1086. Tree Traversals Again (25)

本次AC参考了他人代码：http://blog.csdn.net/jmlikun/article/details/50003781

1086. Tree Traversals Again (25)

时间限制

200 ms

内存限制

65536 kB

代码长度限制

16000 B

判题程序

Standard

作者

CHEN, Yue

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

#include <iostream>
#include <string>
#include <cstdio>
#include <vector>
#include <queue>
#include <algorithm>
#include <stack>
using namespace std;
struct Node
{
int data;
Node * left;
Node * right;
Node():data(0),left(NULL),right(NULL) {};
Node(int x):data(x),left(NULL),right(NULL) {};
};
void Post(Node *T,vector<int>* o)
{
if(T->left)
Post(T->left,o);
if(T->right)
Post(T->right,o);
(*o).push_back(T->data);
}
int main()
{
int n,num;
cin>>n;
string str;
Node *lastpop=NULL,*root = NULL;
stack<Node *>sta;
for(int i=0; i<2*n; ++i)
{
cin>>str;
if(str=="Push")
{
cin>>num;
if(sta.empty()&&lastpop==NULL) //上次没弹出，这次压入，根节点
{
root = new Node(num);
sta.push(root);
}
else if(lastpop)  //上次有弹出，这次压入为右节点
{
lastpop->right = new Node(num);
sta.push(lastpop->right);
}
else//栈不为空且上次未弹出为压入栈顶元素左子树
{
sta.top()->left = new Node(num);
sta.push(sta.top()->left);
}
lastpop = NULL;
}
else
{
lastpop = sta.top();
sta.pop();
}
}
vector<int>outt;//输出序列
Post(root,&outt);
for_each(outt.begin(),outt.end(),[&](int a)
{
static int count = 0;
cout<<a;
count++;
if(count == outt.size())
cout<<endl;
else
cout<<" ";
});
return 0;
}