PAT1086

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

提交代码

就是入栈的时候是前序遍历，出栈中序遍历， 那么用stack存一下放到两个数组里面，之后就是前序中序已知求出这棵树再后序遍历一下就好了。

#include
#include
#include
#include
#include
using namespace std;
struct Node{
struct Node * left;
struct Node * right;
int value;
};
int pre[50];
int in[50];
int flag=1;

void build(Node * &Root , int pbegin,int pend,int ibegin,int iend ){
if(pbegin>pend)return ;
Root= new Node;
Root->value = pre[pbegin];
//cout<value<left=NULL;
Root->right=NULL;
int k=0;
while(pre[pbegin]!=in[k]){
k++;
}
build(Root->left,pbegin+1,pbegin+k-ibegin,ibegin,k-1);
build(Root->right,pbegin+k-ibegin+1,pend,k+1,iend);
}
void travel(Node * Root){
if(Root==NULL)return ;
travel(Root->left);
travel(Root->right);
if(flag){
flag=0;
cout<value;
}
else {
cout<<" "<value;
}
}
int main(){
stacks;
int n;
cin>>n;
int p_index=-1,i_index=-1;
for(int i=1;i<=2*n;i++){
string name;
cin>>name;
if(name=="Push"){
int x;
cin>>x;
s.push(x);
pre[++p_index]=x;
//	cout<<"pre:"<

left->value;
travel(Root);
}