1086 Tree Traversals Again (25)

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

思路 ：

用栈的形式给出一棵二叉树的建立的顺序，求这棵二叉树的后序遍历。

Push的次序即是先序序列，pop的次序即是中序序列

则根据先序序列和中序序列构建一棵树。

C++：

#include "stack"
#include "algorithm"
#include "cstring"
#include "cstdio"
using namespace std;
const int maxn=50;
struct node
{
	int data;
	node* lchild,* rchild;
};
int pre[maxn],in[maxn],post[maxn];
int n;
node* create(int prel,int prer,int inl,int inr){
	if (prel>prer)//先序序列长度小于0
	{
		return NULL;
	}
	node* root=new node;
	root->data=pre[prel];
	int k;
	//在左子树中寻找根结点
	for (k=inl;k<=inr;k++)
	{
		if (in[k]==pre[prel])
		{
			break;
		}
	}
	int numleft=k-inl;//左子树的个数
	//返回左子树的根结点地址，赋值给root的左指针
	root->lchild=create(prel+1,prel+numleft,inl,k-1);
	//返回右子树的根结点地址，赋值给root的右指针
	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;
}