1020. Tree Traversals-优快云博客

本文链接：https://blog.youkuaiyun.com/caicai_zju/article/details/49280875

本文介绍了一种根据给定的后序遍历和中序遍历序列构建二叉树，并输出该树的层序遍历序列的方法。通过递归方式构建二叉树结构，再利用队列实现层序遍历，最终输出符合题目要求的层序遍历序列。

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1020. Tree Traversals (25)

时间限制

400 ms

内存限制

65536 kB

代码长度限制

16000 B

判题程序

Standard

作者

CHEN, Yue

Suppose that all the keys in a binary tree are distinct positive integers. Given the postorder and inorder traversal sequences, you are supposed to output the level order traversal sequence of the corresponding binary tree.

Input Specification:

Each input file contains one test case. For each case, the first line gives a positive integer N (<=30), the total number of nodes in the binary tree. The second line gives the postorder sequence and the third line gives the inorder sequence. All the numbers in a line are separated by a space.

Output Specification:

For each test case, print in one line the level order traversal sequence of the corresponding binary tree. All the numbers in a line must be separated by exactly one space, and there must be no extra space at the end of the line.

Sample Input:

7
2 3 1 5 7 6 4
1 2 3 4 5 6 7

Sample Output:

4 1 6 3 5 7 2

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

struct Node
{
	Node* lchild;
	Node* rchild;
	int data;
}Tree[32];

int T;
//分别保存后序和中序遍历序列 
int PostOrder[32], InOrder[32];

Node* create()
{
	Tree[T].lchild = NULL;
	Tree[T].rchild = NULL;
	return &Tree[T++];
}

queue<Node*> que;//层序遍历

Node *build(int s1, int e1, int s2, int e2)
{
	Node* root = create();
	root->data = PostOrder[e1];//后序遍历的最后一个节点是根节点 
	int rootIndex;
	for(int i = s2; i <= e2; i ++)
	{
		//找到根节点在中序遍历序列中的位置 
		if(InOrder[i] == PostOrder[e1])
		{
			rootIndex = i;
			break;
		}
	}
	
	//递归的建立左子树与右子树 
	if(rootIndex != s2)
		root->lchild = build(s1, s1+rootIndex-s2-1, s2, rootIndex-1);
	if(rootIndex != e2)
		root->rchild = build(s1+rootIndex-s2, e1-1, rootIndex+1, e2);
	
	return root; 
}

int main()
{
	int n;
	while(cin>>n)
	{
		for(int i = 0; i < n; i ++)
			cin>>PostOrder[i];
		for(int i = 0; i < n; i ++)
			cin>>InOrder[i];
		
		T = 0;
		Node* T = build(0, n-1, 0, n-1);
		que.push(T);
		
		while(!que.empty())
		{
			Node* tmp = que.front();
			if(tmp->lchild != NULL)
				que.push(tmp->lchild);
			if(tmp->rchild != NULL)
				que.push(tmp->rchild);
			que.pop();
			if(que.empty())
				cout<<tmp->data<<endl;
			else
				cout<<tmp->data<<" ";
		}
	}
	
	return 0;
}