03-树3 Tree Traversals Again（25 分）

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 $N$ ($\le 30$) 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

用了一个enum顺利解决了问题

#include<iostream>

#include<stack>
using namespace std;
enum State{return_from_left,return_from_right};
struct treenode{
int data;
treenode* left;
treenode* right;
enum State state;
};
using tree=treenode*;
int tag=1;
void preordertraversal(tree st){
if(st){
preordertraversal(st->left);
preordertraversal(st->right);
if(tag--==1)
    cout<<st->data;
    else
    cout<<" "<<st->data;
}
}
int main()
{
int N,data; cin>>N;
tree BT=new treenode();
stack<tree> s;
tree t,st;
st=BT;
string operation;  
cin>>operation>>data;
BT->data=data;
BT->state=return_from_left;
s.push(BT);
    for(int i;i<2*N-1;i++){
    cin>>operation; 
    if(operation=="Push"){
    cin>>data;
    t=new treenode();
    t->data=data;
    t->state=return_from_left;
    s.push(t);
    if(BT->state==return_from_right)
    BT->right=t;
    if(BT->state==return_from_left)
    BT->left=t;
    BT=t;
}
if(operation=="Pop"){
BT=s.top();
s.pop();
BT->state=return_from_right;
}
}
preordertraversal(st);
}