1086. Tree Traversals Again (25)

最新推荐文章于 2021-02-26 14:07:35 发布

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#c++ #二叉树

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本文详细阐述了如何通过给定的二叉树后序遍历序列，构建原始二叉树并输出其后序遍历序列。采用非递归方式利用栈实现遍历过程，提供了解决此类问题的算法流程与步骤。

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1086. Tree Traversals Again (25)

时间限制

200 ms

内存限制

65536 kB

代码长度限制

16000 B

判题程序

Standard

作者

CHEN, Yue

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

用笨笨的方法，先建立对应的二叉树，在后根遍历输出。

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

int N,count=0,count1=0;
struct Node{
	int data;
	Node *left,*right;
};

Node *BuildTree()		//建立对应的二叉树
{
	Node *r=new Node;
	r->left=NULL;
	r->right=NULL;
	int num;
	string str;
	if(count<2*N){
		cin>>str;
		if(str=="Push"){
			cin>>num;
			r->data=num;
			count++;
		}
		else if(str=="Pop"){
			count++;
			return NULL;
		}	
		r->left=BuildTree();
		r->right=BuildTree();
	}
	else					//末尾节点处理
		return NULL;
	return r;
}

void Printf(Node *T)		//输出
{
	if(T){
		Printf(T->left);
		Printf(T->right);
		cout<<T->data;
		count1++;
		if(count1<N)		//末尾不留空
			cout<<" ";
	}
}

int main()
{	
	cin>>N;
	Node *Tree=BuildTree();
	Printf(Tree);
	cout<<endl;
	return 0;
}