PAT(A) 1127. ZigZagging on a Tree (30)

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title: PAT(A) 1127. ZigZagging on a Tree (30)
tags: PAT
categories: PAT甲级
date: 2018-03-13 14:21:14
description:
updated:
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原题目：

原题链接：https://www.patest.cn/contests/pat-a-practise/1127

1127. ZigZagging on a Tree (30)

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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.

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

题目大意

根据中序遍历和后序遍历建立一棵二叉树，然后拉链式输出，可以看到的是，奇数层由左向右输出，偶数层由右向左。

解题报告

1. 建树
2. 找出后序中最后一个元素在中序的位置，由这个位置将中序和后序一分为二，即两个序列左右部分元素个数相同。
3. 以后序最后一个元素为根，将分开的两部分依次按同样方法建立左右子树。
4. 输出，区分每一层的元素个数，判断向左还是向右，存入栈中，当一层进行完，再由栈进入队列中，保证顺序正确。

对于输出，也可以，先层序遍历，把每一层的元素分别保存下来，再根据是奇数层还是偶数层决定顺序输出还是倒序输出。

代码

/*
* Problem: 1127. ZigZagging on a Tree (30)
* Author: HQ
* Time: 2018-03-13
* State: Done
* Memo: 队列 栈 建树
*/
#include "iostream"
#include "queue"
#include "stack"
#include "vector"
using namespace std;

struct Node {
	int data = 0;
	struct Node * left = NULL;
	struct Node * right = NULL;
};

int N;
vector inorder, postorder;

int findPos(int x,int s,int l) {
	for (int i = 0; i < l; i++) {
		if (inorder[s + i] == x)
			return s + i;
	}
	return -1;
}

struct Node * makeTree(int s1, int s2, int n) {
	struct Node *root = NULL;
	int x = postorder[s2 + n - 1];
	int pos = findPos(x, s1, n);
	if (pos != -1) {
		root = new struct Node;
		root->data = x;
		if (pos - s1 > 0)
			root->left = makeTree(s1, s2, pos - s1);
		if (s1 + n - pos - 1> 0)
			root->right = makeTree(pos + 1, s2 + pos - s1, s1 + n - pos - 1);
	}
	return root;
}

void levelOrder(struct Node * root) {
	queue q;
	stack s;
	struct Node * temp;
	q.push(root);
	bool first = true;
	bool left = false;
	int cnt = 0, num = 1, tempnum = 0;
	while (!q.empty()) {
		temp = q.front();
		q.pop();
		if (first) {
			cout << temp->data;
			first = false;
		}
		else
			cout << " " << temp->data;
		if (left) {
			if (temp->left != NULL) {
				s.push(temp->left);
				tempnum++;
			}
			if (temp->right != NULL) {
				s.push(temp->right);
				tempnum++;
			}
		}
		else {
			if (temp->right != NULL) {
				s.push(temp->right);
				tempnum++;
			}
			if (temp->left != NULL) {
				s.push(temp->left);
				tempnum++;
			}
		}
		cnt++;
		if (cnt == num) {
			num = tempnum;
			tempnum = 0;
			cnt = 0;
			left = !left;
			while (!s.empty()) {
				q.push(s.top());
				s.pop();
			}
		}
	}
}

int main() {
	cin >> N;
	inorder.resize(N);
	postorder.resize(N);
	for (int i = 0; i < N; i++)
		cin >> inorder[i];
	for (int i = 0; i < N; i++)
		cin >> postorder[i];
	struct Node * root;
	root = makeTree(0, 0, N);
	levelOrder(root);
	cout << endl;
	system("pause");
}