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:

6Push 1Push 2Push 3PopPopPush 4PopPopPush 5Push 6PopPop

Sample Output:

3 4 2 6 5 1

用笨笨的方法，先建立对应的二叉树，在后根遍历输出。

#include<iostream>#include<string>using namespace std;int N,count=0,count1=0;struct Node{int data;Node *left,*right;};Node *BuildTree()//建立对应的二叉树{Node *r=new Node;r->left=NULL;r->right=NULL;int num;string str;if(count<2*N){cin>>str;if(str=="Push"){cin>>num;r->data=num;count++;}else if(str=="Pop"){count++;return NULL;}r->left=BuildTree();r->right=BuildTree();}else//末尾节点处理return NULL;return r;}void Printf(Node *T)//输出{if(T){Printf(T->left);Printf(T->right);cout<<T->data;count1++;if(count1<N)//末尾不留空cout<<" ";}}int main(){cin>>N;Node *Tree=BuildTree();Printf(Tree);cout<<endl;return 0;}