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 (≤) which is the total number of nodes in a tree (and hence the nodes are numbered from 1 to N). Then 2 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题意：
用栈的形式给出一棵二叉树的建立的顺序，求这棵二叉树的后序遍历
题解：
栈实现的是二叉树的中序遍历（左根右），而每次push入值的顺序是二叉树的前序遍历（根左右），所以该题可以用二叉树前序和中序转后序的方法做～
AC代码：
#include<bits/stdc++.h>
using namespace std;
int n;
struct node{
int data;
node *left,*right;
};
vector<int>pre,in,post;
stack<int>s;
node *buildTree(vector<int>pre,vector<int>in,int pl,int pr,int il,int ir){
if(pl>pr || il>ir) return NULL;
int pos=-1;
for(int i=il;i<=ir;i++){
if(in.at(i)==pre.at(pl)){
pos=i;
break;
}
}
node *root=new node();
//root->left=root->right=NULL;
root->data=pre.at(pl);
root->left=buildTree(pre,in,pl+1,pl+pos-il,il,pos-1);
root->right=buildTree(pre,in,pl+pos-il+1,pr,pos+1,ir);
return root;
}
void postorder(node *root){
if(root){
postorder(root->left);
postorder(root->right);
post.push_back(root->data);
}
}
int main(){
cin>>n;
pre.push_back(-1);
in.push_back(-1);
char c[10];
int x;
for(int i=1;i<=2*n;i++){
cin>>c;
if(strcmp(c,"Push")==0){
cin>>x;
s.push(x);
pre.push_back(x);
}else{
in.push_back(s.top());
s.pop();
}
}
node *root = buildTree(pre,in,1,n,1,n);
postorder(root);
for(int i=0;i<post.size();i++){
cout<<post.at(i);
if(i!=post.size()-1) cout<<" ";
}
return 0;
}