PAT A 1086. Tree Traversals Again (25)-优快云博客

本文介绍了一种利用栈实现二叉树中序非递归遍历的方法，并通过一组特定的栈操作（如压栈和弹栈）来唯一确定一棵二叉树。此外，还给出了构造该二叉树的算法并实现了其后序遍历。

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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

先构造树，然后后续遍历

代码：

#include <iostream>
#include <string>
using namespace std;

struct node	//点
{
	int key;
	node *left,*right;
	node(int k):key(k),left(NULL),right(NULL){};
};

node* buildtree();	//构造树
void postorder(node* p,node* root);	//后续遍历
int n,num_t=0;	//n，已经执行的指令数
string ins;	//指令
int key;	//输入的值

int main()
{
	node *root=NULL;
	cin>>n;

	root=buildtree();
	postorder(root,root);

	return 0;
}

node* buildtree()	//根据堆栈信息构造树，递归
{
	node* r=NULL;
	if(num_t<2*n)
	{
		cin>>ins;
		if(ins=="Push")	//压入，依次处理左右子树
		{
			num_t++;
			cin>>key;
			r=new node(key);
			r->left=buildtree();
			r->right=buildtree();
		}
		else	//弹出
		{
			num_t++;
			return NULL;
		}
	}
	return r;
}

void postorder(node* p,node* root)	//后续遍历
{
	if(p->left!=NULL)
		postorder(p->left,root);
	if(p->right!=NULL)
		postorder(p->right,root);
	cout<<p->key;
	if(p!=root)
		cout<<" ";
}