03-树3 Tree Traversals Again（25 分）
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用了一个enum顺利解决了问题
1 #include<iostream> 2 3 #include<stack> 4 using namespace std; 5 enum State{return_from_left,return_from_right}; 6 struct treenode{ 7 int data; 8 treenode* left; 9 treenode* right; 10 enum State state; 11 }; 12 using tree=treenode*; 13 int tag=1; 14 void preordertraversal(tree st){ 15 if(st){ 16 preordertraversal(st->left); 17 preordertraversal(st->right); 18 if(tag--==1) 19 cout<<st->data; 20 else 21 cout<<" "<<st->data; 22 } 23 } 24 int main() 25 { 26 int N,data; cin>>N; 27 tree BT=new treenode(); 28 stack<tree> s; 29 tree t,st; 30 st=BT; 31 string operation; 32 cin>>operation>>data; 33 BT->data=data; 34 BT->state=return_from_left; 35 s.push(BT); 36 for(int i;i<2*N-1;i++){ 37 cin>>operation; 38 if(operation=="Push"){ 39 cin>>data; 40 t=new treenode(); 41 t->data=data; 42 t->state=return_from_left; 43 s.push(t); 44 if(BT->state==return_from_right) 45 BT->right=t; 46 if(BT->state==return_from_left) 47 BT->left=t; 48 BT=t; 49 } 50 if(operation=="Pop"){ 51 BT=s.top(); 52 s.pop(); 53 BT->state=return_from_right; 54 } 55 } 56 preordertraversal(st); 57 }
