(PAT 1020) Tree Traversals (给出后序中序求二叉树层序)

最新推荐文章于 2022-07-29 22:26:20 发布

alex1997222

最新推荐文章于 2022-07-29 22:26:20 发布

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分类专栏： ACM算法习题

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本文介绍了一种根据后序遍历和中序遍历序列重建二叉树，并输出层序遍历序列的方法。通过递归创建二叉树节点，然后使用队列进行层序遍历，最后输出结果。该方法适用于计算机科学与技术领域的数据结构和算法学习。

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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

解题思路:

思路见模板:https://blog.youkuaiyun.com/alex1997222/article/details/86600086

#include <iostream>
#include <algorithm>
#include <vector>
#include <queue>
using namespace std;
const int MAXN = 60;
struct Node {
	int data;
	Node* lchild;
	Node* rchild;
	Node(int _data) {
		data = _data;
		lchild = rchild = nullptr;
	}
};
int preArrays[MAXN], inArrays[MAXN], postArrays[MAXN];

Node* CreateTree(int postL, int postR, int inL, int inR) {
	if (postL > postR) return nullptr;
	int curPostRoot = postArrays[postR];
	Node* root = new Node(curPostRoot);
	int curInRoot = -1;
	int k;
	for (k = inL; k <= inR; ++k) {
		if (inArrays[k] == curPostRoot) {
			curInRoot = inArrays[k];
			break;
		}
	}
	if (curInRoot == -1) return nullptr;
	
	int numleft = k - inL;  
	root->lchild = CreateTree(postL,postL+numleft-1,inL,k-1);
	root->rchild = CreateTree(postL+numleft,postR-1,k+1,inR);
	return root;
}
void LayerTravas(Node* root) {
	queue<Node*> Bfs_queue;
	Bfs_queue.push(root);
	bool flag = true;
	while (!Bfs_queue.empty()) {
		Node* tempNode = Bfs_queue.front();
		if (flag) {
			cout << tempNode->data;
			flag = !flag;
		}
		else {
			cout <<" "<<tempNode->data;
		}
		Bfs_queue.pop();
		if (tempNode->lchild != nullptr) Bfs_queue.push(tempNode->lchild);
		if (tempNode->rchild != nullptr) Bfs_queue.push(tempNode->rchild);
	}
}

int main() {
	int N;
	cin >> N;
	for (int i = 0; i < N; ++i) {
		cin >> postArrays[i];
	}
	for (int i = 0; i < N; ++i) {
		cin >> inArrays[i];
	}
	Node* troot = CreateTree(0, N - 1,0,N-1);
	LayerTravas(troot);


	system("PAUSE");
	return 0;
}