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
#include<cstdio>
#include<iostream>
#include<stack>
#include<string>
using namespace std;
struct Node{
int data;
Node* lchild;
Node* rchild;
Node(int num):data(num),lchild(NULL),rchild(NULL){}
Node():lchild(NULL),rchild(NULL){}//不要忘了寫上默認構造函數,否則孩子不爲空,判斷會有誤
};
Node* createTree(int nn){
stack<Node*>Node_Stack;
Node* ret = new Node();
string str_tmp;
cin >> str_tmp;
int i_tmp;
cin >> i_tmp;
ret->data = i_tmp;//先讀取第一個元素
Node* curNode = ret;
Node_Stack.push(ret);
while (--nn){//因爲已經少一個了,所以--nn
string type;
cin >> type;
if (type == "Push"){
int num_push;
cin >> num_push;
Node* newNode = new Node(num_push);
if (curNode->lchild == NULL)
curNode->lchild = newNode;
else
curNode->rchild = newNode;
Node_Stack.push(newNode);
curNode = newNode;
}
else if (type == "Pop"){
curNode = Node_Stack.top(); //*特別注意當前節點是pop出來的節點
Node_Stack.pop();
}
}
return ret;
}
void postOrder(Node* root){//後續遍歷輸出
if (root->lchild != NULL)postOrder(root->lchild);
if (root->rchild != NULL)postOrder(root->rchild);
static bool firstFlag = true;//設置標籤位,看是否是第一次輸出,爲了最後的空格格式化,注意這裏需要static
if (firstFlag){
cout << root->data;
firstFlag = false;
}
else
cout << " " << root->data;
}
int main(){
freopen("F://Temp/input.txt", "r",stdin);
int n;
cin >> n;
if (n == 0)return 0;//如果沒有輸入元素,則不用輸出,直接返回
Node* root = createTree(2*n);//有2n組數據輸入
postOrder(root);
return 0;
}