(PAT 1127) ZigZagging on a Tree (二叉树建立+层序遍历)

ゝ一世哀愁。 2022-04-23 09:18 163阅读 0赞

Suppose that all the keys in a binary tree are distinct positive integers. A unique binary tree can be determined by a given pair of postorder and inorder traversal sequences. And it is a simple standard routine to print the numbers in level-order. However, if you think the problem is too simple, then you are too naive. This time you are supposed to print the numbers in "zigzagging order" -- that is, starting from the root, print the numbers level-by-level, alternating between left to right and right to left. For example, for the following tree you must output: 1 11 5 8 17 12 20 15.

![zigzag.jpg][]

### 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 inorder sequence and the third line gives the postorder sequence. All the numbers in a line are separated by a space.

### Output Specification: ###

For each test case, print the zigzagging sequence of the tree in a line. 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: ###

8
    12 11 20 17 1 15 8 5
    12 20 17 11 15 8 5 1

### Sample Output: ###

1 11 5 8 17 12 20 15

解题思路:

先建树,建树过程参见模板

然后层序遍历,保留层号,把每一层的所有元素保留到对应的vector容器中

输出时,层号为奇数,逆序输出,层号为偶数,正序输出

#include <iostream>
    #include <algorithm>
    #include <queue>
    #include <vector>
    using namespace std;
    const int MANX = 35;
    int postSquence[MANX], inSquence[MANX];
    vector<int> nums[MANX];
    struct Bnode {
    	int data;
    	Bnode* lchild;
    	Bnode* rchild;
    	Bnode(int _data) {
    		data = _data;
    		lchild = nullptr;
    		rchild = nullptr;
    	}
    };
    
    Bnode* BbuildTree(int postL, int postR, int inL, int inR) {
    	if (postL > postR) return nullptr;
    	int curPostRoot = postSquence[postR];
    	//确定根的值
    	Bnode* root = new Bnode(curPostRoot);
    	int curInRoot = -1;
    	int k;
    	//在中序中查找根,确定根的位置
    	for (k = inL; k <= inR; ++k) {
    		if (inSquence[k] == curPostRoot) {
    			curInRoot = inSquence[k];
    			break;
    		}
    	}
    	if (curInRoot == -1) return nullptr;
    	//后序遍历
    	int numleft = k - inL;
    	root->lchild = BbuildTree(postL, postL + numleft - 1, inL, k - 1);
    	root->rchild = BbuildTree(postL + numleft, postR - 1, k + 1, inR);
    
    	return root;
    }
    
    int order = 0;
    void LayerPrint(Bnode* root) {
    	queue<Bnode*> bfs_queue;
    	queue<int> bfs_orderQueue;
    	bfs_queue.push(root);
    	bfs_orderQueue.push(1);
    	while (!bfs_queue.empty()) {
    		Bnode* currentNode = bfs_queue.front();
    		order = bfs_orderQueue.front();
    		nums[order].push_back(currentNode->data);
    		bfs_queue.pop();
    		bfs_orderQueue.pop();
    		if (currentNode->lchild != nullptr) {
    			bfs_queue.push(currentNode->lchild);
    			bfs_orderQueue.push(order + 1);
    		}
    		if (currentNode->rchild != nullptr) {
    			bfs_queue.push(currentNode->rchild);
    			bfs_orderQueue.push(order + 1);
    		}
    	}
    }
    int main() {
    	int N;
    	cin >> N;
    	for (int i = 0; i < N; ++i) {
    		cin >> inSquence[i];
    	}
    	for (int i = 0; i < N; ++i) {
    		cin >> postSquence[i];
    	}
    
    	Bnode* root = BbuildTree(0, N - 1, 0, N - 1);
    	LayerPrint(root);
    	for (int i = 1; i <= order; ++i) {
    		if (i % 2 != 0) {
    			reverse(nums[i].begin(), nums[i].end());
    			for (auto itr = nums[i].begin(); itr != nums[i].end(); ++itr) {
    				cout << *itr;
    				if (i != order || itr != nums[i].end() - 1) cout << " ";
    			}
    		}
    		else {
    			for (auto itr = nums[i].begin(); itr != nums[i].end(); ++itr) {
    				cout << *itr;
    				if (i != order || itr != nums[i].end() - 1) cout << " ";
    			}
    		}
    	}
    
    	system("PAUSE");
    	return 0;
    }

[zigzag.jpg]: /images/20220321/dec3df6d3861448492509cfd25e13c58.png