PTA 03-树3 Tree Traversals Again (25 分）30行代码 -__-

03-树3 Tree Traversals Again (25 分)

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

题目大意：知道先序列和中序求后序。但我觉得在构造先序列时就可以输出后序及在递归之后！

 

#include

using namespace std;

int n,k=1;

bool flag = false;

string st[120];

string no;

void DFS(int x){

k = x; //防止函数返回的时候

if(st[x] == "Pop" || x >= 2*n)

return;

string s;

DFS(x+1);

DFS(k+1);

if(st[x] != "Pop"){

if(flag)printf(" ");

cout << s.assign(st[x],5,st[x].length()-5);

flag = true;

}

}

int main(){

cin >> n;

getchar();

int cnt;

stacks;

for(int i = 1; i <= 2*n; i++)

getline(cin,st[i]);

DFS(1);

return 0;

}