PAT(A) 1127. ZigZagging on a Tree (30)
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)
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
- 建树
- 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。
- 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。
- 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<int> 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<struct Node *> q;stack<struct Node *> 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;}elsecout << " " << 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");
}
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