天梯赛 L2-011 玩转二叉树

和上面一道树的题思路一样，只需要换一下递归的顺序即可。

#include <bits/stdc++.h>
using namespace std;
#define endl '\n'
// #define int long long
typedef long long ll;
const int N = 50;
int n;
int a[N],b[N];
struct nod{
	int value;
	nod* l = NULL;
	nod* r = NULL;
};
nod* build(int al,int ar,int bl,int br){
	if(al > ar) return NULL;
	nod* root = (nod*)malloc(sizeof(nod));
	root->value = a[al];
	int x = a[al];
	int p = bl;
	while(b[p] != x){
		p++;
	}
	int len = p - bl;
	root->l = build(al+1,al+len,bl,p-1);
	root->r = build(al+len+1,ar,p+1,br);
	return root;
}
void bfs(nod* x){
	queue<nod> q;
	q.push(*x);
	while(!q.empty()){
		nod tmp = q.front();
		q.pop();
		if(tmp.value != x->value) cout<<" ";
		cout<<tmp.value;
		if(tmp.r != NULL) q.push(*(tmp.r));
		if(tmp.l != NULL) q.push(*(tmp.l));
	}
}
void solve() {
	cin>>n;
	for(int i = 1 ; i <= n ; i++){
		cin>>b[i];
	}
	for(int i = 1 ; i <= n ; i++){
		cin>>a[i];
	}
	nod* head = build(1,n,1,n);
	bfs(head);	//输出
}
int main() {
    ios::sync_with_stdio(0); cin.tie(0); cout.tie(0);
    int tt = 1;
    // cin >> tt;
    while (tt--) {
        solve();
    }
    return 0;
}