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
Pop
Sample Output:
3 4 2 6 5 1
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Create : 2018-03-01 16:22:45
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#include <cstdio>
#include <cstdlib>
#include <iostream>
#include <stack>
#include <cstring>
using namespace std;
struct Tree{
Tree *li;
Tree *ri;
int data;
};
Tree *Tr;
stack<Tree *> q;
int k = 1;
int print(Tree *x){
if (x == NULL)return 0;
print(x->li);
print(x->ri);
if(k==0){
printf(" %d",x->data);
}else{
printf("%d",x->data);
k--;
}
return 0;
}
int main(){
int n;
int temp;
scanf("%d",&n);
string str;
cin >> str;
cin >> temp;
Tree *Tr = (Tree *)malloc(sizeof(Tree));
Tr->data = temp;
Tr->li = NULL;
Tr->ri = NULL;
q.push(Tr);
Tree *root = q.top();
int kk = 0;
while(kk < n){
cin >> str;
if(str=="Push"){
cin >> temp;
Tree *t = (Tree *)malloc(sizeof(Tree));
t->data = temp;
t->li = NULL;
t->ri = NULL;
if(root->li == NULL){
root->li = t;
}else{
root->ri = t;
}
q.push(t);
root = t;
}else{
kk++;
root = q.top();
q.pop();
}
}
print(Tr);
return 0;
}