1086. Tree Traversals Again (25)

An inorder binary tree traversal can be implemented in a non-recursive way with a stack. For example, suppose that when a 6-node binary tree (with the keys numbered from 1 to 6) is traversed, the stack operations are: push(1); push(2); push(3); pop(); pop();push(4); pop(); pop(); push(5); push(6); pop(); pop(). Then a unique binary tree (shown in Figure 1) can be generated from this sequence of operations. Your task is to give the postorder traversal sequence of this tree.Figure 1Input Specification:Each input file contains one test case. For each case, the first line contains a positive integer N (<=30) which is the total number of nodes in a tree (and hence the nodes are numbered from 1 to N). Then 2N lines follow, each describes a stack operation inthe format: "Push X" where X is the index of the node being pushed onto the stack; or "Pop" meaning to pop one node from the stack.Output Specification:For each test case, print the postorder traversal sequence of the corresponding tree in one line. A solution is guaranteed to exist. All the numbers must be separated by exactly one space, and there must be no extra space at the end of the line.Sample Input:

6
Push 1
Push 2
Push 3
Pop
Pop
Push 4
Pop
Pop
Push 5
Push 6
Pop
Pop

Sample Output:

3 4 2 6 5 1

根据输入获得获得中序遍历和前序遍历，之后递归建立二叉树，再后续遍历获得结果！

#include<iostream>#include<vector>#include<cstring>#include<stack>using namespace std;vector<int>vecPre, vecInorder,vecPost;struct Node{Node *lChild;Node *rChild;int ID;};int findRootPos(int istart, int iend, int pstart){int pos = istart;for (int i = istart; i <= iend; ++i){if (vecInorder[i] == vecPre[pstart])pos = i;}return pos;}void createTree(int pstart,int pend,int istart,int iend,Node *&root){if (istart > iend || pstart > pend)return;if (!root){root = new Node();root->ID = vecPre[pstart];}int pos = findRootPos(istart,iend,pstart);createTree(pstart+1,pstart+pos-istart,istart,pos-1,root->lChild);createTree(pend-iend+pos+1,pend,pos+1,iend,root->rChild);}void postOrder(Node *root){if (root){postOrder(root->lChild);postOrder(root->rChild);vecPost.push_back(root->ID);}}int main(){int N, id;char input[5];//vector<int>vecorder;//1 for push,0 for popstack<int> s;Node *root = NULL;scanf("%d", &N);for (int i = 0; i < 2 * N; ++i){scanf("%s", input);//cout << strcmp(input, "Pop") << endl;if (strcmp(input, "Pop") == 0){int temp = s.top();s.pop();vecInorder.push_back(temp);}else{scanf("%d", &id);s.push(id);vecPre.push_back(id);}}createTree(0, N - 1, 0, N - 1, root);postOrder(root);for (int i = 0; i < vecPost.size()-1; ++i){printf("%d ",vecPost[i]);}printf("%d",vecPost[vecPost.size()-1]);return 0;}