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
由后序和中序确定二叉树。
然后就随便就能输出来了。
1 #include <bits/stdc++.h>
2 using namespace std;
3 int n;
4 struct Node
5 {
6 int val;
7 Node *left, *right;
8 }*root;
9 int inorder[50], post[50];
10 struct mat{
11 int val, ord, pri;
12 };
13 bool cmp(mat &a, mat &b){
14 return a.ord < b.ord;
15 }
16 vector<mat> vt;
17 vector<int> v;
18 Node *create(int x, int ll, int rr){
19 Node *node = (Node*)malloc(sizeof(Node));
20 node->val = post[x];
21 node->left = NULL, node->right = NULL;
22 int mid = 0;
23 for(int i = ll; i <= rr; i++){
24 if(inorder[i] == post[x]){
25 mid = i;
26 break;
27 }
28 }
29 if(ll < mid){
30 node->left = create(x-rr+mid-1, ll, mid-1);
31 }
32 if(rr > mid){
33 node->right = create(x-1, mid+1, rr);
34 }
35 return node;
36 }
37
38 void output(Node *root, int x, int y){
39 if(root != NULL){
40 output(root->left, x<<1, y+1);
41 vt.push_back({root->val, x, y});
42 output(root->right, (x<<1)+1, y+1);
43 }
44 }
45
46 int main(){
47 cin >> n;
48 for(int i = 0; i < n; i++){
49 cin >> inorder[i];
50 }
51 for(int i = 0; i < n; i++){
52 cin >> post[i];
53 }
54 root = create(n-1, 0, n-1);
55 output(root, 1, 1);
56 sort(vt.begin(), vt.end(), cmp);
57 for(int i = 0; i < vt.size(); i++){
58 if(vt[i].pri%2 == 0){
59 if(vt[i].pri != vt[i-1].pri){
60 int j = i-1;
61 while(vt[j].pri == vt[i-1].pri){
62 printf("%d ", vt[j].val);
63 j--;
64 }
65 }
66 printf("%d%c",vt[i].val, i != n-1?' ':'\n');
67 }
68 }
69 int j = vt.size()-1;
70 if(vt[j].pri%2 == 1){
71 while(vt[j].pri == vt[n-1].pri){
72 v.push_back(vt[j].val);
73 j--;
74 }
75 for(int i = 0; i < v.size(); i++){
76 printf("%d%c", v[i], i == v.size()-1?'\n':' ');
77 }
78 }
79 return 0;
80 }