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

分手后的思念是犯贱 2022-04-23 01:04 84阅读 0赞

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.csdn.net/alex1997222/article/details/86600086][https_blog.csdn.net_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;
    }

[https_blog.csdn.net_alex1997222_article_details_86600086]: https://blog.csdn.net/alex1997222/article/details/86600086