03-树3. Tree Traversals Again

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:

6Push 1Push 2Push 3PopPopPush 4PopPopPush 5Push 6PopPop

Sample Output:

3 4 2 6 5 1

#include <string>#include <iostream>#include <stack>using namespace std;const string PUSH("Push");const string POP("Pop");typedef struct Node{int data;Node* left;Node* right;Node(int d):data(d), left(NULL), right(NULL){}}Node;void PostOrderTraverse(Node *root){Node* temp = root;Node* pre = NULL;stack<Node*> S;int flag = 0;while(temp || !S.empty()){if(temp){S.push(temp);temp = temp->left;}else{temp = S.top();if(temp->right && temp->right != pre)temp = temp->right;else{if(!flag){flag = 1;cout<< temp->data;}elsecout<<" "<<temp->data;S.pop();pre = temp;temp = NULL;}}}cout<<endl;}int main(){int n, data;string act;stack<Node*> S;Node* root = NULL, *pre = NULL;int l = 1, r = 0;cin >> n;//First, build the tree , root of tree is *root.for(int i=1; i <= 2*n; i++){Node* temp;cin >> act;if(act == PUSH){cin >> data;temp = new Node(data);if(i == 1){root = temp;}S.push(temp);if(pre){if(l == 1)pre->left = temp;elsepre->right = temp;}l = 1;pre = temp;}else if(act == POP){pre = S.top();S.pop();l = 0;}}PostOrderTraverse(root);system("pause");return 0;}