FBI樹（先遞歸構建二叉樹，再後序遍歷之）

#include <iostream>
#include <algorithm>
#include <math.h>
using namespace std;
#define N 1200
char arrr[N];
int arr[N];
typedef struct S
{
	char type;
	S *lchild;
	S *rchild;
}tree;
void make_tree(tree*, int,int);
char judge(int,int);
void back_traverse(tree*);
int len;


int main()
{
	int n;
	cin >> n;
	len = pow(2, n);
	//cout << n << endl;
	for (int i = 1; i <= len; i++)
	{
		cin >> arrr[i];
		arr[i] = arrr[i] - '0';
	}
	tree *root = new tree;
	root->type = judge(1, len);
	make_tree(root, 1, len);
	back_traverse(root);

	//system("pause");
	return 0;
}

void make_tree(tree* root, int l, int r)
{
	if (l == r)
	{
		root->lchild = NULL;
		root->rchild = NULL;
		root->type = judge(l, l);
		return;
	}
	else
	{
		int mid = l + (r - l - 1) / 2;
		//cout << mid << endl;
		tree *z = new tree;
		z->type = judge(l, mid);
		root->lchild = z;
		tree *p = new tree;
		p->type = judge(mid + 1, r);
		root->rchild = p;
		make_tree(z, l, mid);
		make_tree(p, mid + 1, r);
		return;
	}
}

char judge(int l,int r)
{
	int b0 = 0;
	int b1 = 0;
	for (int i = l; i <= r; i++)
	{
		if (arr[i] == 0)
			b0 = 1;
		if (arr[i] == 1)
			b1 = 1;
	}
	if (b0 == 1 && b1 == 0)
		return 'B';
	else if (b0 == 0 && b1 == 1)
		return 'I';
	else
		return 'F';
}

void back_traverse(tree* root)
{
	if (root == NULL)
		return;
	back_traverse(root->lchild);
	back_traverse(root->rchild);
	cout << root->type;
	return;
}