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
1 #include<cstdio> 2 #include<iostream> 3 #include<algorithm> 4 #include<queue> 5 #include<vector> 6 using namespace std; 7 typedef struct NODE{ 8 struct NODE* lchild, *rchild; 9 int data; 10 int lev; 11 }node; 12 int post[31], in[31], N; 13 node* create(int inL, int inR, int postL, int postR, int level){ 14 if(inL > inR){ 15 return NULL; 16 } 17 node* root = new node; 18 root->data = post[postR]; 19 root->lev = level; 20 int mid; 21 for(mid = inL; mid <= inR; mid++){ 22 if(in[mid] == root->data) 23 break; 24 } 25 int len = mid - inL; 26 root->lchild = create(inL, mid - 1, postL, postL + len - 1, level + 1); 27 root->rchild = create(mid + 1, inR, postL + len, postR - 1, level + 1); 28 return root; 29 } 30 void levelOrder(node* root){ 31 queue<node*> Q; 32 Q.push(root); 33 int nowLev = root->lev; 34 int reverTag = 1; 35 int cnt = 0; 36 vector<node*> vec; 37 while(Q.empty() == false){ 38 node* temp = Q.front(); 39 Q.pop(); 40 if(temp->lev == nowLev){ 41 vec.push_back(temp); 42 }else{ 43 nowLev = temp->lev; 44 if(reverTag == 1){ 45 reverTag = 0; 46 for(int i = vec.size() - 1; i>= 0; i--){ 47 cnt++; 48 printf("%d ", vec[i]->data); 49 } 50 }else{ 51 reverTag = 1; 52 for(int i = 0; i < vec.size(); i++){ 53 cnt++; 54 printf("%d ", vec[i]->data); 55 } 56 } 57 vec.clear(); 58 vec.push_back(temp); 59 } 60 if(temp->lchild != NULL) 61 Q.push(temp->lchild); 62 if(temp->rchild != NULL) 63 Q.push(temp->rchild); 64 } 65 if(reverTag == 0){ 66 for(int i = 0; i < vec.size(); i++){ 67 cnt++; 68 if(cnt == N) 69 printf("%d", vec[i]->data); 70 else printf("%d ", vec[i]->data); 71 } 72 }else{ 73 for(int i = vec.size() - 1; i >= 0; i--){ 74 cnt++; 75 if(cnt == N) 76 printf("%d", vec[i]->data); 77 else printf("%d ", vec[i]->data); 78 } 79 } 80 } 81 int main(){ 82 scanf("%d", &N); 83 for(int i = 1; i <= N; i++){ 84 scanf("%d", &in[i]); 85 } 86 for(int i = 1; i <= N; i++){ 87 scanf("%d",&post[i]); 88 } 89 node* root = create(1, N, 1, N, 1); 90 levelOrder(root); 91 cin >> N; 92 return 0; 93 }
总结：
1、题意：中序后序建树，再用层序输出，输出时一行逆序一行正序。
2、可以在建树时顺便将层次信息也加入节点。在层序遍历时使用一个vector，在当前层数相同时，仅仅把本该访问的节点存入vector，当层数发生改变时，输出vector内所有元素（设置一个计数器，如果上次是正序输出，则本次逆序输出）； 或者正常层序遍历，再多用一个栈+一个队列，当需要正序输出该层时，节点入队列，需要逆序则节点入栈，层数改变时输出栈中或队列中全部节点，并清空。