1127 ZigZagging on a Tree (30分)

最新推荐文章于 2020-10-29 15:19:46 发布

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本文介绍了一种解决二叉树锯齿形层次遍历问题的方法，通过构建二叉树并进行广度优先搜索，实现从根节点开始，按层次交替从左到右和从右到左打印节点值。输入包括节点总数、中序和后序遍历序列，输出为锯齿形层次遍历序列。

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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.
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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.

本菜鸡只会用建树的方法做：建树->求每一层的节点顺序->分层从前往后或从后往前输出

#include<iostream>
#include<vector>
#include<queue>
using namespace std;
typedef struct node {
	int data, level;
	node *l, *r;
}*tree;
int n, post[40], in[40], depth = 0;
vector<int>res[40];
tree createT(int inS, int inE, int postS, int postE) {
	if (inS > inE)return NULL;
	tree t = new node();
	t->data = post[postE];
	int i;
	for (i = inS; i <= inE; i++) {
		if (in[i] == post[postE])break;
	}
	t->l = createT(inS, i - 1, postS, postS + i - 1 - inS);
	t->r = createT(i + 1, inE, postE + i - inE, postE - 1);
	return t;
}
void bfs(tree root) {
	queue<tree>q;
	q.push(root);
	root->level = 1;
	while (!q.empty()) {
		auto temp = q.front();
		if (temp->level > depth) {
			depth = temp->level;
		}
		q.pop();
		res[temp->level].push_back(temp->data);
		if (temp->l != NULL) {
			temp->l->level = temp->level + 1;
			q.push(temp->l);
		}
		if (temp->r != NULL) {
			temp->r->level = temp->level + 1;
			q.push(temp->r);
		}
	}
}
int main() {
	cin >> n;
	for (int i = 0; i < n; i++)cin >> in[i];
	for (int i = 0; i < n; i++)cin >> post[i];
	tree t = createT(0, n - 1, 0, n - 1);
	bfs(t);
	cout << res[1][0];
	for (int i = 2; i <= depth; i++) {
		if (i % 2 == 0) {
			for (int j = 0; j < res[i].size(); j++) {
				cout << " " << res[i][j];
			}
		}
		else if (i % 2 == 1) {
			for (int j = res[i].size() - 1; j >= 0; j--) {
				cout << " " << res[i][j];
			}
		}
	}
	return 0;
}