【PAT】A1086 Tree Traversals Again (25分)

题目

• non-recursive 非递归

解决

思路

• push序列=先序序列，样例：123456
• pop序列=中序序列，样例：324165
• 题目其实就是根据先序序列和中序序列输出后序序列

实现

Code1

#include<cstdio>
#include<queue>
#include<stack>
#include<cstring>
using namespace std;
const int maxn = 50;

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

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] == pre[preL]){
			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;
}

int num = 0;
void postorder(node* root){
	if(root == NULL){
		return;
	}	
	postorder(root->lchild);
	postorder(root->rchild);
	printf("%d", root->data);
	num++;
	if(num < n) printf(" ");
}

int main(){
	scanf("%d", &n);
	char str[5];
	stack<int> st;
	int x, preIndex = 0, inIndex = 0;
	for(int i=0; i<2*n; i++){
		scanf("%s", str);
		if(strcmp(str, "Push") == 0){
			scanf("%d", &x);
			pre[preIndex++] = x;
			st.push(x);
		}else{
			in[inIndex++] = st.top();
			st.pop();
		}
	}
	
	node* root = create(0, n-1, 0, n-1);
	postorder(root);
	return 0;
}