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 1
Input 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 in the 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
#include <stack> #include <iostream> #include <fstream> #include <sstream> using namespace std; #define Empty (-1) template<class T> struct TreeNode { T data; TreeNode<T> *left; TreeNode<T> *right; TreeNode(const T d = T(-1)) :data(d), left(nullptr), right(nullptr) {} }; template<class T> TreeNode<T>* createTree() { fstream file("data.txt"); int nodeNum = 0; file >> nodeNum; if (nodeNum == 0) return nullptr; stack<TreeNode<T>*> stack; TreeNode<T> *root = nullptr; TreeNode<T> *cur = nullptr; bool lastActionIsPop = false; //记录上一次操作是否是POP for (int i = 0; i < nodeNum *2; i++) { string action; file >> action; if (action == "Push") { T item; file >> item; if (root == nullptr) { root = new TreeNode<T>(item); cur = root; stack.push(cur); } else { TreeNode<T> *tmp = new TreeNode<T>(item); if (lastActionIsPop) cur->right = tmp; else cur->left = tmp; cur = tmp; stack.push(cur); } lastActionIsPop = false; } else if (action == "Pop") { cur = stack.top(); //return reference in the top stack.pop(); //none retval lastActionIsPop = true; } } file.close(); return root; } template<class T> void postorderTraversal(TreeNode<T> *root, string &res) { if (root == nullptr) return; postorderTraversal(root->left, res); postorderTraversal(root->right, res); stringstream os; os << root->data; res += (os.str() + " "); } int main() { string res = ""; TreeNode<char> *root = createTree<char>(); postorderTraversal<char>(root, res); cout << res.substr(0, res.size() - 1); //不输出最后面的空格 system("pause"); return 0; }