1020 Tree Traversals （25 分） 题解

1020 Tree Traversals （25 分）

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

这题思路很简单，没有那没多花里胡哨的东西，但是如果没有接触过的话短时间可能很难写出来。首先就是根据中序和后序序列递归构建二叉树，再对这个二叉树进行层次遍历，实际上就是对这个二叉树进行广度优先遍历，这个地方需要借助队列来实现。

代码如下:

#include <iostream>
#include<vector>
#include<queue>

using namespace std;

struct Node
{
	int key;
	Node *leftNode;
	Node *rightNode;
};

Node * createTree(vector<int> postOrder, vector<int> inOrder)  //根据后根序列和中根序列创建二叉树
{
	if (postOrder.size() == 0)
	{
		return NULL;
	}
	int key = postOrder[postOrder.size() - 1];
	int index = -1;
	for (int i = 0; i < inOrder.size(); i++)
	{
		if (inOrder[i] == key)
		{
			index = i;
			break;
		}
	}
	vector<int> leftPost, leftIn, rightPost, rightIn;
	for (int i = 0; i < index; i++)
	{
		leftPost.push_back(postOrder[i]);
		leftIn.push_back(inOrder[i]);
	}
	for (int i = index; i < inOrder.size()-1; i++)
	{
		rightIn.push_back(inOrder[i+1]);
		rightPost.push_back(postOrder[i]);
	}
	Node *root=new Node;
	root->key = key;
	root->leftNode = NULL;
	root->rightNode = NULL;
	root->leftNode = createTree(leftPost, leftIn);
	root->rightNode = createTree(rightPost, rightIn);
	return root;
}
void BFS(Node *root)  //BFS对树进行广度优先遍历
{
	queue<Node> que;
	que.push(*root);
	while (que.empty() != 1)
	{
		Node node=que.front();
		if (que.size() == 1 && node.leftNode == NULL && node.rightNode == NULL)
		{
			cout << node.key;
		}
		else
		{
			cout << node.key << " ";
		}
		que.pop();
		if(node.leftNode!=NULL)
		{
			que.push(*node.leftNode);
		}
		if (node.rightNode != NULL)
		{
			que.push(*node.rightNode);
		}
	}
}

int main()
{
	int N;
	cin >> N;
	vector<int> postOrder(N), inOrder(N);
	for (int i = 0; i < N; i++)
	{
		cin >> postOrder[i];
	}
	for (int i = 0; i < N; i++)
	{
		cin >> inOrder[i];
	}
	Node *root = createTree(postOrder, inOrder);  //创建二叉树
	BFS(root);                                    //对树广度优先遍历
	system("pause");
	return 0;
}