PAT 1086 Tree Traversals Again [stack] [先序+中序建树] [后序遍历]

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

------------------------------------这是题目和解题的分割线------------------------------------

和PAT 1020 Tree Traversals差不多。不过这道题的先序序列给的十分隐晦，push进去的是先序，pop出来的是中序。

#include<cstdio>
#include<string>
#include<iostream>
#include<stack>

using namespace std;

struct node
{
	int data;
	node *left,*right;
};

int n,in[35],pre[35],d = 0;

//后序遍历 
void postOrder(node* T)
{	
	if(T==NULL) return;
	postOrder(T->left);
	postOrder(T->right);
	printf("%d",T->data);
	if(d!=n-1) printf(" "); //末尾没有空格 
	d++;
}

//已知先序和中序求后序 
node* createT(int preL,int preR,int inL,int inR)
{
	if(preL>preR) return NULL; //先序序序列长度小于等于0 
	node* root = new node;
	root->data = pre[preL]; //根节点是先序序列的第一个 
	int k;
	for(k=inL;k<=inR;k++) //中序序列找到根结点 
		if(in[k]==root->data) break;
	int leftNum = k - inL; //左子树的结点个数
	//左子树的先序序列和中序序列 
	root->left = createT(preL+1,preL+leftNum,inL,k-1);
	//右子树的先序序列和中序序列 
	root->right = createT(preL+leftNum+1,preR,k+1,inR);
	return root;
}

int main()
{
	int i,x,cnt = 0,num = 0;
	stack<int> st;
	string s;
	cin>>n;
	int m = 2*n; //入栈出栈的操作一共2*n次 
	getchar();
	while(m--)
	{
		cin>>s;
		//如果是push操作 
		if(s=="Push")
		{
			cin>>x; //读取入栈元素 
			st.push(x);
			pre[cnt++] = x; //先序序列记录 
		}		
		else
		{
			in[num++] = st.top(); //出栈元素 
			st.pop(); //中序序列记录 
		}	 
	}
	node* T = createT(0,n-1,0,n-1);
	postOrder(T);
	return 0;
}