PAT Advanced 1086 Tree Traversals Again (25) [树的遍历]

题目

An inorder binary tree traversal can be implemented in a non-recursive way with a stack. For example, suppose that when a 6-node binary tree (with the keys numbered from 1 to 6) is traversed, the stack operations are: push(1); push(2); push(3); pop(); pop(); push(4); pop(); pop(); push(5); push(6); pop(); pop(). Then a unique binary tree (shown in Figure 1) can be generated from this sequence of operations. Your task is to give the postorder traversal sequence of this tree.

Input Specification:

Each input file contains one test case. For each case, the first line contains a positive integer N (<=30) which is the total number of nodes in a tree (and hence the nodes are numbered from 1 to N). Then 2N lines follow, each describes a stack operation in the format: “Push X” where X is the index of the node being pushed onto the stack; or “Pop” meaning to pop one node from the stack.

Output Specification:

For each test case, print the postorder traversal sequence of the corresponding tree in one line. A solution is guaranteed to exist. All the numbers must be separated by exactly one space, and there must be no extra space at the end of the line.

Sample Input:

6

Push 1

Push 2

Push 3

Pop

Pop

Push 4

Pop

Pop

Push 5

Push 6

Pop

Pop

Sample Output:

3 4 2 6 5 1

题目分析

已知非递归中序（借助栈）遍历树的操作流程，求后序序列

注：操作流程中push操作依次组成前序序列，又可以借助push和pop操作得到中序序列，题目即转换为中序+前序->后序

解题思路

思路 01

1. 中序+前序建树
2. 递归后序遍历树

思路 02（最优）

1. 中序+前序直接转后序序列

知识点

1. while(~scanf("%s",s)) {} //等价于scanf("%s",s)!=EOF

两者作用是相同的

~是按位取反

scanf的返回值是输入值的个数

如果没有输入值就是返回-1

-1按位取反结果是0

while(~scanf("%d", &n))就是当没有输入的时候退出循环

EOF，为End Of File的缩写，通常在文本的最后存在此字符表示资料结束。

EOF 的值通常为 -1

Code

Code 01

#include <iostream>
#include <vector>
#include <stack>
using namespace std;
vector<int> pre,in;
int index,n;
struct node {
	int data;
	node * left;
	node * right;
};
node * create(int preL,int preR,int inL,int inR) {
	if(preL>preR)return NULL;
	node * now = new node;
	now->data=pre[preL];
	int k=inL;
	while(k<inR&&in[k]!=pre[preL])k++;
	now->left=create(preL+1, preL+(k-inL), inL, k-1);
	now->right=create(preL+(k-inL)+1, preR, k+1, inR);
	return now;
}
void postOrder(node * root) {
	if(root==NULL)return;
	postOrder(root->left);
	postOrder(root->right);
	printf("%d",root->data);
	if(++index<n)printf(" ");
}
int main(int argc,char * argv[]) {
	int id;
	string s;
	stack<int> sc;
	scanf("%d",&n);
	int len=n<<1;
	for(int i=0; i<len; i++) {
		cin>>s;
		if(s=="Push") {
			scanf("%d",&id);
			pre.push_back(id);//preorder
			sc.push(id);
		}
		if(s=="Pop") {
			in.push_back(sc.top());
			sc.pop();
		}
	}
	node * root = create(0,n-1,0,n-1);
	postOrder(root);
	return 0;
}

Code 02（最优）

#include <iostream>
#include <vector>
#include <stack>
#include <cstring>
using namespace std;
vector<int> pre,in,post;
int n;
void postOrder(int preL,int preR,int inL,int inR){
	if(inL>inR)return;// preL>preR也可以
	int k=inL;
	while(k<inR&&in[k]!=pre[preL])k++;
	postOrder(preL+1, preL+(k-inL), inL, k-1);//先存放左子节点
	postOrder(preL+(k-inL)+1, preR, k+1, inR);//后存放右子节点
	post.push_back(pre[preL]); //再存放父节点
}
int main(int argc,char * argv[]) {
	int id;
	char s[5];
	stack<int> sc;
	scanf("%d",&n);
	while(~scanf("%s",s)) { //等价于scanf("%s",s)!=EOF
		if(strlen(s)==4) {
			//Push
			scanf("%d",&id);
			pre.push_back(id);
			sc.push(id);
		} else {
			//Pop
			in.push_back(sc.top());
			sc.pop();
		}
	}
	postOrder(0,n-1,0,n-1);
	for(int i=0;i<post.size();i++){
		printf("%d",post[i]);
		if(i!=post.size()-1)printf(" ");
	}
	return 0;
}