标签：begin 中序 wing ica art 层序遍历 cin struct think
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.
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.
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.
8
12 11 20 17 1 15 8 5
12 20 17 11 15 8 5 1
1 11 5 8 17 12 20 15
首先通过后续遍历和中序遍历重构二叉树
然后在层序遍历的基础上，进行元素逆序
1 #include <iostream> 2 #include <vector> 3 #include <queue> 4 #include <algorithm> 5 using namespace std; 6 struct Node 7 { 8 int v; 9 Node *l, *r; 10 Node(int a = -1) :v(a), l(nullptr), r(nullptr) {} 11 }; 12 int n; 13 vector<int>inOrder, posOrder, zigOrder; 14 Node* reCreatTree(int inL, int inR, int posL, int posR) 15 { 16 if (posL > posR) 17 return nullptr; 18 Node* root = new Node(posOrder[posR]); 19 int k = inL; 20 while (k<=inR && inOrder[k] != posOrder[posR])++k;//找到根节点 21 int numL = k - inL;//左子树节点个数 22 root->l = reCreatTree(inL, k - 1, posL, posL + numL - 1); 23 root->r = reCreatTree(k + 1, inR, posL + numL, posR - 1); 24 return root; 25 } 26 void zigOrderTravel(Node* root) 27 { 28 if (root == nullptr) 29 return; 30 queue<Node*>q; 31 q.push(root); 32 zigOrder.push_back(root->v); 33 bool flag = false; 34 while (!q.empty()) 35 { 36 queue<Node*>temp; 37 vector<int>tt; 38 while (!q.empty()) 39 { 40 Node* p = q.front(); 41 q.pop(); 42 if (p->l != nullptr) 43 { 44 temp.push(p->l); 45 tt.push_back(p->l->v); 46 } 47 if (p->r != nullptr) 48 { 49 temp.push(p->r); 50 tt.push_back(p->r->v); 51 } 52 } 53 if (flag) 54 reverse(tt.begin(), tt.end()); 55 zigOrder.insert(zigOrder.end(), tt.begin(), tt.end()); 56 flag = !flag; 57 q = temp; 58 } 59 } 60 int main() 61 { 62 cin >> n; 63 inOrder.resize(n); 64 posOrder.resize(n); 65 for (int i = 0; i < n; ++i) 66 cin >> inOrder[i]; 67 for (int i = 0; i < n; ++i) 68 cin >> posOrder[i]; 69 Node* root = reCreatTree(0, n - 1, 0, n - 1); 70 zigOrderTravel(root); 71 for (int i = 0; i < zigOrder.size(); ++i) 72 cout << (i > 0 ? " " : "") << zigOrder[i]; 73 return 0; 74 }
PAT甲级——A1127 ZigZagging on a Tree【30】
标签：begin 中序 wing ica art 层序遍历 cin struct think
原文地址：https://www.cnblogs.com/zzw1024/p/11478533.html
