1127 ZigZagging on a Tree （30 分)
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：给定二叉树的中序和后序遍历的数列，蛇形逐层打印二叉树。
思路：中序+后序，建立二叉树；（建树的具体实现方法和思路见我之前的文章：后序+中序，先序+中序，建立二叉树）建树的过程顺便记录每个节点所处的深度以及树的总深度；接着层序遍历，将每个节点的所属层数对应二维数组ans的行，依次push back~~最后将ans蛇形输出，奇数层（我是从0开始计数的）从左往右，偶数层反之。
1 #include<iostream> 2 #include<vector> 3 #include<unordered_map> 4 #include<queue> 5 using namespace std; 6 typedef struct node* BT; 7 struct node { 8 int value, level = 0; 9 BT left = NULL, right = NULL; 10 }; 11 int treeDepth = 0;//树的深度 12 unordered_map<int, int> mp;//映射元素在中序数组里的位置 13 vector<int> Inorder, Post, ans[30]; 14 BT buildTree(int InLeft, int InRight, int PostIndex, int depth); 15 void levelOrder(BT t); 16 void printLtoR(vector<int> &v); 17 void printRtoL(vector<int> &v); 18 int main() 19 { 20 int N, i; 21 scanf("%d", &N); 22 Inorder.resize(N); 23 Post.resize(N); 24 for (i = 0; i < N; i++) { 25 scanf("%d", &Inorder[i]); 26 mp[Inorder[i]] = i; 27 } 28 for (i = 0; i < N; i++) 29 scanf("%d", &Post[i]); 30 BT tree = NULL; 31 tree = buildTree(0, N - 1, N - 1, 0); 32 levelOrder(tree); 33 for (i = 0; i <= treeDepth; i++) { 34 if (i % 2 == 0) printRtoL(ans[i]); 35 else printLtoR(ans[i]); 36 if (i <= treeDepth - 1) 37 printf(" "); 38 } 39 return 0; 40 } 41 void printRtoL(vector<int> &v) 42 { 43 int n = v.size(); 44 for (int i = n - 1; i >= 0; i--) { 45 printf("%d", v[i]); 46 if (i > 0) 47 printf(" "); 48 } 49 } 50 void printLtoR(vector<int> &v) 51 { 52 int n = v.size(); 53 for (int i = 0; i < n; i++) { 54 printf("%d", v[i]); 55 if (i < n - 1) 56 printf(" "); 57 } 58 } 59 void levelOrder(BT t) 60 { 61 queue<BT> Q; 62 Q.push(t); 63 while (!Q.empty()) { 64 BT tmp = Q.front(); 65 ans[tmp->level].push_back(tmp->value); 66 Q.pop(); 67 if (tmp->left) 68 Q.push(tmp->left); 69 if (tmp->right) 70 Q.push(tmp->right); 71 } 72 } 73 BT buildTree(int InLeft, int InRight, int PostIndex, int depth) 74 { 75 if (InLeft > InRight) return NULL; 76 if (treeDepth < depth) treeDepth = depth; 77 BT t = new node(); 78 t->level = depth; 79 t->value = Post[PostIndex]; 80 t->right = buildTree(mp[Post[PostIndex]] + 1, InRight, PostIndex - 1, depth + 1); 81 int rtreeNum = InRight - mp[Post[PostIndex]];//当前节点的右子树节点个数 82 t->left = buildTree(InLeft, mp[Post[PostIndex]] - 1, PostIndex - rtreeNum - 1, depth + 1); 83 return t; 84 }