PAT A1127 ZigZagging on a Tree (30point(s))

Scroll Down

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.

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

解答

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

vector<int> in, post;

struct node {
	int key, depth;
	node *lchild, *rchild;
};

node *createTree(int inl, int inr, int postl, int postr) {
	if (inl > inr) return NULL;
	int i;
	node* root = new node();
	root->key = post[postr];
	for (i = inl; i <= inr; i++) 
		if (in[i] == root->key) break;
	int leftNum = i - inl;
	root->lchild = createTree(inl, i - 1, postl, postl + leftNum - 1);
	root->rchild = createTree(i + 1, inr, postl + leftNum, postr - 1);
	return root;
}
void bfs(node* root) {
	queue<node*> q;
	vector<vector<node*>> v(1);
	q.push(root);
	int currentCnt = 1, nextCnt = 0, i = 0, depth = 0;
	while (!q.empty()) {
		node* tmp = q.front();
		v[depth].push_back(tmp);
		q.pop();
		i++;
		if (tmp->lchild) {
			q.push(tmp->lchild);
			nextCnt++;
		}
		if (tmp->rchild) {
			q.push(tmp->rchild);
			nextCnt++;
		}
		if (i == currentCnt) {
			depth++;
			currentCnt = nextCnt, i = 0, nextCnt = 0;
			v.resize(v.size() + 1);
		}
	}
	printf("%d", v[0][0]->key);
	for (int i = 1; i < v.size(); i++) {
		if (i % 2) 
			for (auto it = v[i].begin(); it != v[i].end(); it++)
				printf(" %d", (*it)->key);
		else
			for (auto it = v[i].rbegin(); it != v[i].rend(); it++)
				printf(" %d", (*it)->key);
	}
}

int main()
{
	int n;
	cin >> n;
	in.resize(n), post.resize(n);
	for (int i = 0; i < n; i++) cin >> in[i];
	for (int i = 0; i < n; i++) cin >> post[i];
	node *tree = createTree(0, n - 1, 0, n - 1);
	bfs(tree);
	return 0;
}