PAT (Advanced Level) 1127. ZigZagging on a Tree (30)
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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:812 11 20 17 1 15 8 512 20 17 11 15 8 5 1Sample Output:
1 11 5 8 17 12 20 15
这道题今晚终于AC了，相比前面两道看不懂的25分的题目，这道题就是传说中的，看得懂但是不会做的类型。。。。
所以我在想，除非我做过类似的题目，不然我是不可能在那么短的考试时间里面写出来并且AC的。。。
这道题，暴力算确实可以，而且我也百度了，看见有些大佬是用双向链表，有的说用vector<vector<int> > a;来解决，鉴于我目前对vector掌握的还不是很熟练，所以我想了一晚上，终于想到用栈结合队列，然后把他们一起放进BFS里面来输出~
想到用栈，是因为，题目要求是反方向的Z字型输出，然后刚开始也是考虑在create树的时候，记录他们每个点（node）的层值，
然后判断奇偶性来输出，但是后来觉得为什么不用BFS输出呢？
比如只要在奇数层的时候（从1开始），判断当前队列是否还有数字，如果还有就说明是上一层偶数层的数字还没有完全输出，此时要耐心“等候”，也就是要将他们的左右孩子们读入栈里面，直到判断到队列已经空啦，队列要求要加入新的数字的时候，我们再将栈里面的数值“一口气”全部输出~~在偶数层也是类似的道理，但是要注意在将奇数层的数字加入队列的时候，要先加入右孩子，偶数层要先加入左孩子~~（这是因为反方向Z字决定的性质）
下面就是代码~~
#include<cstdio>#include<queue>#include<stack>#include<string.h>#include<algorithm>#include<math.h>using namespace std;const int maxn = 35;int n;int f = 0;int xx = 0;int pos[maxn], in[maxn];struct node {int data;int layer;node* lchild;node* rchild;};node* create(int posL, int posR, int inL, int inR, int ll) {node* root = new node;if (posL > posR) {return NULL;}root->data = pos[posR];root->layer = ll;int k;for (k = inL; k <= inR; k++) {if (in[k] == pos[posR]) {break;}}int num = k - inL;root->lchild = create(posL, posL + num - 1, inL, k - 1, ll + 1);root->rchild = create(posL + num, posR - 1, k + 1, inR, ll + 1);return root;}void BFS(node* root) {stack<node*> sss;queue<node*> q;q.push(root);while (!q.empty()) {node* top = q.front();q.pop();if (f != n - 1) {printf("%d ", top->data);f++;}else {printf("%d\n", top->data);break;}if ((top->layer) % 2 != 0) {if (top->rchild != NULL) { sss.push(top->rchild); }if (top->lchild != NULL) { sss.push(top->lchild); }if (q.empty()) {while (!sss.empty()) {node* temp = sss.top();sss.pop();q.push(temp);}}}else {if (top->lchild != NULL) { sss.push(top->lchild); }if (top->rchild != NULL) { sss.push(top->rchild); }if (q.empty()) {while (!sss.empty()) {node* temp = sss.top();sss.pop();q.push(temp);}}}}}int main() {scanf("%d", &n);for (int i = 0; i < n; i++) {scanf("%d", &in[i]);}for (int i = 0; i < n; i++) {scanf("%d", &pos[i]);}node* root = create(0, n - 1, 0, n - 1, 1);BFS(root);return 0;}0 0
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