证一层只输出一个结点
void BFS(node* root) {
	queue<node*> q;
	root->layer = 1;
	q.push(root);	
	while (!q.empty()) {
		node* now = q.front();
		q.pop();
		int lay = now->layer;
		if (hashTable[lay] == false) {
			printf("%d", now->data);
			hashTable[lay] = true;
		}
		
		if (now->lchild) {
			now->lchild->layer = now->layer + 1;
			q.push(now->lchild);
		}
		if (now->rchild) {
			now->rchild->layer = now->layer + 1;
			q.push(now->rchild);
		}
	}
}

int main() {
	freopen("input.txt", "r", stdin);
	int n;
	scanf("%d", &n);
	for (int i = 0; i < n; i++) {
		scanf("%d", &in[i]);
	}
	for (int i = 0; i < n; i++) {
		scanf("%d", &pre[i]);
	}
	node* root = create(0, n - 1, 0, n - 1);
	BFS(root);

	return 0;
}