PAT (Advanced Level) Practice 1086 Tree Traversals Again (25分) （寻找先序遍历（+建树+）后序遍历）

1.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.

Figure 1

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

2.题目分析

竖着看所有给出的数字，你会发现这些数字的序列正好是树的先序遍历，这样用栈模拟得到中序遍历，先序+中序（建树）求后序

3.代码

 不建树：

#include<iostream>
#include<stack>
#include<string>
#include<cstring>
#include<vector>
using namespace std;

int n;
bool space = false;
void out(vector<int>in, vector<int>pre, int root, int ins, int ine)
{
	if (ins > ine)return ;
	int i = ins;
	while (in[i] != pre[root])i++;
	if (i <= ine)
	{
	out (in, pre, root + 1, ins, i - 1);
	out(in, pre, root + i + 1 - ins, i + 1, ine);
	printf("%s%d", space == false ? "" : " ", pre[root]); space = true;
	}
}
int main()
{
	int a;
	string temp;
	scanf("%d", &n);
	vector<int>in, pre;
	stack<int>list;
	while (1)
	{
		cin >> temp;
		if (temp == "Push")
		{
			cin >> a;
			pre.push_back(a);
			list.push(a);
		}
		else
		{
			in.push_back(list.top());
			list.pop();
		}
		if (in.size() == n&&list.size() == 0)break;
	}
	out(in, pre, 0, 0, n - 1);
}

建树：

#include<iostream>
#include<stack>
#include<string>
#include<cstring>
#include<vector>
using namespace std;
typedef struct node *tree;
struct node
{
	int data;
	tree left;
	tree right;
};
int n;
bool space = false;
void postordertraverse(tree t)
{
	if (!t)return;
	postordertraverse(t->left);
	postordertraverse(t->right);
	printf("%s%d", space == false ? "" : " ", t->data); space = true;
}
tree create(tree t,vector<int>in,vector<int>pre,int root,int ins,int ine)
{
	if (ins > ine)return NULL;
	if (!t) 
	{
		t = (tree)malloc(sizeof(struct node));
		t->left = t->right = NULL;
		t->data = pre[root];
	}
	int i = ins;
	while (in[i] != pre[root])i++;
	if (i <= ine)
	{
		t->left=create(t->left,in, pre, root + 1, ins, i - 1);
		t->right = create(t->right, in, pre, root + i +1 - ins, i + 1, ine);
	}
	return t;
}
int main()
{
	int a;
	string temp;
	scanf("%d", &n);
	vector<int>in,pre;
	stack<int>list;
	while (1)
	{
		cin >> temp;
		if (temp == "Push")
		{
			cin >> a;
			pre.push_back(a);
			list.push(a);
		}
		else
		{
			in.push_back(list.top());
			list.pop();
		}
		if (in.size() == n&&list.size() == 0)break;
	}
	tree t = NULL;
	t=create(t, in, pre, 0, 0, n - 1);
	postordertraverse(t);
}