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

### 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

### 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。

### 代码

  1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 110 /* * 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"); }