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
解题思路：由几个Push的操作可以组成树的先序遍历结果，而Pop的结果是树的中序遍历的结果，利用先序和中序即可构建一颗唯一的树，然后在后序遍历即可求解。
#include<iostream>
#include<vector>
#include<stack>
#include<cstdio>
#include<cstring>
using namespace std;
struct Tree{
int data;
Tree* left;
Tree* right;
Tree(){
left=NULL;
right=NULL;
}
};
vector<int>preorder;
vector<int>inorder;
int n,cur;
int find(int key){
for(int i=0;i<n;i++){
if(inorder[i]==key){
return i;
}
}
return -1;
}
Tree* create(int start,int end){
if(start>end)return NULL;
Tree* t=new Tree();
t->data=preorder[cur];
cur++;
int index=find(t->data);
if(start!=end){
t->left=create(start,index-1);
t->right=create(index+1,end);
}
return t;
}
int flag=0;
void postorder(Tree* t){
if(!t)return;
postorder(t->left);
postorder(t->right);
if(flag==0){
printf("%d",t->data);
flag=1;
}else{
printf(" %d",t->data);
}
}
int main(){
scanf("%d",&n);
stack<int>s;
int i,j;
char str[5];
int val;
for(i=0;i<n*2;i++){
scanf("%s",str);
if(str[1]=='u'){
scanf("%d",&val);
s.push(val);
preorder.push_back(val);
}else{
val=s.top();
s.pop();
inorder.push_back(val);
}
}
Tree* t=create(0,n-1);
postorder(t);
printf("\n");
return 0;
}