1086. Tree Traversals Again (25)
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 PopSample Output:
3 4 2 6 5 1
这道题首先根据输入建立一个二叉树,然后后序遍历这个二叉树即可。代码如下:
#include <iostream>
#include <algorithm>
#include <cmath>
#include <vector>
#include <cstring>
using namespace std;
typedef struct node{
int element;
node* left;
node* right;
bool popped;
node(){
left=NULL;
right=NULL;
popped=false;
}
}node;
node* root;
node* push(node* x,int element)
{
if(x==NULL)
{
x=new node;
x->element=element;
}
else if(x->popped)
x->right=push(x->right,element);
else
x->left=push(x->left,element);
return x;
}
bool pop(node* x)
{
if(x==NULL)
return false;
if(x->popped)
{
if(!pop(x->right))
return false;
}
else if(!pop(x->left))
x->popped=true;
return true;
}
void postorder(node* x)
{
if(x==NULL)
return;
postorder(x->left);
postorder(x->right);
if(x!=root)
cout<<x->element<<" ";
else
cout<<x->element;
return;
}
int main(void)
{
int N,ele;
cin>>N;
int step=2*N;
string raw;
cin>>raw>>ele;
getchar();
root=new node;
root->element=ele;
for(int i=1;i<step;i++)
{
cin>>raw;
if(raw[1]=='u')
{
cin>>ele;
root=push(root,ele);
}
else
pop(root);
}
postorder(root);
}